Home/Blog(This is a follow-up to my post from a little over a year ago, but with some exprimental results).
I sometimes see the claim (usually in comments on a blog post) that infrared radiation cannot warm a water body, because IR only affects the skin surface of the water, and any extra heating would be lost through evaporation.
I have tried to point out that evaporation, too, only occurs at the skin of a water surface, yet it is a major source of heat loss for water bodies. It may be that sunlight is more efficient, Joule for Joule, than infrared due to the depth of penetration effect (many meters rather than microns). But I would say it pretty clear that any heat source (or heat sink) like evaporation which only affects the skin is going to affect the entire water body as well, especially one that is continually being mixed by the wind.
Partially water-filled Styrofoam cooler with a temperature sensor in the bottom.
As seen in the second photo, over one of the coolers I put a piece of aluminum flashing painted white with high IR emissivity paint to block IR emission from the “coldest” part of the sky (directly overhead), but small enough to not restrict air flow around it as evaporation (and sensible heat transfer while the water was warmer than the air) cooled the water in both containers:
Two partially water-filled styrofoam coolers, with one partially shielded with aluminum flashing painted with high-emissivity white paint.
My FLIR i7 IR imager said that the sky “effective temperature” directly overhead was about 7 deg F, and the temperature of the aluminum sheet (viewed from below) was close to 80 deg. F, indicating that the presence of the sheet should reduce the radiative energy loss from the water, thus keeping the water warmer than if the sheet was not there (which is what the greenhouse effect does to the Earth’s surface).
Of course, these IR imagers do not provide a good estimate of the broadband IR effective emitting temperature of the sky because they are tuned to the 7.5 to 14 micron wavelength range, which avoids some of the greenhouse gas emission/absorption from the sky. So the true broadband-IR emission from the sky would be at a higher effective temperature.
For example, the most recent ground-based radiometer data from Goodwin Creek, MS the day before my experiment (June 28) showed early nighttime downwelling IR fluxes of about 360 W/m2, which corresponds to an effective emitting temperature of about 48 deg. F, considerably warmer than the 7 deg. F atmospheric window measurement I made with my imager.
Also, my backyard has numerous trees (and my house) blocking the lower sky elevations, with only about 30-40% of the sky visible above.
Nevertheless, the metal sheet in the above photo will still block a portion of the colder sky from view by the warmer water, and it should cause the water in the cooler on the right to be a little warmer than the unshielded water on the left.
So, let’s see what happened to the temperatures:
Overnight temperature data (taken every 5 mins) for water in the coolers, and air temperature between the coolers. A 1-hour trailing average is also shown in the second graph.
As can be seen in the 5 minute temperature data overnight, the cooler with the IR shield stayed a little warmer. The relative faster cooling of the unshielded cooler was slowed when high-level clouds moved in around 1:30 a.m. (as deduced from GOES satellite imagery).
I used this simple energy balance model to deduce that the shielded cooler had about 6 W/m2 less cooling than the unshielded one, to account for the relative weaker temperature drop of 0.4 deg F over about 8 hours.
I’d like to try this again when the air mass is not so humid (surface dewpoints were in the mid to upper 60s F). Also, putting Saran wrap over the water surfaces would eliminate the primary source of heat loss – evaporation.
Note that the temperature effect is small in such an experiment because the atmosphere is already mostly opaque in the infrared, so adding a shield whose emitting temperature is a little higher than what the sky is already producing won’t cause as dramatic of an effect. But under the right conditions (warm water covered with Saran wrap, a very dry atmosphere, and unobstructed view of the sky) much larger differential impacts on water temperature should be possible.
But Roy, both basins are cooling. The shielded one is cooling a bit slower than the unshielded one. That’s not “LW infrared radiation warming a water body”. That’s less energy escaping the shielded water surface per unit time than the unshielded one. That’s a very different thing. In fact, it’s fundamentally different.
We all agree that a cooling water surface (or any outside, non-heated surface) will cool more slowly during the night under humid and/or cloudy conditions. However, if you want that water surface (or body of water) to actually warm (have its temperature rise in absolute terms), then you need the Sun. You need SW.
It doesn’t matter whether you call it “reduced rate of cooling”, or “warming”, the result is the same: a higher temperature.
When an object warms, it doesn’t know whether it’s energy source has increased, or its energy sink has decreased. The effect is the same.
Indeed, the effect is the same. But it is not the LW doing the warming. The LW is doing the cooling. The SW is what’s doing the warming.
I’m not talking about the effect at all. OF COURSE a better insulated, constantly heated object will be warmer than a lesser insulated, constantly heated object. But it is not the insulating layer doing the warming. It’s the heater.
Yes, it’s semantics. But it’s important semantics. Because the way you describe it confuses people who don’t know the physics behind the different mechanisms at work (which are fundamentally different, in fact, opposite in direction). Which is most people …
ALWAYS stick with “reduced cooling” when talking about LW, NEVER use the terms “heating” or “warming”; those are reserved for SW. And you’ll be fine.
Why not conduct the experiment indoors with a heat lamp? Controlling for visible/uv, humidity and temperature would be easy. The temp difference isn’t very convincing and it would be interesting to see how repeatable it is (indoors).
–Eric Barnes says:
June 30, 2015 at 7:43 PM
Why not conduct the experiment indoors with a heat lamp? Controlling for visible/uv, humidity and temperature would be easy. The temp difference isn’t very convincing and it would be interesting to see how repeatable it is (indoors).–
A heat lamp is IR, but it’s mostly shortwave IR. Or about 1/2 of Sunlight is shortwave IR. Shortwave IR does pass thru the top mm of water- most of it will reach around a meter under water. Red light [of visible light] is similar to Near IR and other SW IR, and red light can reach to about 2 meters under water- red colored things more two meters under water [which lit by sunlight appears black. Blue and UV travel tens of meters under water.
General IR in terms climate refers to long wave IR. Everything glows in long wave IR.
Thanks for the information.
This seems like it would work. Not terribly expensive…
http://www.acubest.com/en/d-4.html
Maybe we could chip in and buy one for Dr. Spencer. 🙂
So if the atmosphere was not filled with water vapor (and a tiny bit of CO2 🙂 ), it would seem that Solar IR could heat water. Solar IR being absorbed by our GHG atmosphere and re-emitted as true LW IR.
https://en.wikipedia.org/wiki/Sunlight#/media/File:Solar_spectrum_en.svg
We are so fortunate to have all these GHG’s in the atmosphere.
🙂
“General IR in terms climate refers to long wave IR. Everything glows in long wave IR.”
Actually, it would appear that Sunlight at sea level is Short Wave Infrared according to Wikipedia at least…
https://en.wikipedia.org/wiki/Infrared
https://upload.wikimedia.org/wikipedia/commons/e/e7/Solar_spectrum_en.svg
Please correct me if I’m not reading that correctly.
–General IR in terms climate refers to long wave IR. Everything glows in long wave IR.”
Actually, it would appear that Sunlight at sea level is Short Wave Infrared according to Wikipedia at least…
https://en.wikipedia.org/wiki/Infrared
https://upload.wikimedia.org/wikipedia/commons/e/e7/Solar_spectrum_en.svg
Please correct me if I’m not reading that correctly.–
Yes, correct.
All significant amount of sunlight is shortwave. The Sun, a huge ball plasma does emit longwave IR, but we are 149.6 million km from it, and is very weak at this distance.
Or said differently much more energy is radiated in the shortwave spectrum of UV, Visible light, and the Shortwave Infrared.
As wiki says:
“In terms of energy, sunlight at Earth’s surface is around 52 to 55 percent infrared (above 700 nm), 42 to 43 percent visible (400 to 700 nm), and 3 to 5 percent ultraviolet (below 400 nm).”
So sun is shown to emit on that graph 250 to 2500 nm.
And long wave IR is not on graph, and it’s beyond 3000 nm.
Long wave Ir is 8000 to 15,000 nm. But everything beyond 2500+ would be about 1% of sun’s energy- or all beyond +2500 nm is insignificant, reason is because it’s weaker at the sun, and then all the radiation has weaken over the 149.6 million km distance it’s traveled. So the long wave barely able to be measurable. Though there graphs available which do show the entire spectrum which reaches Earth distance.
But I don’t have a link to it at hand [probably forever lost in all the bookmarks:)
Oh there is graph I found, but not one I was thinking of.
http://www.windows2universe.org/sun/spectrum/multispectral_sun_overview.html
Note the exponentiation quantities the power of the energy on left of graph
So it goes from X-rays to radio [beyond Longwave IR]
And this graph is mid way down the page.
And 1 µm is equal to 1000 nm
“It doesn’t matter whether you call it “reduced rate of cooling”, or “warming”, the result is the same: a higher temperature.”
Poorly worded, or “climate science”?
A “reduced rate of cooling” is not the same as “warming”. “Cooling” does not result in increased temperatures, except, of course, in “climate science”.
(Is my calendar wrong? Is this April 1st?)
geran, you need to learn some physics.
Dr. Roy, that is very good advice. I would freely give that advice to anyone. If more folks understood physics, our country would not have wasted $100 billion on “climate change” in the last 5 years.
The essential aspects of physics, related to Earth’s temperature, are thermodynamics, heat transfer, and quantum physics. A sound understanding of these three individual sciences indicates that the IPCC AGW/GHE/CO2 nonsense is bogus science. (Common sense should have taught folks about the bogus science, and also there is the last 20 years of temps. You do monitor temps, don’t you?)
You have indicated you do not have a solid background in these sciences. Your training in meteorology likely taught you the GHE. You have believed in it for numerous years. Likely, you will always believe in it. But, I am willing to try, nevertheless.
I believe you are smart and honest. I believe I could convince (deprogram) you in about 4 hours. If you can set aside 4 hours on your next trip to my “neck of the woods”, I will offer to pay for your lunch, at the restaurant of you choosing. Listen to the facts, ask questions, bring your calculator, as you said, “learn some physics”.
You don’t want to be a “lukewarmer” when you can be a hard-core, extreme skeptic!
“A “reduced rate of cooling” is not the same as “warming”.”
GHGs reduce the rate at which IR escapes to space. The surface is trying to “cool” all the time by releasing infrared. The effect of increased GHGs is to reduce the pace at which the surface cools.
In this case it is *exactly* the same as warming.
NOPE!
But, if all the imagined GHE can do is “slow the cooling”, then it has NO ability to “heat the planet”.
If all that GHGs can do is slow the rate of cooling, provided that there is sufficient time for planet Earth to give up its ‘heat’ before the next bout of incoming solar irradiance, then GHGs can do nothing of substance; they may render the coldest time of the day say at 03:56hrs rather than at say 03:52 hrs. But such a difference is insignificant.
But those who subscribe to the K&T energy budget cartoon and AGW take no account of the daily cycle and the fact that incoming solar is not a 24 hour event whereas outgoing radiation is a 24 hour ongoing event.
I mostly agree with you, but ain’t you confusing insulation and heating, and latent heat and LW radiation from CO2 (different post)?
I’ve heard you speak last year in Las Vegas, and while comparing what you said then and what you write on your blog, hmm .. Doesn’t seem to be consistent.
My research show CO2 doesn’t heat the planets atmosphere, in fact it is cooling the atmosphere, i have reached the same conclusion as Fred Singer, CO2 cools the atmosphere (in theory, of course).
Does your research support F. Singer?
Fred Singer and I agree, as far a I know. Greenhouse gases are radiatively active, and so they can either cause a temperature rise (surface and lower atmosphere) or a temperature fall (most of the upper atmosphere).
Or, they can merely delay the flow of energy through the system by acting as a sort of “hybrid” thermal-optical delay line. Optical delay lines are well understood (but rarely applied). If the delay is orders of magnitude (a few tens of milliseconds versus about 86 million milliseconds in each day) smaller than the “period” of the incoming energy it cannot affect the “average” temperature.
This “delay line” effect causes the atmosphere to warm up ever so slightly faster after sunrise, shifts the peak daytime temperature ever so slightly sooner in the day and causes the gases in the atmosphere to cool down ever so slightly slower after sunset. This is classic electronic circuit response time effects, well understood for a century or more.
The “average” voltage in an electronic circuit can; “rise”, “fall”, or merely “rise/fall” “faster/slower” when conditions in the circuit change. Rise/Fall are NOT the only possibilities.
Cheers, KevinK
— Roy W. Spencer, Ph. D. says:
June 30, 2015 at 4:15 PM
Fred Singer and I agree, as far a I know. Greenhouse gases are radiatively active, and so they can either cause a temperature rise (surface and lower atmosphere) or a temperature fall (most of the upper atmosphere).–
One thing I am pretty certain about is that greenhouse gases don’t cool the upper atmosphere.
In simple terms I would think Venus proves this.
I do think latent heat can be transported to higher elevation and cause net radiant losses.
But again transportation of latent heat is not related to radiant properties of greenhouse gases. H20 can be a gas, and as a gas it’s a greenhouse gas. Liquid water is not a gas, and transforming into a liquid, is not a radiant aspect of gases.
But anyhow, the transports of latent heat to upper atmosphere is NOT cooling- rather it’s warming the upper atmosphere, and just because things which are warmed can radiate more energy, does mean cooling. A fire does not cool, though a fire can make things to radiate more heat.
@Roy…”Fred Singer and I agree, as far a I know…”
Singer is an electrical engineer. Lindzen, who is an expert in atmospheric physics does not agree with your radiation-based theories. He claims the GFE theory is generally recognized to be over-simplified.
http://www-eaps.mit.edu/faculty/lindzen/230_TakingGr.pdf
Singer made a fool of himself by defending the claim that heat transfer from a cooler atmosphere to a warmer surface that warmed it does not contradict the 2nd law. In that sense, you and he seem to agree.
I wont go into it again, but suffice it to say that not nearly enough GHGs exist in the atmosphere to create a significant warming and those that are there cannot transfer heat back to the surface, due to the 2nd law.
In this article you claim a radiative measurement of 300+ w/m^2. What you don’t seem to understand is that IR is not heat. IR from a cooler medium (the atmosphere) cannot raise the kinetic energy in a warmer medium (the surface), where that KE is heat.
That’s likely what is wrong with your theory. IR from a cooler atmosphere will not raise the temperature of a warmer ocean. Short wave solar energy will, to a point, because it has the intensity and it comes from a warmer source.
You told geran that he needs to learn physics, which strikes me as the pot calling the kettle black. If you cannot explain the difference between heat and IR then you don’t understand the physics behind heat and heat transfer.
Talking about blocking heat transfer, or slowing it down, is sheer nonsense. One of your colleagues in meteorology, who is also a physicist, Craig Bohren, claimed that the heat trapping theory is a metaphor at best and, at worst, plain silly.
Bohren has written a book on atmospheric radiation.
@KevinK…”…they can merely delay the flow of energy through the system by acting as a sort of “hybrid” thermal-optical delay line”.
What energy? Are you referring to electromagnetic energy or thermal energy, they are different entities?
Also, what comprises your delay line? Are you talking about the 1% of the atmosphere deemed to be GHGs?
GHGs absorb IR, which is not heat. They do warm when they absorb the IR, which raises the kinetic energy level in the molecular bonds. Please note, however, that the direction of heat transfer is from a warmer surface to cooler GHGs. The reverse process is prohibited by the 2nd law.
If you consider the massive IR flux leaving the Earth’s surface, which has a cooling effect on the surface since there is a reduction in the kinetic energy of surface atoms/molecules due to IR photon emission, and compare that to the 1% of atmospheric gases that are GHGs, how can you possibly expect a delay in surface generated IR?
By the time the GHGs aborbs their loads of IR, the entire surface has cooled. It doesn’t care what is in the way in the atmosphere, it cools despite any of that. What you are suggesting is that a massive heat generating processes will slow down because of a trivial amount of GHGs in the atmosphere.
Perhaps in a different environment your opacity may have relevance but not in an atmosphere that is 99% nitrogen/oxygen and 1% GHGs.
The only way you could slow the flow of surface IR down is to bring atmospheric temperatures close to the temperature of the surface so an equilibrium is established.
The surface does transfer thermal energy to the atmosphere but mainly by conduction. It makes sense that the 99% of the atmosphere which is made up of oxygen and nitrogen molecules contributes most to the warming of the atmosphere.
Lindzen agrees that atmospheric warming is largely due to convective currents that transport heat poleward from the tropics
The temperature of the atmosphere, which is a measurement of the heat in the atmosphere, is based on the average kinetic energy of it’s molecules, which are 99% oxygen and nitrogen. C02, especially ACO2, contributes very little.
–GHGs absorb IR, which is not heat. —
I would say it this way IR absorbed solids, and liquid is
converted to heat. Though solid or liquid have to be cooler
to absorb and convert to heat. Or the Matter needs a pattern
to take a pattern. Gases are kinetic energy- they will take any kind of kinetic energy. The speed of one molecule is not the temperature of a gas, the temperature of a gas is average velocity of a group of molecules. So a molecule at zero velocity can make another molecule faster, or fast molecule could not impart much kinetic to another molecule.
So whole thing of not being to warm something warmer, does not apply to a gas molecule- and therefore to any gas molecules. But when a gas molecule absorbs radiant energy- that is not kinetic energy. One thing which warms gas, is more gas in volume and more kinetic energy. So radiant energy does not warm a gas molecule. And if a gas molecule electron can be bumped to higher level it can accept the photon energy which can do this. The temperature of gas is unrelated to whether and gas molecule can accept a photon’s energy. Nor does it matter the velocity of a molecule or a group of molecules. So gases of different average velocities
[faster warmer, slower cooler] can accept photons from any source, as long as it bumps up the electron. But a molecule
like solid or liquid, does heat up when accepts energy of photon- the electron goes to higher orbit, then goes to lower orbit, and emits a photon. So it’s energized, but not warmed.
–They do warm when they absorb the IR, which raises the kinetic energy level in the molecular bonds.–
And and gas has no molecular structure, unlike liquids and solids [and unlike plasma].
Shouldn’t the basin with the shield be colder to prove DWLIR is “warming” the water.
The uncovered water is subject to the full brunt of the atmosphere IR raining down on it, while the shielded water should stay nice and cool protected from this 300+ w/m heat element in the sky.
IR is not accounting for the current sea surface temperatures in any serious manner in my opinion.
I am with the argument icecap.com makes for IR and how insignificant it is. Joe D’ Aleo of Weatherbell Inc. heads icecap.com
His thinking.
Type of prediction
Ocean warming
Model prediction
Warming caused by direct heating of thermal radiation at 15 microns.
Actual measurements
Warming of about 0.06 C over 50 years.
More here.
Comment
The absorption coefficient for liquid water as a function of wavelength is given at http://www.lsbu.ac.uk/water/vibrat.html (see the figure near the end). Thermal infrared in the Earth’s atmosphere is around 10 to 20 microns where the absorption coefficient (A) is about 1000 cm-1. The transmission in liquid water (T) equals exp(-A*L) where L is the depth of penetration. For the case where 1/e or 27% of the incident photons remain unabsorbed, with A=1000 cm-1, the L= 1/1000 cm = 1/100 mm. 98% of the incident photons will be absorbed within 3 times this distance. So one can see from the figure, than practically no infrared photons penetrate beyond 3/100 mm. When I said all the photons are absorbed in the top millimeter of the water, I was being very generous. A more precise estimate of A is 5000 cm-1 at 15 microns where carbon dioxide is emitting radiation, so even 0.03 mm is extremely generous. Since the liquid water is such an effective absorber, it is a very effective emitter as well. The water will not heat up, it will just redirect the energy back up to the atmosphere much like a mirror.
It is worth mentioning for A = 5000 cm-1 at 15 microns, the implied water emissivity is 0.9998 implying that of the incident radiation only 0.02% of it will be absorbed. The emitted radiation will closely follow a blackbody emission curve whereas the incident flux from carbon dioxide is confined to a band centered at 15 microns. The implication of this is that much of the radiation emitted will escape directly to space through the IR windows, so it is a negative feedback. The initially absorbed energy cannot be transferred to the ocean depths by conduction (too slow), by convection (too small an absorption layer), or by radiation (too opaque). It must escape by the fastest way possible meaning upwards radiation away from the water. I don’t see why anyone is having problems understanding basic physics.
The only way to explain the ocean heating in depth is for the solar radiation to change and decreasing clouds, as measured by ISCCP, indicate increasing solar radiation is occurring right where the ocean heating is reported to be occurring. The Willis paper does not even mention the ISCCP data that has a similar geographic distribution to the water warming. Simply put, where clouds decrease in amount, the water warms. It has nothing to do with carbon dioxide. A handy plot of the ISCCP results can be found as Figure 3 at http://www.worldclimatereport.com/index.php/2006/01/11/jumping-to-conclusions-frogs-global-warming-and-nature/ Clouds have large natural variations going up and down entirely independent of any greenhouse effect. The climate models do not predict these variations and apparently Willis and others are unaware of these variations.
Score
1-24-4
Salvatore, I’m afraid you have made a couple serious mistakes in your statements. Your assignment for tonight is to find them. 😉
Roy: It is worth noting that the thin layer of water that absorbs DLR also emits OLR. If DLR photons of a particular wavelength only penetrate 1 or 10 um, then any OLR photon of the same wavelength emitted 10 or 100 um below the surface would never reach the surface. Since the atmosphere is almost always cooler than the ocean below, the “skin” layer loses more energy than it gains at thermal infrared wavelengths. As you note, the skin layer is where evaporation also transfers heat to the atmosphere. Those who believe that the skin layer of the ocean would boil away if DLR existed ignore both OLR and evaporation. DLR exists and the skin layer is still colder than the bulk water below.
The ocean below the skin layer is warmed by the sun during the day. If heat wasn’t continuously transported upward to the skin layer from below, the skin layer would soon freeze and the water below would warm indefinitely! Unless mixing by wind makes it unnecessary, every night the surface of the ocean cools enough for surface water to sink and be replaced by warmer water from below.
Frank
Please rethink your description with 10 or 100 layers.
OLR emits in ALL directions equally. If it is absorbed in the next layer, that layer in turn will emit uniformly etc.
The water absorption/radiation (ignoring evaporation) would then be similar to the atmosphere absorbing/radiating.
” Since the atmosphere is almost always cooler than the ocean below”.
Clarify.
Yes for outer atmosphere, stratosphere, troposphere.
However, the near ocean air often warmer in day, cooler at night.
David Hagen: You raise some good points and I’m not sure I have all the answers. I’d appreciate sources that correct any misunderstandings I may have.
Do water molecules 1 cm below the surface of the ocean emit thermal infrared photons just like molecules 1 um below the surface (or on the surface), whose photons escape to the atmosphere? As best I can tell, all water molecules must follow the same “rules”. However, since we can’t easily measure the flux of photons through less than a mm through a liquid or solid, it is difficult to distinguish between conductive and radiative transfer of heat under these conditions. As best I can tell, the blackbody radiation that is emitted by solids and liquids has been traveling through the material following the same principles as radiation traveling through a gas (the Schwarzschild eqn). The only difference is that equilibrium between absorption and emission is reached over much shorter distances, which is why these substances usually emit like blackbodies. (Scattering/reflection as radiation crosses the interface can produce emissivity=emissivity less than 1.) The coatings on low emissivity glass have been analyzed using the Schwarzschild equation.
So I assume that the bulk of OLR photons escaping from the ocean to the atmosphere do originate in the top 10 um of the water – the same layer than absorbs DLR. I think most people would call the flux of heat below the top 10 um “conduction” (or convection) even though there must be large numbers of photons being exchanged in addition to energy transfer by molecular collisions. I’d love to hear from any authority on this subject, because I’ve never found any authoritative source that confirms these deductions. At the moment, therefore I see no problem with my understanding of the top 10 cm of the ocean as being 1000 layers 10 um thick following the same physics as the top 10 um layer. (One exception is surface tension in the top layer.)
David wrote: “However, the near ocean air [is] often warmer in day, cooler at night.”
As best I can tell, air temperature near the surface of the ocean and far away from land is tightly coupled with SST. Diurnal variation in air temperature is only about 0.5 degC.
http://journals.ametsoc.org/doi/full/10.1175/1520-0442(2004)017%3C0930%3ATDCAID%3E2.0.CO%3B2
This coupling must occurs through the “skin” layer and the above paper doesn’t pay attention diurnal changes near the skin layer.
http://ghrsst-pp.metoffice.com/pages/sst_definitions/index.htm
The boundary layer (about 1 km thick) over the ocean that is turbulently mixed by wind is somewhat cooler and drier (80% relative humidity) than expected for being in equilibrium with the ocean surface because some mixing with the free atmosphere occurs at the top of the boundary layer. I don’t expect the air to be much warmer during daytime or cooler at night.
It is my understanding (though I can’t document it right now) that the average DLR photon arriving at the surface of the earth is emitted from perhaps 1 km above the surface. With the lower emissivity of air and cooler temperature 1 km above the surface, the skin layer must lose more in terms of infrared radiation by OLR than it gains by DLR.
David the skin layer is always cooler than the water below it except immediately following something that breaks the surface tension like a breaking wave. Even in that case the temperature difference is reestablished in ten seconds. Net energy transfer is thus always in the upward direction whether by conduction or radiation.
That can be stated as a general truism: The sun heats the ocean and the ocean heats the atmosphere.
If for any reason – manmade or not – the air above an ocean is warmer at one time relative to another, won’t it simply be that the ocean will lose less heat to it (if the ocean is warmer than the air), or gain more heat from it (if the ocean is cooler than the air)? In other words, forget for a moment CO2, infrared, etc., and just think about air temperature compared to the temperature of the water below it: isn’t it mandatory that the two temperatures will tend to converge, ceteris paribus?
NONE of this is to say that manmade CO2 contributions to the atmosphere actually do produce a net result in air temperature sufficient to make any meaningful difference in ocean temperatures over any meaningfully short period of time; that is an entirely different question (and one that I personally answer in the negative, i.e., CO2 increase has no large, relatively quick net impact on either air or ocean temperatures, say I). I’m simply saying that it seems a bit goofy to suggest that even if increased CO2 does increase air temperatures, no change in water temperature could additionally result. If the air stays warmer relative to some baseline for the next thousand years, it sure seems like the ocean temperatures should get a little warmer, also.
The real question (as usual) is not whether CO2/IR can have ANY effect on ocean temps, but HOW MUCH? An experiment to ascertain THAT is the one that needs to be designed and carried out.
I should have said that what I was calling “goofy” was the idea implcit in this post’s title, “Can Infrared Radiation Warm a Water Body?”
If the question is changed to, “By just how much can infrared radiation warm a water body?”, I will retract my charge of goofiness.
While I’m at it, I’ll also throw in that I generally agree with earlier comments that Dr. Spencer’s experiment probably doesn’t advance anyone’s argument much, pro or con. I’m a big fan of his site and his work, and I understand and appreciate his doing this experiment on his own time and coinage. I would say, however, that more work is needed to get his point across more effectively on this one.
I echo the sentiment above. This site and Dr. Spencer pursue knowledge and advancement of understanding. While the participants here may argue vociferously on the topics, it is a healthy debate which we all benefit from. If only our taxpayer funded agencies engaged this important dialogue as openly as here, mankind could move to the future with the benefit of enlightenment vs. the agenda driven superstitions that clog current dialogue.
Thank you Dr. Spencer
If you read the study I sent from icecap.com you would realize IR radiation is a very small player when it comes to warming up the oceans.
Salvatore, I actually agree that CO2 *might* be a minor player (less than 50% warming, and with low climate sensitivity causing very little warming anyway). But as provoter correctly states, it’s a matter of just how much warming increasing CO2 causes.
…and how much CO2 has increased due to warming.
Before trying the Saran Wrap, it would be a good idea o make sure it is transparent to long wave IR. I’ve found that many plastics were opaque to LWIR, and seem to recall that Saran Wrap was one of them.
Living in an area where RH can vary from 10% to 100%, I can attest to night time temperatures being colder when RH is low as opposed to when RH is approaching 100%.
Saran wrap transmits about 90% of the IR incident upon it. I’ve verified this with my IR imager.
Hmm, so it appears that my recollection was incorrect…
Another way of doing the experiment would be placing one of the of the coolers under a pop-up – that way both will have similar evaporation environments. OTOH, getting rid of evaporation means getting rid of a very pesky uncontrolled variable.
–Saran wrap transmits about 90% of the IR incident upon it. I’ve verified this with my IR imager.–
There is difference between IR, and longwave IR.
I don’t think Saran wrap is 90% transparent to Longwave IR.
Though I would agree it’s probably 90% transparent to IR.
Saran wrap is PVC. Here is an absorbance spectrum for PVC:
http://www.ftir-polymers.com/soon_soubory/image015.gif
Unfortunately, I don’t know the thickness of their material for that plot. I’ve run ~23 micron Saran wrap on an FTIR and it was quite clean except for the ~3000 cm-1 region, ~1500 cm-1 region, and the etalons for internal reflection.
-Scott
–Unfortunately, I don’t know the thickness of their material for that plot. I’ve run ~23 micron Saran wrap on an FTIR and it was quite clean except for the ~3000 cm-1 region, ~1500 cm-1 region, and the etalons for internal reflection.–
I think in wavelength, ~3000 cm-1 is about 3.3 µm which mid-wavelength infrared, according to wiki:
https://en.wikipedia.org/wiki/Infrared
So that not longwave IR.
And ~1500 cm-1 in wavelength is
6.6 µm which also is mid-wavelength infrared, and per that graph it should be transparent [though could be impurities or something].
As I read it, 3.6 to 6.8 µm is transparent, 3.3 µm
is absorbs some.
And 7 to 16 has some of spectrum completely absorbed and over that range as an average of say 30% or more is absorbed.
So your graph does not indicate that in terms of longwave IR
which is in range 8 to 15 µm [or wavenumber: 1250 to 666.6]
that it’s not 90% transparent.
But for entire IR [which not all included in that graph] it’s about 90% [or more] transparent.
no, I think got that wrong, 1500 is looks like would be absorbed- so 3.6 to 6.7 µm would be transparent.
I have been into this important issue previously here:
http://www.newclimatemodel.com/our-saviour-the-hydrological-cycle/
The ocean skin is always colder than both the air above and the water below because evaporation has a strong net cooling effect.
Evaporation takes up nearly 5 times as much energy than is required to provoke it at 1 bar atmospheric pressure. That is a consequence of the latent heat of vaporisation being a very energy hungry process.
There is no way that any surplus energy can be left over from the evaporative process in order to either warm the oceans or reduce their rate of cooling.
In reality, the energy content of the oceans controls Earth’s air temperature and the ability of the oceans to retain energy derived from insolation is determined by the weight of atmospheric mass bearing down on the ocean surfaces.
http://www.newclimatemodel.com/the-setting-and-maintaining-of-earths-equilibrium-temperature/
If there were no oceans then Earth’s surface temperature would be determined by the kinetic energy required at the surface to hold the weight of the atmosphere up against the force of gravity in hydrostatic balance:
http://www.public.asu.edu/~hhuang38/mae578_lecture_06.pdf
Steven Wilde,
you are right about latent energy as an “energy hungry” process. Evaporation is a forcible draft of available energy and this “draft” determines the temperature profile and the thermodynamics of the sea surface, including the nightly convection/overturning of the near surface water.
To claim that wind will “mix” incident LWIR energy into deeper water and so warm SST claims in effect that wind warms the ocean.Wind does not warm but cools the ocean by increasing evaporation, a cooling process. Any mixing simply facilitates the “draft” of warmer subskin energy to the cooler surface, where evaporation occurs.
Did you switch probes to verify the results were repeatable?
But, this is “climate science”, so if you get the results you want you get to quit!
did you notice the water contained that ended up warmer at the end, actually started out cooler?
Also, then you twist the wire probes together, they indicate the same temperature to the readout precision (0.1 deg.)
Twisting the wire together is not the same as reversing the probes. But, like I said, this is “climate science”. The guidelines of the scientific method are thrown out.
Does not a simple thought experiment answer the original question.
IF the water surface is warm and the sky above it cold and clear, then both IR emission and evaporation from the surface would exceed downward IR radiation warming the surface, and the water would not warm, but cool.
But if the water was cold, thereby retarding both evaporation and IR loss, but the air/sky above was warm such that downward IR was high, then the water would warm.
Of course, this neglects heat loss by conduction, but there are situations that would be small.
Actually, warm air does not easily transfer heat to colder water, because atmospheric convection works against it.
“If the ocean were colder than the atmosphere (which of course happens) the air in contact with the ocean cools, becoming denser and hence more stable, more stratified. As such the conduction process does a poor job of carrying the atmosphere heat into the cool ocean.”
Summary from
http://eesc.columbia.edu/courses/ees/climate/lectures/o_atm.html
yes, but that’s not the point. The point is that if you REDUCE the rate of cooling from a warm surface to a cool atmosphere, the warm surface will have a higher temperature.
Temperature change stops when the rates of energy loss and energy gain become equal — thus, temperature is not a function of the rate of energy input alone. For the same energy input, you could have a stable temperature at 100K or 1000K.
@Roy…”The point is that if you REDUCE the rate of cooling from a warm surface to a cool atmosphere, the warm surface will have a higher temperature”.
Roy, old chap, sometimes you have a convoluted way of stating a point. I appreciate your comments in general but I simply cannot understand your points when it comes to thermodynamics.
Why do you relate surface temperature to an external and mysterious phenomenon that controls it? The only way you could control surface cooling is to wrap the Earth in a thick insulator. But that would also prevent solar energy warming the surface.
Presuming that GHGs in the atmosphere act as such an insulator just ain’t physics. What is there in ACO2, given it’s concentration of about 1/1000 of 1% of atmospheric gases, based on a CO2 density of 390 ppmv, that could possibly slow down surface cooling?
In fact, what is there in an atmospheric concentration of 1% of atmospheric gases that represents GHGs, that could account for greenhouse activity?
As Gerlich and Tscheuschner stated in their paper on the greenhouse theory, if CO2 behaved anything like it is claimed to behave it would be a new super-insulator.
In an earlier comment, you alluded to CO2 having a 50% effect on warming. I have no idea what you meant by that so I wont jump to conclusions. Gavin Schmidt at that other site, claims a warming effect from ACO2 of between 9% and 25%, depending on the humidity. He supplies no evidence whatsoever to back that claim. I think he grabbed it out of a hat.
Lindzen claims in the paper I posted earlier that a doubling of CO2 would have no more effect on atmospheric temperature than a few tenths of a degree C. Based on a 390 ppmv density for all CO2, ACO2 has a density of about 4% of that value. That is admitted by the IPCC, that ACO2 is only a small fraction of natural CO2 in the atmosphere.
Where is this insulator coming from to slow down surface cooling?
Gordon 1:29am: “Where is this insulator coming from to slow down surface cooling?”
See the Feynman experimental evidence presented above Gordon – in the non-zero difference in the red and grey curves in the top post. No pretending.
G&T spent something like 115 pages completely missing an atmospheric layer has distinct sources of opacity (e.g. various absorbing gases) defined by their grey opacity since each gas has a unique mass extinction coefficient depending on pressure, temperature and mass density of each gas in the mixture. This experimental science is not missed in Bohren’s text book on atm. radiation that Gordon previously mentions.
Any differential in the evaporation rate between a surface open to the sky and a surface overlain by any material is due to higher humidity developing below the material which then suppresses the rate of further evaporation by reducing the speed of upward convection.
It is not correct to assert that the flow of air is not reduced beneath an overlying material however small.
The smaller the material at a given wind strength the smaller the effect will be but nonetheless the warming effect is due to increased humidity beneath the material and not downward IR.
Every cloud is floating on its own column of higher humidity which is why the surface beneath it warms. Nothing to do with downward IR.
@Stephen Wilde…”Nothing to do with downward IR”.
Stephen…even if you had oodles of down-dwelling IR, there is no reason to presume it will cause heating in a surface it contacts.
If you have two surface at a distance, one hotter than the other, and the hotter surface has warmed the cooler surface, there is no reason to think that IR radiated by the cooler surface will have any heating effect on the warmer surface whatsoever.
The radiation laws of Kircheoff and Bolzmann have nothing to do directly with heat transfer, which is governed by the 2nd law of thermodynamics. The 2nd law states clearly that heat can only be transferred from a warmer surface to a cooler surface.
Some people are claiming that the net IR flow between the bodies determine heat transfer and that is plain wrong. The explanation is at an atomic level. Atoms with a certain average kinetic energy, which is heat, also have electrons residing at certain orbital quantum levels. Raising or lowering those electron energy levels requires precise energy input/output at precise frequencies.
For example in the hydrogen atom the Balmer series of spectral emissions lines shows that hydrogen emits energy only at specific frequencies in the visible spectrum. Each emission line is related to a precise change in electron orbital energy levels.
I am hypothesizing, till someone supplies me information to prove me wrong, that IR emissions from a cooler surface, especially one warmed by a warmer surface, cannot affect the energy levels in surface atoms of the warmer body because they lack the intensity and the frequency.
I wish someone could straighten that out for me.
I don’t think there is any law claiming that any surface intercepting an IR field has to warm it. The conditions have to be right otherwise the IR has no effect.
The “water with the shield” is ever so slightly warmer because the foil canopy itself is closer to the water and is slightly warmer that the air-layer above the other.
Surely Dr. Spencer, surely you see that? Stay with the primary relevant principles, temperature differentials, not IR. The difference is so close that IR is shown to be irrelevant, not non-existent in both cases, but irrelevant.
Thanks for showing this is true.
Surely it seems you do! Sorry Roy, I posted without reading slowly and a second time. My bad.
For all essence, the canopy is acting just like a tiny cloud over one side in this small experiment. The -6 W/m2 being prevented from leaving the covered water due to the temperature difference, σAε(t1^4-t0^4), is likewise leaving +6 W/m2 going upward above and from the canopy’s upper surface in comparison to the uncovered water surface temperature, with a little thermal mass in the way. Both emitting upward the same to space? Seems so.
The uncovered emitted more with nothing in the way but the top surface of the canopy also emitted more because the canopy is slightly warmer that the uncovered water surface.
Dr. spencer, do I have that correct?
Symmetry in physics can make fools of all of us many times.
I’ve created some “Greenhouse warming test cases” http://scottishsceptic.co.uk/2015/06/28/greenhouse-warming-test-cases-do-you-agree/ which I would love people to look at and give their views.
The reason for this, is that I’m never sure whether people are just trying to describe the same thing in a different way (warming = not cooling), or whether there is a fundamental difference in view that affects what people predict will happen.
As for the experiment – nice and simple.
Suggested change to satisfy the “it can’t warm”.
Perhaps if you filled both tubs with an equal weight of ice then you could repeat the experiment in a way where it cannot be argued it is not warming up.
OK, I did a simple experiment that very quickly proved this idea that IR can’t heat water is bs.
I turned on the grill and put a pan of cold water under the grill and guess what – it got hot it was 65C within about three minutes and much hotter than the air around. So there was no place for this heat to be come from, except the grill. A very small amount of visible red light was also produced, but no more than a 20ma LED so <20mw.
Busted.
Works for me. Although, the grill may be emitting a different spectrum (hotter) than the night sky.
I hope so otherwise we are all toast.
But seriously, it’s dull to bright red, which means the element temperature is around 700-1000C and the visible is about 1millionth the power of the IR.
Now written up: http://scottishsceptic.co.uk/2015/07/01/5-minute-experiment-shows-ir-heats-water-from-above/
The structure of the grill enclosure prevents convection and air circulation so humidity rises, evaporation slows and the temperature rises.
It wouldn’t work beneath an openn freely convecting sky.
Experiment busted.
Stephen,
Read his experiment. He put the water BELOW the grill. It wasn’t INSIDE the grill.
So: local air temperature rises, decreasing relative humidity and increasing evaporation rate. Thus, your cooling mechanism should be working even stronger than normal. Debunking busted.
mpainter’s objection below does have some merit. To prove mpainter wrong, he’d want to rerun the experiment in a container that is either highly insulating or reflects IR light.
-Scott
A _pan_ of cold water? Hmmm. Obviously the grill warmed the metal pan which warmed the water via conduction. Nothing to do with IR.
Mpainter, I did an experiment on the stove with a pan of hot water and two thermometers: a contact one, and an IR one (8-14 um blackbody cal). The pan plus the hot water measured 124 deg F with both thermometers. So either the water is transparent to IR, or it is emitting IR as a black body would. To resolve which, I IR measured the pan of water through a plastic bag of cold water. The contact thermometer in the cold water and the IR thermometer both read 84 deg F. That seems to indicate that the water is opaque to IR. (The bag has only a slight attenuation – the 124 deg F pan of hot water reads 114 deg F looking through the bag alone). If the water is opaque then the 124 deg F I saw in the first step must be the water, not the pan below it. Likewise Sceptic’s pan shouldn’t have been heated directly by the grill, but by the IR heating the water which heated the pan.
Jimc,
Great info. The challenge is that the atmosphere is known to be considerably cooler than the water. How do you propose measuring cooler object IR transferring to warmer objects?
The above experiment already shows that. I’ve shown IR from a hotter body heats water, Roy’s shown that IR from a less cold one keeps it warmer.
We’ve shown that both warmer surfaces than the ambient whether colder (roy) or warmer (me) cause water to be warmer than it would without those surfaces being present.
There’s nothing at all I’ve seen that even vaguely hints that the normal theories of IR heat transfer work exactly as the theory predicts.
Sceptic, I think you mean “There’s nothing at all I’ve seen that even vaguely hints that the normal theories of IR heat transfer DON’T work exactly as the theory predicts.”
Also, items at different temperatures emit IR spectra that overlap. The receiver of the IR photons doesn’t know the temperature of the source.
The proper experiment involves a dark heat gun (one whose coils stay “dark”) and a bucket of water at room temperature. You will see that LWIR does not heat water.
Ok, done. I’ve heated up room-temp water with a CO2 laser (a “dark gun”). LWIR did heat the water.
-Scott
Mpainter, the rule I’ve always heard is that a good (or poor) emitter of radiation is a good (or poor) receptor of it. I know water is a good (near blackbody) emitter in the 8-14 um range – I’ve measured it.
@Maike Haseler…”I turned on the grill and put a pan of cold water under the grill and guess what – it got hot…”
Awa the noo, and dinny bae meddlin’ in the scullery!!
You took a radiant source of at least 1500 watts at a narrow heating angle, which is a far different situation than in the atmosphere. Even solar energy can’t supply 1500 watts in such a narrow heating angle at the surface.
I don’t think the argument is that IR can’t heat water I think the debate is about how deep atmospheric IR radiation can penetrate the ocean.
If your pan of water was a moderate sized lake, and you applied your 1500 watt oven to it, what effect would that have?
The reason your water rose in temperature so rapidly was it’s proximity to a very hot IR source. That does not exist in the atmosphere.
@Mike…sorry for the typo of Maike.
Completely beside the point, but I’m baffled by the behavior of the shield. It’s a good emitter, and thus a good absorber, of IR. It is being illuminated from below by 80 deg F IR and from above by 7 deg F IR, yet emits 80 deg F IR below (and above?). Where does this extra energy come from? Shouldn’t it be emitting something between 7 and 80 deg F IR in both directions? Is it following ambient air (not a good conductor) temperature?
Next, the presence of the shield at 80 deg F and the blank sky at 7 deg F proves that the shielded water will be warmer than the unshielded water without even doing the experiment. (If the water has any IR absorption at all.)
the shield is probably a little under the ambient air temperature, for the reason you suggest. But I can’t accurately measure its temperature…it will be much close to the air temperature than the “sky temperature” which was probably more like 50 deg. F.
Hi Roy,
Jimc stated: “I’m baffled by the behavior of the shield.” I ask: Why is it that no one else beside he and you have commented directly about the shield? It is heart of the entire experiment. This is a rhetorical question which I do not expect anyone to answer. It would be enough if they only thought about a possible answer.
You wrote: “I put a piece of aluminum flashing painted white with high IR emissivity paint to block IR emission from the “coldest” part of the sky (directly overhead), but small enough to not restrict air flow around it as evaporation (and sensible heat transfer while the water was warmer than the air) cooled the water in both containers:”
I ask you, now expecting an answer, why did you not paint a sheet of extruded Styrofoam with high IR emissivity paint to block IR emission from the “coldest” part of the sky? Another question I would like to have you answer is: How is it that the atmosphere cools below the temperatures of the water. For a cloudless atmosphere has nothing to evaporate?
Have a good day, Jerry
Hi Roy,
See I got my periods and question marks mixed up.
Have a good day.
Clouds aren’t gases.
Greenhouse effect theory says only greenhouse gases cause
warming.
Clouds both have a cooling effect and reduce the amount of cooling.
As others said greenhouse gas can’t cause warming, so that is strike one against the theory- it’s not accurate so it has no right to calling itself science.
The only thing on earth which actually warms large area of earth is the ocean, because it brings heat to a cooler area, though winds can have more limited regional effect upon warming a land region and latent heat can bring warmth into region.
So greenhouse gases are about reducing the rate in which one has radiant losses, but as I said clouds also do this.
And there other factors which results in less cooling- and most significant is the ocean. As sunlight penetrate the ocean surface and the heat of sunlight is trapped and most of sunlight reaching the earth surface is absorbed/trapped
by the ocean. And both direct and indirect sunlight penetrates the ocean surface.
So noon with sun at zenith, as wiki says:
” If the extraterrestrial solar radiation is 1367 watts per square meter (the value when the Earth–Sun distance is 1 astronomical unit), then the direct sunlight at Earth’s surface when the Sun is at the zenith is about 1050 W/m2, but the total amount (direct and indirect from the atmosphere) hitting the ground is around 1120 W/m2. In terms of energy, sunlight at Earth’s surface is around 52 to 55 percent infrared (above 700 nm), 42 to 43 percent visible (400 to 700 nm), and 3 to 5 percent ultraviolet (below 400 nm)”
So unless reflected by the surface of ocean, somewhere around 90% of this shortwave energy penetrates the ocean surface, and this heat generated remains for hours, days, centuries.
Also if there is clouds and sun is not at zenith one gets indirect sunlight- a large portion of sunlight is indirect sunlight. So sunlight is diffused by the mass of the atmospheric, and by other elements in the atmosphere, and such diffused light doesn’t matter much in regard to the ocean, because the ocean itself diffuses sunlight. So diffused sunlight it receives is no different than the diffused sunlight the ocean makes.
And this *may* be one factor of why solar ponds can reach a a under the surface temperature of 80 C. And in terms making the highest temperatures of land surface, this diffused sunlight could be less effective in raising the surface temperature.
So it seems from the start the greenhouse theory is invalid,
but this does not mean that there is no radiant effect of greenhouse gases, rather it’s a question of how much.
Um…what??
Clouds are condensation of water vapor into little droplets of water which suspended in the atmosphere. Droplets are liquid water and not a gas.
So greenhouse gases are not liquids or solid [ice]. Unless one has some special need to alter the scientific definition of gases, liquids, solids- or plasma.
The greenhouse effect theory says that only gases called greenhouse gases can increase the average temperature of a planet.
Or the actual greenhouse effect of a actual greenhouse which warming by trapping air, according to the greenhouse effect theory, does not warm a planet.
Or here:
“Greenhouse Effect
The greenhouse effect refers to circumstances where the short wavelengths of visible light from the sun pass through a transparent medium and are absorbed, but the longer wavelengths of the infrared re-radiation from the heated objects are unable to pass through that medium. The trapping of the long wavelength radiation leads to more heating and a higher resultant temperature. Besides the heating of an automobile by sunlight through the windshield and the namesake example of heating the greenhouse by sunlight passing through sealed, transparent windows, the greenhouse effect has been widely used to describe the trapping of excess heat by the rising concentration of carbon dioxide in the atmosphere. The carbon dioxide strongly absorbs infrared and does not allow as much of it to escape into space.
A major part of the efficiency of the heating of an actual greenhouse is the trapping of the air so that the energy is not lost by convection. Keeping the hot air from escaping out the top is part of the practical “greenhouse effect”, but it is common usage to refer to the infrared trapping as the “greenhouse effect” in atmospheric applications where the air trapping is not applicable. ”
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/grnhse.html
So the greenhouse effect of a greenhouse is not applicable
to the greenhouse effect theory.
Or the staggering amount of heat within earth atmosphere, is not applicable to greenhouse effect theory, as it’s related to trapped convectional heat.
[Or perhaps, similarly to below, the idea is there would not be this warm atmosphere without the radiant effects of greenhouse gases. Despite the fact that were earth average temperature to be -18 C [or -40] there still would be an enormous amount of heat in the atmosphere.]
Supporters of greenhouse effect theory may be well aware that say the Gulf Stream warms a large part of Earth. But Greenhouse effect theory does not allow for the Gulf stream to warm Earth. Only in sense that radiant greenhouse gases prevent the gulf stream from cooling could the Gulf stream be increasing the average temperature of the planet-according to the Greenhouse Effect Theory.
And likewise many believers of Greenhouse Effect are aware that droplets in clouds cause warming of large region of Earth, but again it’s just the radiant effect of greenhouse gases which could cause clouds to increase the average temperature of Earth. This does not include latent heat either [or again the latent heat is cause be greenhouse gases.
And it’s believed that if not for CO2 [or some other non-condensing greenhouse gas] earth would have average temperature of of -18 C. So primary reason for 33 C of warming on Earth is due to CO2.
So as for:
“Um…what??”
Perhaps you be so kind of to direct me to the greenhouse effect theory, that you think is correct- or at least worthy
of refuting. Also like to know who the author of the theory is- the theory which actually correct. And due a fondness of tradition of person rather than committee which is the author.
Good idea. Where is your neck of the woods? BTW i learned the GHE on my own as a post-doc. It really never got mentioned in my meteorology courses.
you know my email. DFW area. Give me about a week’s notice.
HiRoy,
Thank you for the experiment, but couldn’t the shield have blocked some downdrafts of cold night time air from conveying energy from the water’s service? How conclusive do you believe the results to be?
Have a great day?
JohnKl,
That is a valid point, though the effect is likely minor. Roy thinks in terms of meteorology and set up his experiment as such. I’ve proposed a more controlled experiment here:
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194070
This is similar to the experiment run here:
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194017
But the difference is that this experiment would use insulated vessels and also has a control. I’m guessing no experiment will satisfy many of the commenters here, but you’ve seemed pretty reasonable in posts I’ve seen in the past.
-Scott
Hi Scott,
Thanks for the reply and links. The effect may indeed be minor but so appears to be the temperature differential with and without the shield. It seems an experiment devised in a vacuum could reduce to negligible any convective noise.
Thanks again and have a great day!
Hi Scott,
My statement should have read “…an experiment performed in a vacuum could reduce to negligible any convective noise.”
Thanks and have a great day
i have done quite a bit of research on this subject specifically & I think that everyone is lost trying to do experiments – I have set up a group as a database on facebook & collected every study or article on the subject- from every source that I could find the link is here
https://www.facebook.com/groups/1509057762686691/
After months of research & collating data this is what I have found out- the atmosphere is 1-3 degrees on average cooler than the ocean – the ocean skin layer is cool – 0.6 to 1 degree COOLER than the ocean & takes 12 secs to recover as shown here in this peer review & others
“Abstract
Measurement of the long-wave infrared radiation from the top 0.1 mm of the evaporating ocean demonstrates the existence of a cool surface layer characterized by departures of as much as 0.6 degrees C from the “surface temperature” found by conventional methods. Being very thin, the layer cools sufficiently rapidly to reestablish itself in less than 12 seconds after disruption by a breaking wave.”
http://www.ncbi.nlm.nih.gov/pubmed/17841610
Now physics tells me that IR cannot heat the ocean if it cant penetrate further than the cool skin layer which is cooler than the ocean – the cool skin layer shows that IR is heating up the already cold ocean surface layer to get it hot enough to evaporate & it so cold because of evaporation absorbing all the heat from the surrounding area
I don’t care how many experiments or whatever you have to say – you cant deny physics & these figures show me that it is impossible to heat the oceans with IR
You would think that would be the end of it, but the zealots will claim a cold wind warms the ocean next.
Some men you just can’t reach – Cool Hand Luke
They dont want to know anything good about climate change – they will do anything to preserve their religion
Marc,
Where people go wrong:
1. They fail to grasp what such a minute interval cannot transmit heat downward, that the IR is returned to the atmosphere within a few seconds as latent energy or radiation.
2. That energy flows from the warmer subskin to a cooler surface night and day, hence no heat can be conducted downward.
People also declare, as Roy has, that “wind will mix” the absorbed IR energy “down” and so the sea is warmed by wind. Well, no, the wind cools the sea surface and there is no room for disputing this fact.
Yes if the skin layer is colder than the ocean … heat MUST flow from the warmer water below up & any wind & wave
mixing will only make the ocean cooler
Marc Facer,
From my reading of the Roy Spencer experiment with the water tubs it has nothing to do with IR heating the water. Look at his graphs and tell me the direction of the water temperature over time?
In both cases the water is cooling. The water at the beginning of the experiment is around 81F and ends at 72F. In both cases the water is cooling. The rate of cooling is the only thing changing. Neither condition is WARMING the water! The shield is not warming the water. Carbon Dioxide with backradiation does not warm the ocean. The ocean still cools with no solar input. The rate of cooling is what changes, then with the water slightly warmer when the sun rises again it can reach a higher temperature until it reaches a new equilibrium.
Where does the meme come from that GHE is producing this direct warming of the surface? I have asked many and know one yet has explained it. Do you know?
First of all my post has nothing to do with what Roy has posted & I don’t care to comment on what he found
Secondly the IR has to heat the surface to bring it to a temperature so it can evaporate – that temperature is still lower than the oceans surface
Please look at my post & show me where I am wrong in the physics & explain to me how IR can heat the ocean
Hi Roy,
Thanks for running this experiment. I’ve spent far too much time trying to convince people that IR can warm bodies of water. Unfortunately, their ignorance of the terminology used in scientific/engineering heat transfer and their misinterpretation of both models & experimental results are almost impossible to get around. Even showing that heat transfer (even absent of advection) is far faster than the evaporation rate won’t change their minds. Sometimes I think that these people are paid by Greenpeace etc to just pretend to be skeptics and then do dumb things to make the rest of us look bad through guilt by association.
Oh, and here’s a much easier (and fun!) experiment to show people that LWIR can heat water–use a CO2 laser cutter. We have a 40 W laser in my lab and I’ve used it in the past to seriously heat up some water (to boiling). It’s so cut and dry proof of what you’re saying, though people still won’t accept it. Maybe if we put a thermometer at the bottom of the water container? When they ignore that example, then you know they won’t change their minds.
-Scott
“We have a 40 W laser in my lab and I’ve used it in the past to seriously heat up some water (to boiling).”
Scott, do you understand how a laser works? Do you understand there are no lasers in the atmosphere?
I guess not.
geran,
Same standard approach I get from all of you people. No refutation, just proof by assertion.
Though I prefer to avoid snark, I try to match the approach applied by the people with which I converse, therefore: do you understand that the photons from a laser do not know whether they’re emitted from a COS laser or from CO2 molecules in the atmosphere? I guess not.
LWIR is LWIR, doesn’t matter the source.
-Scott
Reasonable Scott vs the others
Others (definition): Have never taken a physics level thermodynamics or physics class in their lives. Read some good sounding articles on the internet and now are the “experts” and all the hundreds of years of study on the topic are wrong and they know it. Others rarely link to any articles but are the absolute end of knowledge on the topic. Believe in science to the degree it does not conflict with their own beliefs, regardless of how illogical they may be, but when it conflicts then it is a conspiracy funded by the evil greens to brainwash people into a false belief of science.
I compliment Roy Spencer for actually doing an experiment! Others can say what they wish, the experts should come up with their own experiments to prove what they believe but they never will.
Norman,
Here’s another well-controlled experiment one could do that might be easier if one doesn’t have access to a CO2 laser:
Get a thin-film resistive heater and place it reasonably close above an insulated, open water container. Make sure it’s not touching the water and there’s enough room for convective heat/mass transfer above the water. I’d say 1-2 cm would be a good distance. Turn on the heater and let it sit at 80 C or so for an hour. Measure the change in water temperature. Then do the same with an identical setup but with the heater off…this is a control.
By their reasoning, the water below the “on” resistive heater will be colder than the water below the “off” resistive heater. I would be willing to bet my entire June paycheck (which I receive tomorrow) that they are wrong.
-Scott
Hi Norman ,
Personally I have nothing against physics. However, I do resist pseudoscience academics and politicians spouting feigned concern for nature while actively pursuing genocide. This includes paranoia of a runaway greenhouse effect resulting in Venusian temperatures from simply consuming hydrocarbons on ones way to work, and Malthusian scare-mongering including the notion by many greens that billions must be eliminated for the planet to be sustainable.
In the past I have documented false predictions and the expressed homicidal intentions of various green Nazi’s. To deny the empirical evidence and expressed intentions of many in the green movement appears at best ignorant and at worst dishonest.
Jacque Cousteau,
“In order to stabilize world population we must eliminate 350,000 people per day.”
Bill Gates,
“the world today has 6.8 billion people. That’s heading up to about nine billion. Now if we really if we do a really great job on new vaccines, health care, reproductive health services, we could lower that by press 10 or 15 percent.”
Have a great day!
Hi Norman,
My last post contained a statement apparently from Bill Gates that should have stated:
“…we could lower that by perhaps 10 or 15 percent.”
Have a great day!
JohnKl,
I do understand that some feel the world is getting too crowded and see this as a threat and may have hijacked Climate Science to use it as an emotional steering wheel. I have also learned information is not as significant as the package it is wrapped in, the presentation of it. Some people are more susceptible to how the information is presented in determining if it is true or not. I would call this group the “Alarmist camp”.
The overpopulation group may be right. It would be hard to say. If every available person had as many children as they could possibly produce you may run out of food at some point.
I think the intrusion of political motivation in Climate Science is a very bad thing for science. Reason is destroyed.
I see fanatics on both sides blaring out their views but never quieting their minds to hear the science.
I look at Roy Spencer as a more classical scientist. Stay calm, look for evidence, try to understand the data, stay clear of the absolutes.
My objective is to keep the science alive. With people like mpainter and geran it is a most difficult task. They spout things off like complete experts but never link to a source. When you explain how they are not understanding things they are unable to grasp it and just go right along saying the same thing over and over like stuck in a loop or something unable to break free and think. Cult programmed by the goofy dudes at PSI.
Poster Curt comes and and spends many posts with some deep thermodynamics concepts that are spot on with textbook explanations available on the web and they look at him as a moron that understands nothing. Really sad. I think you have a good questioning mind and it is pleasant to reply to your posts. I will not get all my ideas correct but I will think about them and do further study to correct the errors.
If science is going to maintain its position as impartial truth seeker then only an open mind will guide the way.
You have a great day!
Scott, my “snark” is only an attempt to keep you from making a fool of yourself. Do you know what the “A” in LASER refers to? The atmosphere does not contain any LASERs.
(No “snark” intended.)
Scott seriously ?? A laser is not the same thing now is it ?? We don’t deny that IR heats the ocean surface at all – the cool skin later – see my post just above yours
Evaporation clearly overwhelms the heating effect
The atmosphere is 1-3 degrees on average cooler than the ocean – the ocean skin layer is cool – 0.6 to 1 degree COOLER than the ocean
If IR cannot get further than the cool skin layer – please show the physics by which IR can heat the ocean or the cool skin layer for that matter
Marc,
Please inform me of how the photons from a CO2 laser are different from photons from free CO2 molecules in the atmosphere.
I am apparently ignorant.
-Scott
I didnt say that photons were different – but a laser is very different to what comes out of the sky & I did not deny that IR does not heat now did I – now go look at what I have shown you & please answer the question
Marc Facer says:
June 30, 2015 at 10:17 PM
So how is a laser “very different” if the photons aren’t different? I’m apparently ignorant. Please enlighten me.
Here are the questions you have asked me:
I see no other question marks.
If you want me to show physics by how IR can heat below, I can do that. But I want solid numbers from you first b/c every one of the individuals with your beliefs seems to have different numbers and changes the numbers/rules every time I put in the effort to write some serious calculations. Here is the information I need:
1) What is your assumed evaporation rate of the ocean?
2) What is your assumed thickness of the “cool skin layer”
I can show multiple mechanisms by which the oceans are heated and/or have reduced cooling via IR, but I need a non-moving target. I’ve already shown by experiment that IR can heat water below the skin layer, but people aren’t buying that so I’ll show via theory instead.
-Scott
Hi Scott,
Not to be a curmudgeon but co2 lasers emit coherent IR. Excuse me for not using the term “photon” but I don’t suffer from wave particle/duality and ensuing cognitive dissonance.
Coherent microwave oven emissions boil water and heat food as do IR emitting instruments, but I’ve yet to see IR emitting clouds boil away the oceans beneath.
However, as I’ve mentioned in previous threads direct visible spectrum solar radiation penetrates clear transparent sea water readily, eventually being absorbed by particulate matter, biomass, and/or the ocean floor. Unlike typical atmospheric IR, direct sunlight contains more than sufficient energy to bridge the quantum gap and significantly increase temperature and not merely reduce the rate of cooling. Trying to find an IR pin in a mountain of visible spectrum needles seems brave. Good luck you’ll need it.
Have a great day!
I tried to point out that the intensity of a laser is quite different to what comes from the sky but for some reason you missed it – my initial post is above yours if you want information |
If you want to discuss in depth your quite welcome to discuss it all here on my facebook database which has all known information that I have been able to collate about the subject – I dont think that this is the place to do that
https://www.facebook.com/groups/1509057762686691/
By the way lets keep with the physics first & answer the question that has been posed to you twi8ce now
The atmosphere is 1-3 degrees on average cooler than the ocean – the ocean skin layer is cool – 0.6 to 1 degree COOLER than the ocean
If IR cannot get further than the cool skin layer – please show the physics by which IR can heat the ocean or the cool skin layer for that matter
Marc,
I don’t have Facebook.
And what question have you posted “twi8ce now”? There isn’t even a question in your post.
To keep with the “physics first” I need some pre-defined numbers from your side so you can’t just change them on the fly. 1) What is your assumed evaporation rate of the ocean. 2) What is your “cool skin layer” thickness?
-Scott
Marc Facer says:
June 30, 2015 at 10:40 PM
Please link to where you said the laser intensity is different. I did a search for “intensity” after you typed this message and the only place it came up was this one time, so I’m calling BS.
BTW, why would intensity matter. Would evaporation continue to “clearly overwhelm” the heating effect?
-Scott
Marc,
Since light intensity matters for your proposed mechanism, at what intensity is the switch from cooling to heating? I can’t see how intensity matters in your mechanism, so I must be ignorant. Please be quantitative so I can understand.
-Scott
–Oh, and here’s a much easier (and fun!) experiment to show people that LWIR can heat water–use a CO2 laser cutter. We have a 40 W laser in my lab and I’ve used it in the past to seriously heat up some water (to boiling). It’s so cut and dry proof of what you’re saying, though people still won’t accept it. Maybe if we put a thermometer at the bottom of the water container? When they ignore that example, then you know they won’t change their minds.–
If put ice under an inch of water, and laser directed toward ice under the inch of water, does laser melt the ice?
gbaikie says:
July 1, 2015 at 8:32 PM
Hmm, I’ve never tried this. The laser would not make a direct “strike” on the ice if it was held 1″ below. Also, the ice would melt by itself anyway. My best guess is that since the water warms up then the ice would melt faster. It’s a lot easier to just measure the water temperature before and after for both the laser-exposed water and a control vial of water that wasn’t exposed.
-Scott
As for downwelling thermal infrared being absorbed in an extremely thin surface skin at the top of the water: This sets up an extremely high temperature gradient from this “skin” to what is next below, and that would cause extremely rapid heat downward conduction. This would go on recursively until the warmed upper layer with a temperature gradient gets thick enough and its temperature gradient gets low enough for continued downward expansion of this layer to become slow. At that point, as I see this, most of this layer is not subject to evaporative cooling.
It cant move heat downward – The atmosphere is 1-3 degrees on average cooler than the ocean – the ocean skin layer is cool – 0.6 to 1 degree COOLER than the ocean due to evaporation
This helps cool the ocean
This explains it all
http://www.klimaatfraude.info/oceaanopwarming-of-zeespiegelstijging-door-co2-is-niet-mogelijk_193094.html
“Abstract
Measurement of the long-wave infrared radiation from the top 0.1 mm of the evaporating ocean demonstrates the existence of a cool surface layer characterized by departures of as much as 0.6 degrees C from the “surface temperature” found by conventional methods. Being very thin, the layer cools sufficiently rapidly to reestablish itself in less than 12 seconds after disruption by a breaking wave.”
http://www.ncbi.nlm.nih.gov/pubmed/17841610
Marc Facer says:
June 30, 2015 at 10:32 PM
I have left nothing out & your comment is irelavant – but at night its the same temp – but still irelavant
Answer the question – what would the temperature difference be without the presence of IR irradiance?
You can’t point to a dT in the presence of IR light and make a statement about what the IR light does to that dT without having the number for that dT in the absence of the IR light. It’s called a control/blank and the most fundamental part of empirical science.
-Scott
Scott I have answered the question – I have given you all the information & I have suggested that you move this elsewhere – I have provided links to further information & you have ignored everything that I posted – now you are demanding information – I tried to be nice – now follow what I said – if you don’t have or want to go to facebook then fine don’t bother – if you dont want to read the information provided then fine – just stop harassing me !! I have already pointed out that I do not wish to discuss it further & clog up the forum with irelavant bullshit – so please have some respect for others & stop being a twit – I have asked you a question & you claim to need more info – well you don’t – now please move to my forum / database or STFU & leave me alone
Marc Facer says:
June 30, 2015 at 11:15 PM
No you have not. The question is – what would the dT between the surface layer and the bulk ocean be in the absence of IR irradiance? You’ve only given a value in the presence of IR irradiance. You need both to be able to make a direct claim that IR does not warm or inhibit cooling.
Harassing you? Wow. I ask for a control, the most basic part of empirical science, and that’s harassment? Or I ask for specific numbers so I can show physics that overcome them…that’s harassment?
Looks like you can’t put up so you throw a tantrum and start name calling. Here’s what’s funny about this discussion–IT IS COMPLETELY ON TOPIC. I’ve given up interacting with others on some of the other threads here because it was off topic there and clogging the thread…but now it’s completely on topic. I’ll compile several mechanisms as to why you’re wrong tomorrow and post them (almost bedtime here). I’ve posted these before on other threads on this site. I’ve scrolled through the rest of the comments and found at least some semblance of a number from you. That one number should be enough to completely wipe the floor with you. You still won’t believe me, I’m sure. But maybe a few people with some sense will read it and understand. Norman, who posted above, will at least enjoy it. Probably Roy too if he’s not too busy.
-Scott
I don’t see how this experiment proves anything. LWR cannot penetrate water more than skin deep and even with wave action any warming will be shallow and temporary. LWR from CO2 won’t even reach the sea as it is only emitted and absorbed at a wavelength of 15 microns which only happens at -80C (my understanding). Saturation of this band occurs in 100m so a doubling of CO2 means saturation in 50m.
So this experiment shows water cools slower with a reflector above it. So what was the air temperature and humidity level? This slower cooling was caused by radiation from moist air, not CO2.
I applaud Dr. Spencer’s effort, but must concur with your assessment. The air / surface layer is being warmed by the water. This warmer air is convecting upward. The cover slows this upward airflow process and causes a slight temperature variance.
Since the water with the aluminum is warmer, one could argue that a thin piece of aluminum is generating more DWLIR than the entire atmosphere. Since based on the Trenberth diagram, DWLIR delivers more to the surface than the sun, it is hard to grasp how a thin piece of inert aluminum is a stronger energy provider than the sun.
I am sure the S-B equation can prove this though /sarc
Could someone (more that one would be even better) with an IR camera see if they can take a picture showing whether IR reflects off of a calm water surface or not similar to how this example commented on over at watts that shows IR reflected off of tarnished copper plate?
https://wattsupwiththat.files.wordpress.com/2015/06/thermal-reflection.png
Much of the disagreements seem to have reflection/no-reflection at the very core.
I should have said more precisely “… whether IR reflects off of a calm seawater surface or not …” for possibly the present of free flowing electrons/ions may be necessary compared to distilled pure water.
Dr. Richard Feynman explicitly made it clear in his lectures in New Zealand decades ago that all reflections are really absorption/re-emissions events (from glass, metals, water) and that the refection occurs not at the very interface/surface but from deeper atoms within the reflecting matter by symmetric interaction with free electrons as shown by his theory of quantum electrodynamics (QED) that, as far as I know, still stands. He limited his lecture to visible frequencies but I still wonder on the infrared aspect. If I am correct all microwaves bounce (reflect) right off of such surfaces… what of the frequencies inbetween.
We need an infrared photo it seems for a glimpse.
Hi Wayne,
I tried to respond to this comment of July 1 at 1:52pm, but it ended up near the end of all the responses at this time.
Have a good day, Jerry
Dr Specer
A couple things.
You say the atmosphere is largely opaque to infrared. No it isn’t. Somewhat more than 50% of the energy reaching the surface is IR, whether you like it or not.
Your temperature graphs show a continuously falling temperature. This is not the same as warming, unless you redefine the meaning of the word.
Place a portion of the water in a vacuum flask, and it will cool even more slowly. No warming, just slower cooling.
Look at the operation of a properly constructed solar pond. Temperatures above 80 C can be achieved – at the bottom of the pond, not the top. Contrary to what “everybody knows”, warm water does not always rise, any more than hot air will always rise. There are many other factors involved, but simple observation will suffice to verify the final outcome.
Your experiment merely shows that a body of water allowed to cool, apparently follows Newton’s Law of Cooling.
I’m not sure why a slight insulating effect is promoted as “warming”. It isn’t. The temperatures of the surface of the Moon, after the same period of exposure to the same Sun, exceed anything achievable on the surface of the Earth exposed to the same sunlight. The reason is that the Moon has no atmosphere – no insulation, if you will. Unlike the Earth, no energy from the Sun is prevented from reaching the surface. Hence higher temperatures.
No GHE. Your satellite data will show this, in time,
Dr Spencer. Sorry. Fat finger re your name.
Last night I read the comments and particularly those saying Roy had not proved IR could warm, so I quickly worked out a very simple experiment to show IR does heat water and one many people can easily replicate and I’ve detailed it on my blog:
http://scottishsceptic.co.uk/2015/07/01/5-minute-experiment-shows-ir-heats-water-from-above/
Please try to keep up, Mike. No one is arguing that a “heat transfer” (ANY heat transfer) can’t heat/warm water.
In your experiment, the grill is your heat source. In the Earth system, the Sun is the surface’s heat source, while the atmosphere is its heat sink.
What should be obvious to everyone (Roy and you included) is that insulation never heats/warms the insulated object. It cools it. Only it cools it less effectively than the outside surroundings.
The atmosphere NEVER EVER warms or heats the surface of the Earth. It cools it. UNLESS the air above the surface is actually warmer than the surface.
Mike Haseler,
One can warm water with any wavelength, ranging from the shortest to the longest. A 10 cm RADAR will warm the water in your body rather nicely, until you die. Or a microwave oven.
Or try a UV laser. No IR, but it will warm water, or cut tungsten, if you prefer.
One problem is that the top layer of your water absorbs IR energy quite efficiently, and expands. It cannot sink through the more dense water. If heat “diffuses” downwards, the water warms, and rises. It cannot sink. Physically impossible.
This is why, in the absence of geothermal energy, deep water has the densest (usually the coldest) water at the bottom.
No GHE. None. No experiment has ever shown a GHE. Anybody who tries, fails. Ordinary physics explains everything you can observe at the macro level, with few exceptions. The Mpemba effect has me scratching my head, I must admit.
Oh well!
The skin layer is cooler than the ocean by up to a degree & water does not absorb IR – it cannot go further than the skin layer – the cool skin layer aids cooling
Marc Facer says:
July 1, 2015 at 9:14 AM
Ok, let me get this straight:
Water does not absorb IR, but the IR doesn’t get past the skin layer. Where does it go then?
-Scott
Its absorbed or reflected at the skin layer – it heats the skin layer but the skin layer is always cooler than the ocean – If you read my first post as I said originally & read the supplied literature in the link here which I supplied originally – then we wouldn’t be arguing here – you would understand –
The physics wont allow the cooler skin layer to heat the ocean – it is impossible – listen to what I am saying – read the link – then tell me how it can heat the ocean
http://www.klimaatfraude.info/oceaanopwarming-of-zeespiegelstijging-door-co2-is-niet-mogelijk_193094.html
Evaporation overwhelms the heating effect – the skin layer is always cooler than the ocean – its effectively drawing heat from the ocean & the IR reflected down
Marc,
I you would have the years of experience I do in both engineering heat transfer and in optical spectroscopy for monitoring chemical reaction kinetics, then you would “understand” too. I actually design and build devices that work with IR and water. If my stuff didn’t work, I’d be out of a job.
It doesn’t matter that the evaporation “overwhelms” the heating effect from the photon absorption. The energy balance is the sum of all of them. Say evaporation = -90 and photon absorption = +10, then you end up with -90, which is less energy loss than you would have in the absence of IR absorption.
I have personally shot CO2-based LWIR into water from above and observed the water to heat up. No one has provided any explanation as to how this is possible. You are the only one to even offer something, which was “intensity”. How quantitative.
-Scott
The question here is: Can infrared radiation [emitted from the atmosphere] warm a water body [the ocean]?
I say yes, and have a challenge for those who say no.
First, I think we need to clarify some terminology: What does “warm” mean?
Clearly, warm is a verb here and means “to make warmer” i.e. “to cause the object to have a higher temperature”.
So, warmer than what? Given that the planet is very close to a steady state, at least on suitable time scales, I think the answer can only be “warmer than it otherwise would be”. If you have a different answer, please be explicit and provide some justification.
On average, the surface of the earth absorbs about 165 W/m^2 of short wave radiation and 345 W/m^2 of long wave radiation. These are numbers that can be measured and do not depend on any theoretical treatment of the greenhouse effect. The surface emits 398 W/m^2 (also measurable) upward the balance is carried upward as latent and sensible heat. Although these are global averages and actual numbers vary from time to time and place to place, they should serve as a useful starting point for my challenge.
Now, imagine that the character ‘Q’ from Start Trek TNG wanders by and decides to mess with us by causing the atmosphere to become transparent to IR.
My challenge is this: What happens to the temperature of the surface ocean, measured at the conventional depth of 3 meters? A qualitative answer is fine, but you must conserve energy.
In case this is not clear: As a result of Q’s mischievous action, the downward IR at the Earth’s surface drops to zero.
The ocean is warmer than the atmosphere by up to 3 degrees on average – The skin layer is cooler on average than the ocean by up to a degree & water does not absorb IR – it cannot go further than the skin layer – The skin layer is cooler due to evaporation – there is no possible way that the cold surface heats the ocean
Marc Facer,
“there is no possible way that the cold surface heats the ocean”
I would agree with your statement. This is definitely an affirmation of the Second Law of Thermodynamics.
I think a lot of trouble on this thread, IMHO, is semantics. The IR is not warming the ocean in the classical sense of the word which would mean adding energy and in turn increasing the molecular speed and hence the temperature.
The snag in the understanding is that any energy hitting the surface will slow down the fastest possible cooling rate of no returning energy.
The fastest cooling rate for a surface is no energy hitting the surface, all radiating away at the rate set by temperature and emissisivity. But if you have energy hitting the surface of the ocean at the same time it is cooling, it will lose energy at a slower rate (Energy Out minus Energy In determines cooling rate).
If you have 390 watts/meter^2 leaving and zero return you lose this much energy (joules/sec so it is a rate dependent property).
If you still radiate at 390 but have 100 return then you are losing energy at the rate of 290 watts/meter^2 and your surface will not cool as fast. It can’t since less energy is leaving.
You go all the way up and if you have the surface radiate 390 watts/meter^2 but have 390 watts/meter^2 return. What happens. The surface is not getting warmer so the evaporation rate does not change. Now it will lose energy at the rate of evaporation and still cool but will cool much slower than the first situation which would lose energy both by evaporation and radiation loss. It would be the fastest cooling rate.
Norman,
“The fastest cooling rate for a surface is no energy hitting the surface,”
Nice that someone gets the point.
Yes,
Norman’s explanation is excellent. Well done.
-Scott
Hi Norman,
Thank you for responding to my Previous post above. As to this post, your claims appear well-reasoned and we’ve discussed then before. However, I believe it remains an open question as to whether EM radiation that doesn’t change a molecule or atoms quantum state gets absorbed it all and stored as internal energy. Assuming for the sake of argument it does, Claes Johnson makes a valid point that of the 240 w/m^2 absorbed by the Earth’s surface only 40 w/m^2 of re-emitted IR makes it out directly from the Earth’s surface to space unimpeded by the atmosphere. The remaining aperture to be closed by additional co2 proves too small to matter.
Claesjohnson.blogspot.com/2015/03/a-basic-model-of-greenhouse-effect-with.html?m=1
Let me know your thoughts and…
Have a great day!
It is adding energy – evaporation simply overwhelms & the heating effect is playing catchup – there is heat coming from the ocean & the IR reflected down from above – it heats the skin layer – but it wont heat the ocean
http://www.klimaatfraude.info/oceaanopwarming-of-zeespiegelstijging-door-co2-is-niet-mogelijk_193094.html
Almost every day I bike to work into the wind. Because my momentum “overwhelms” the wind then it does not slow me down using your reasoning. It all makes sense now–the wind slowing me down was all in my head! Thanks for clearing that up, I’m sure biking will be easier now.
-Scott
Mike M. says, July 1, 2015 at 9:14 AM:
“The question here is: Can infrared radiation [emitted from the atmosphere] warm a water body [the ocean]?
I say yes, and have a challenge for those who say no.
First, I think we need to clarify some terminology: What does “warm” mean?
Clearly, warm is a verb here and means “to make warmer” i.e. “to cause the object to have a higher temperature”.
So, warmer than what? Given that the planet is very close to a steady state, at least on suitable time scales, I think the answer can only be “warmer than it otherwise would be”. If you have a different answer, please be explicit and provide some justification.”
*Sigh*
No, Mike. When you “make something warmer”, you make its temperature rise. You’re HEATING it. The Sun heats. A stove heats. When you “make something warmer than it otherwise would be” pertaining to insulation, you do NOT make its temperature rise. You make its temperature drop. You’re COOLING it. Only less effectively than what the outside surroundings would. A jacket or a blanket cools the heated object, only less than the cool air outside. The atmosphere cools the surface of the Earth, only less effectively than space.
This is NOT warming by LW, Mike. This is reduced LW cooling. That is a totally different thing. In fact, the opposite thing.
Why is this so hard to fathom?
Kristian,
Thanks for the well informed explanation. Hope the larger group of thinkers on this thread think and consider what you are saying. The GHE does not warm the Earth’s surface directly. The input of solar radiation is doing the energy adding, the excess joules and hence the warming. The GHE is similar to an insulator. The energy returning to the surface (be it water or earth) from the atmosphere is less than what the surface is radiating away. The GHE (in no solar) will lead to a cooler planet. The concept is that by slowing down cooling with a continuous energy supply you will end with more energy overall and a warmer surface.
Roy Spencer’s real world experiment does demonstrate this quite well. The GHE does not warm but it does slow cooling.
But too few can grasp the simple idea that energy hitting the surface of the water will slow the cooling rate and will lead to warmer temperature overall when the sun shines because less energy left the system.
Kristian,
Let me see if I’ve got this right. My nice warm winter coat works by cooling me?
Yup. Only it cools you LESS EFFECTIVELY than the cold air outside your winter coat. This is not so hard to grasp, Mike.
Kristian,
It is pretty clear that we are not speaking the same language. So please have patience while I try to figure out how to translate. Another question to test my understanding of what you are saying:
The average temperature of the Earth has been very nearly constant for many millennia. Therefore, the sun does not warm the Earth. Is that right?
Yes, of course it does. It warms/heats us wherever and whenever it shines. There/then temperatures rise, after all. However, it only shines half the time. So on average it is simply MAINTAINING our dynamic steady state temperature.
Of course it warms/heats the Earth. Wherever and whenever it shines. However, it only shines half the time in any particular place. So on average it only maintains our (dynamic) steady state temperature.
I guess you have noticed that temps rise when the Sun shines. It is normally warmer during the day than during the night, wouldn’t you agree?
So does that mean the sun cools us after ~2-3 pm (peak daily temp)?
-Scott
–So does that mean the sun cools us after ~2-3 pm (peak daily temp)?–
Ocean are still warming. Oceans are most of planet surface.
Clouds are being warmed, snow is being melted. Plants are gaining energy and growing.
Scott says, July 2, 2015 at 8:26 AM:
“So does that mean the sun cools us after ~2-3 pm (peak daily temp)?”
No. The atmosphere (and space) does. The Sun always heats us. But after ~2-3 pm, the cooling exceeds the heating.
Tricky?
So then an absorbed photon always warms, doesn’t matter the source.
Tricky?
-Scott
Scott,
Does the absorbed photon always warm? As in back radiation?
If an ir photon increases the kinetic energy at the water/air interface molecules, will evaporation increase as well?
mpainter,
The absorbed photon will always increase the energy of the system when it hits. In fact, it is a quantized energy increment.
The energy increase can induce additional evaporation under some conditions. However, it usually will not do so directly because of where the two events occur. You guys have yet to tell me where the evaporation occurs relative to the photon absorption.
-Scott
“No. The atmosphere (and space) [cools]. The Sun always heats us. But after ~2-3 pm, the cooling exceeds the heating. Tricky?”
Scott, how do you go from this to this:
“So then an absorbed photon always warms, doesn’t matter the source. Tricky?”
?
Scott,
There are many AGW types who hold that the back radiation absorbed at the ocean’s surface does not cause evaporation. Norman holds that view.
Do you concur in that view?
The same way you went from from this:
To this:
The temperature is cooling, not rising. No different than IR absorption causing the temperature of something to reduce more slowly.
-Scott
mpainter says:
July 2, 2015 at 1:36 PM
I do not strictly agree with that view. The adsorbed radiation should lead to increased evaporation. Where I differ from you is that your group assumes the additional evaporation induced by the absorbed radiation will result in enough evaporation to counteract the gain from the absorbed photons.
I have already shown/mentioned two cases where this is not the case–the case of 100% RH and the case in the lab with the CO2 laser. I can show mathematically that the gained radiation will increase the temperature more than evaporative losses if someone will engage me on the depth at which evaporation can occur. No one has yet.
-Scott
Scott says, July 2, 2015 at 1:37 PM:
“The same way you went from from this:
“No, Mike. When you “make something warmer”, you make its temperature rise.”
To this:
“The Sun always heats us. But after ~2-3 pm, the cooling exceeds the heating.”
The temperature is cooling, not rising. No different than IR absorption causing the temperature of something to reduce more slowly.”
Are you pulling my leg, Scott? You cannot be serious!
You asked me: “So does that mean the sun cools us after ~2-3 pm (peak daily temp)?”
I answered: “No. The atmosphere (and space) does. The Sun always heats us. But after ~2-3 pm, the cooling exceeds the heating.”
So after ~2-3 pm the surface cooling to the ATMOSPHERE and SPACE exceeds the surface heating from the SUN, because at this point the Sun is dropping towards the horizon (heating rates slowing down) while the sfc temps are at their highest, so cooling most rapid.
To this you reply: “So then an absorbed photon always warms, doesn’t matter the source.”
???
Kristian,
My challenge does not depend on the meaning of “warmer”. It is:
Imagine that the character ‘Q’ from Start Trek TNG wanders by and decides to mess with us by causing the atmosphere to become transparent to IR so that the downward IR flux from the atmosphere drops to zero. What happens to the temperature of the surface ocean, measured at the conventional depth of 3 meters? A qualitative answer is fine, but you must conserve energy.
Care to tell me how you would answer the question?
Mike,
They’ve been dodging a similar question from me too. I don’t think we should hold our breath.
-Scott
Mike, on Mars there is no rGHE as defined. The avg gl sfc temp is 210K, EXACTLY the same as the apparent planetary BB temperature in space.
Still, every cubic metre of Martian atmosphere above the surface holds ~28 times as much CO2 as on Earth.
Why is there an apparent rGHE on Earth and not on Mars?
Because the Martian atmosphere is FAR too thin. It can hardly retain any heat.
The atmospheric warming mechanism is a MASSIVE one, not a radiative one. Radiative properties are but a tool. The actual driving force is the atmospheric mass.
If you made a massive atmosphere completely transparent to EMR, then this atmosphere could never effectively rid itself of its absorbed energy/heat transferred to it NON-radiatively every single day from the surface. An intolerable, untenable situation for a planet to be in. So either the atmosphere will have to go. Or it HAS TO become radiatively active. There is no in between.
It is as if no one has ever flown in a plane before. As you rise away from the tremendous mass at the surface (atmospheric pressure), the temperature goes down. It is not because the CO2 is below you, it is because the weight/density is less.
All that gas over our heads is heavy when you view it in aggregate. It is not GHE it is weight.
Kristian,
I asked this question on the prior thread from Dr. Spencer, but it was lost in the noise. If you took our atmosphere and moved it 5km above the surface, but it retained all the same properties of emission to space, albedo, and the like and it still had the same heat capacity and average temperature. The difference being that atmospheric pressure would be zero and the earth could only cool radiatively since the atmosphere does not touch the earth to convect/evaporate.
Thus, the 161 w/m^2 reaches the surface like now but the mass of the atmosphere had no effect, only DWLIR would be hitting the surface along with the sun.
I contend that the earth would be significantly cooler and the DWLIR would fail to warm it sufficiently. We would have significantly lower temperatures even though the same K&T forces were (sun + IR) hitting the surface. The conclusion being a cold atmosphere can’t radiatively warm the surface.
Thoughts?
FTOP says:
July 3, 2015 at 1:04 AM
Kristian,
–I asked this question on the prior thread from Dr. Spencer, but it was lost in the noise. If you took our atmosphere and moved it 5km above the surface, but it retained all the same properties of emission to space, albedo, and the like and it still had the same heat capacity and average temperature. The difference being that atmospheric pressure would be zero and the earth could only cool radiatively since the atmosphere does not touch the earth to convect/evaporate.
Thus, the 161 w/m^2 reaches the surface like now but the mass of the atmosphere had no effect, only DWLIR would be hitting the surface along with the sun.
I contend that the earth would be significantly cooler and the DWLIR would fail to warm it sufficiently. We would have significantly lower temperatures even though the same K&T forces were (sun + IR) hitting the surface. The conclusion being a cold atmosphere can’t radiatively warm the surface.
Thoughts?–
Instead put 100 by 100 km wide glass box which 10 km high.
Put above Moon surface.
Each square meter has 10,000 kg of Nitrogen and Oxygen Earth mixture- plus it’s greenhouse gases. So gas has density of 1 kg per cubic meter.
Btw, Mass of about 100 billion tons, which is 1/6th the weight.
Anyways has walls 100 feet tall around perimeter of box, and pillar supporting in the middle.
So got interior space of 100 feet high, by 100 km square.
You have that vacuum or much easier in terms of support if filled with some kind of gas.
So how does it affect temperature on lunar surface?
So instead 1360 watts, you get 1000 watts per square meter when sun is at zenith. So if vacuum the surface would warm to about 70 C, and the big box above it can not increase this temperature. If filled with helium, the helium would be [at least mostly] heated from the 70 C surface. Since Moon has slow rotation, one has plenty of time with sun at Zenith and helium would heat up to air temperature of 70 C.
And any object in there would also heat to 70 C. Humans with a air mask “could be” quite comfortable with this temperature. They need about 3 psi pressure or more to breathe without a pressure suit. But since air could be dry, human can evaporate and keep quite cool- need to drink a fair amount water. But everything one touches would be as hot a pavement on Earth under sun at zenith.
Of course human can convect or radiate heat, and if one increased to the pressure to 14.7 psi, the hot air would heat up people faster- but just means they need more water.
Of course if humidity increases, then humans are going to be in sauna like conditions- hottest saunas in Finland are about 100 C- and if at that temperature it takes a bit to to get use to, and you not comfortable or could be dangerous
if very long exposure. But at 70 with low humidity, you should not much problem- and drink enough water it could be considered healthy.
In terms of effect of big box above. The container walls could heat up, but the gas by itself doesn’t heat up much- if any. So box above would be heated from the 70 C basement.
And require a lot time to heat up. But if it got warm during the lunar day [say from the basement], it would help keep the basement warm at night.
**Of course human can convect or radiate heat..**
Should be: “Of course human can not convect or radiate heat..”
Ok, say pick various locations 3 meter below the surface.
Let’s start with Arctic ocean winter, and has be dark for 1 month. And let’s say there is already 1/2 meter of frozen ice
at the surface.
so it 3 meter of water under the ice. Now without “Q” interfering, do think the greenhouse effect is warming the water?
I don’t think so, and probably -20 C or colder above the ice- with or without “Q”. And with or without “Q” the 3 meter of water may or may not freeze in coming months, but water is probably already around 0 C.
So now, lets go to equator, pick number of 120 W- thousands of km from any continent, and maybe an island with tens of km.
And water is say 25 C [77 F] at 3 meters.
To lower the temperature by 1 K, requires a loss of 4.181 KJ
per kg. So 3 meter deep by 1 by 1 meters is 3000 kg of water.
So 12,543,000 joules of heat must be lost to lower the 3 meter depth water by 1 C.
Earth radiate on average 240 watts per square meters
and suppose to get about 160 watts of direct sunlight per square meter averaged over entire globe. The tropics both recieve more than 240 watts per square meter and radiates more than 240 watts. From space the tropics is blazing in Longwave IR. But what the number is, we can roughly assume that before “Q” did his thing, it was more or less balanced.
And what I am suppose to do is determine how much of a imbalance would occur from “Q” snapping his fingers and saying, “make it so”.
A starting point might be to guess what the Alarmist could imagine what what happen- as baseline to go from.
So we have the believer chart:
http://www.drroyspencer.com/wp-content/uploads/greenhouse-gas-and-KT-diagram.jpg
Input is roughly 340, 100 reflected, 80 absorbed by sky, 160 absorbed by surface.
So first we need a rule clarification from “Q”, a lot of the 80 absorbed by sky has to do with the H20 molecule.
Does the absorbing of the sacred molecule related to it’s absorption of the radiant light of the sun?
So does a portion of the 80 instead reach the surface?
Also greenhouse effect theory is not clear in regards to whether the clouds are greenhouse effect. Which could relate to how much sunlight is reflected and also could have to do with how much is absorbed by the sky.
So boiling down do have increase in amount of direct sunlight reaching the surface- if so the tropics should gain considerably from “Q”‘s action.
Meanwhile if is assume 240 watt in somehow absorbed by earth
and for moment 240 watt leaves earth. But we focus, now on details as depicted in holy chart.
So starts with 80 from evaporation, and then 17 thermals for the hang gliders. Next is the fun part, so have 356 on average emitting for the surface with 40 of the 356 on average taking express trip to space. Then the back radiation returning 333 to surface on average. Can’t say I know whether tropic get above average in terms of back radiation, but science tropic has average temperature of about 25 C, and it appears to be blazing from Space, that it
it’s 356 total is much higher than average.
Or if go back to dark arctic with temperature somewhere around -20 C, it should be below average in regards to the 356 average, nor would get much help in terms of the 333 average bad radiation.
Depending on the wealth distribution plan, the Tropic might be like the Greeks- the tropic could get net gain from the scheme. Or if not getting a lot more than 333 but dumping out much more than 356, they might profit from “Q” new arrangement.
Perhaps we look for an area where we know there will be going to be big losers from “Q” action.
Take Canada for instance, or Russia. How about Japan since they are island and has ocean near most of it. So right in middle of Sea of Japan. News:
“Average temperatures have risen in a range from 0.7 to 1.6 Celsius (35 Fahrenheit), depending on the area. World ocean temperature rose on average 0.5 Celsius from 1900 to 2006, the Japan Meteorological Agency said in a report issued yesterday.
“We assume the sea temperature change is the result of global warming,” said Yasushi Takatsuki, a weather agency official. The JMA report itself didn’t cite possible causes for the higher ocean temperatures.”
http://www.bloomberg.com/apps/news?pid=newsarchive&sid=a7JNH4PeHgV8
Gosh seems bloody cold, and the worried about it getting warming. Hopeful “Q” is here to save their day.
I better check something else, wiki:
“As a result of the enclosed nature of the sea, its waters form clearly separated layers which may show seasonal and spatial dependence. In winter, the temperature is almost constant with the depth in the northern part of the sea. However, in central-southern parts, it may be 8–10 °C down to 100–150 m, 2–4 °C at 200–250 m, 1.0–1.5 °C at 400–500 m and then remain at about 0 °C until the bottom. ”
So this dinky chunk of shallow waters is quite varied.
So it’s winter in artic, so winter in sea Japan, so maybe 3 meter down it’s 2 C.
Now you can cut the world in half, near tropic is warm, above 40 degree latitude is the cold half. And the middle of sea of Japan is around 40 degree North latitude.
So we in middle of sea of japan at the border of the cold half and the warm half.
So going to pick the average values. So have 240 in and 240 out, and 356 minus 40 up and 333 down. Or 316 up and 333 down. And whole idea is that the 333 number will change from what “Q” did.
Now still don’t have ruling about the clouds. It’s winter and probably cloudy. How much is the 333 from clouds.
Let’s just say there are no cloud so we don’t get any of the 333 watts of back radiation. So sunny in day and clear skies at night. And we in middle of Sea of Japan in 2 C water, and to lower it by 1 C, need to lose 12,543,000 joules of heat.
So it’s noon, say Nov 1, and what is the the angle of the sun? So fall equinox, was 22 September. So we were month darkness in arctic, so has to be say Nov 20th or so.
But at Fall equinox, the sun would be 40 degree above the horizon with 12 hours of daylight, and Jan 20th it would be 23 degree lower, so 17 degree above the horizon.
So roughly in the middle, 28 degrees by horizon. And so sunlight is about 700 watt per square at noon, and this is when “Q” decides to deprive us of the grace of back radiation. So it’s crunch time.
Now if the was a cloud in the sky that blocked the sun which was showering about 700 watt per square meter, it wouldn’t be the end of the world, so if deprived of 333 watts per square meter of magic, the world would likewise not end at noon in the sea of Japan. One could say that when the sun was out, it’s possible no notices losing the 333 watts. People might think it should be warmer than it seems to be- that’s possible. But then again they may be too busy to think about it.
But the water above the the 3 meter, doesn’t think, it must respond, and it’s a bit cold, but sun out so still should warming. And this goes on for another hour, and nothing much happens.
But have to get back it it. And winter is coming.
*But have to get back to it. And winter is coming.*
Now rather than idea of stop getting the 333 watts of back radiation in middle of sea Japan. We going to say what “Q” actually does stop the back radiation from preventing 333 watt from radiating from the surface of the ocean which is above the 3 meters under the ocean.
This require New Math. I am interested in how long it takes to cool the top 1 mm of water of the ocean above the 3 meter of depth.
So 1 mm of one square meter is 1 kg of sea water. And we will say the 1 mm of water is 4 C.
So 1 kg of water requires 4.210 KJ per 1 K.
So that is 4210 joules per degree, times 4 is 16840 watts seconds of loss. 16840 divided by 333 is 50.5 seconds.
So 1 minute after “Q” does his thing, the water above the 3 meter depth becomes 0 C.
And then to freeze 1 kg of water require a loss of 334 kJ/kg.
So 334,000 joules divided by 333 requires 1003.0 second or
16.71 minutes.
In less than 20 minute a 1 mm layer of ice appear above the 3 meter.
The sun is shining at 700 watts per square meter.
And birds stop singing.
And obviously the water all over Japan is also freezing.
Now what “Q” actually did was stop some book keeping in ivory tower of Valhalla. But what undoing this book keeping number- which didn’t really cause heat- did, was to cause the surface to unleash enormous amount of real heat.
“Q” master plan was to incinerate the sky.
Or least kill the birds.
“Q” hated birds, ever since he heard they were descended from dinosaurs.
OK. I surrender. Silly of me to think I could reason with these people.
*Eye roll*
Why is there no rGHE as defined on Mars when every cubic metre of ‘air’ above the surface holds ~28 times as much CO2 as on Earth, Mike?
Hint: Overall (bulk) sfc air density on Mars, 0.02 kg/m^3; on Earth, 1.225 kg/m^3.
I would like to know from Dr. Spencer, what percentage of temperature change does he think variations in IR radiation and Solar radiation each account for when it comes to the sea surface ocean temperatures?
Thanks.
How does downward IR goes about “making up the heat”.
Given the fact it can’t penetrate the surface much beyond it’s own wavelength. Given the fact the the thermal conductivity of water is very poor. Given the fact that turbulent mixing in wave structures takes place mostly further down inside waves, well beyond the distance IR gets to.
Salvatore,
How does your body warm up when standing in front of an IR source (e.g., fire).
Given that fact that the IR penetrates no farther in you than in water. Given the fact that the thermal conductivity of a person is very similar to that of water. Given the fact that there is less mixing at the surface of the human body than in the ocean.
Clearly, it must be the placebo effect and I’m just imagining that I warm up in front of an IR source.
-Scott
Salvatore,
So what would happen the the sea surface temperature if the atmosphere were to become transparent to IR?
I predict that you will dodge the question.
The atmosphere couldn’t become transparent to IR, Mike. Your premise is invalid.
Hypothetically one can of course posit anything, whatever your heart desires. That doesn’t mean your thought experiment has any bearing on reality, on any real-world situation.
An atmosphere having a mass is also radiatively active. By necessity. The two go hand in hand. You can’t have one without the other. Such an atmosphere wouldn’t survive in a real universe.
Why?
Because a massive atmosphere would always automatically and quite naturally have energy/heat transferred to it (convectively) from the solar-heated planetary surface beneath it, every single day and regardless of whether that atmosphere absorbed IR or not. But it couldn’t effectively get rid of this energy/heat, because this can only happen through radiation to space. So energy/heat would pile up. And pretty fast at that.
There is no steady state in sight for a planetary system like this. The atmosphere would inflate until it started eroding into space. Either it would have to go completely, or it would have to somehow become IR active. To save the situation. To make it equilibrate. No atmosphere could function properly and in a stable manner like this. In fact, no thermodynamic system could.
And that’s why we do not see ANY massive atmospheres anywhere that is not also radiatively active.
– – –
So what we will have to compare is 1) a planet with no atmosphere altogether and 2) a planet with a massive atmosphere in place.
A quote from Dr. Spencer in the past that kind of answers my questions.
NEVERTHELESS…
It might well be that solar radiation is more efficient (on a Watt per Watt basis) than IR radiation at changing ocean temperature. In other words, that IR warming of a water body is more likely to be lost through evaporation, since its warming effect does occur only at the surface, and so that energy is more likely to be lost through evaporation than absorbed solar radiation would be.
The effect would be greatest during low wind conditions, when vertical mixing of the water is weakest.
From what little I know of ocean modelling, this potential effect – which I presume is what people are talking about — is not taken into account. I suspect it would have to be parameterized, because it occurs on such a small scale. As far as I know, a Watt of IR is treated the same as a Watt of solar in the energy budget of the top ocean layer in a model.
I would like to hear what others know about this issue. I suspect it is something that would have to be investigated with a controlled experiment of some sort.
How absurd it is to imagine that the GHE warms the ocean. The idea assumes that water molecules, having just removed energy from the ocean, are then going to raise the SST. This notion carries the assumption that warming is the net effect of evaporative cooling.
How absurd the notion that all IR photons have to induce evaporation. The notion assumes that the water molecules, having just gained energy from absorption of radiation, would then somehow choose to always evaporate instead of just warming up a fraction of a degree. This idea carries the assumption that energy loss is the net effect of energy gain.
And really odd that the water molecules on my body when I’m covered with sweat and stand in front of an IR heater warm up instead of cool off. How do they know they’re on me and not at the surface of the ocean?
-Scott
Scott:
“How absurd the notion that all IR photons have to induce evaporation. The notion assumes that the water molecules, having just gained energy from absorption of radiation, would then somehow choose to always evaporate instead of just warming up a fraction of a degree. This idea carries the assumption that energy loss is the net effect of energy gain.”
Exactly:
See my link above to real world measurements.
** Scott says:
July 1, 2015 at 8:02 PM
How absurd the notion that all IR photons have to induce evaporation. The notion assumes that the water molecules, having just gained energy from absorption of radiation, would then somehow choose to always evaporate instead of just warming up a fraction of a degree. This idea carries the assumption that energy loss is the net effect of energy gain.
And really odd that the water molecules on my body when I’m covered with sweat and stand in front of an IR heater warm up instead of cool off. How do they know they’re on me and not at the surface of the ocean?**
A IR heater mostly emits Shortwave IR, or near Infrared, and red/orange light. So:
“There are various good reasons why nichrome is the most popular material for heating elements: it has a high melting point (about 1400°C or 2550°F)”
http://www.explainthatstuff.com/heating-elements.html
Now look for blackbody at round number of 1000 C. Here we go:
https://www.newport.com/Blackbodies/377964/1033/info.aspx
So peaks at 2.5 µm. And long wave is 8 to 15 µm
So +99% of heat is not long wave IR.
Then replace IR heater with a fireplace…effect is still the same.
-Scott
Scott, my poor abused, and confused, amigo, a fireplace easily has temperatures over 1000 ºF. Do you know where the troposphere gets that hot? Oh, I forgot, there are LASERs in the atmosphere….
(No SNARK intended.)
I amend the statement, in part, which I made above:
“People also declare, as Roy has,that “wind will mix” the absorbed IR energy “down”…”
In fact, Roy said something different.
–I sometimes see the claim (usually in comments on a blog post) that infrared radiation cannot warm a water body, because IR only affects the skin surface of the water, and any extra heating would be lost through evaporation.
I have tried to point out that evaporation, too, only occurs at the skin of a water surface, yet it is a major source of heat loss for water bodies.–
I would say that evaporation is significant when there is a large difference between air and water temperature. And I think in your experiment one would gotten more evaporation loss if you did not use pool water with salt added. Mainly the salt, though perhaps pool water has chemicals with work something like salt [I don’t know].
So seems clear to me that salt water did much more to stop evaporation losses at bottom of the container than anything
to do with the shield. But with the pool water with salt, it seems one still had a significant amount of evaporation loss.
In terms of numbers, it appears the water at bottom lost about 4 C. If you repeated with same conditions with fresh water it might in instead cool by 5 to 6 C.
I seems to me that IR would be ineffective at warming the ocean, but would say if you had 300 watts per square meter
of IR, that this would cause much more evaporation.
So without IR with your experiment, you probably loss less than 100 watts per square meter to evaporation.
I don’t know how many square meters is was, 1/10th of square
meter, so 10 watts. And eight hours, 3600 seconds per hour,
so 28800 watts seconds.
So evaporation is lost of 2,270 kJ/kg, would be 2270 joules per gram. So in first 8 hours lost, about 12 gram of water. And if there was 300 watts per square meter of IR
one loss 12 + 36 grams of water due to evaporation.
And if instead it was fresh water, one loss another say 12 grams.
In terms of measurement 1 mm of square meter is 1 kg, 1/10
of meter is 1 mm is 100 gram. So difference of losses would hard to measure.
I assume there was not much wind, more wind would give more evaporation, and/or would tend to negate the effects of the salt. Or one think of salt as providing a lid, and particularly effective in term of measured the water temperature at the bottom. Or over time periods of say 1 hour, colder water at top, would drop to the bottom- allowing warmer water to evaporate more quickly and the colder water being measurably cooler at the bottom.
Also I would say the salt water was factor in retaining heat better than the cooler’s Styrofoam material. I think the container were good in sense they trapped a pocket of still air above the water, and styrofoam was most effective in this area above the waterline.
But what the experiment most does is demonstrate how ocean cool at night.
But getting back to IR, if there was around 300 watts of back radiation it would “work” 24 hours a day. So we would get an enormous amount of evaporation that we aren’t getting globally.
Hi Wayne,
Relative to Richard Feynman’s lecture, are you familiar with what he taught physics students at Caltech about light scattering by clouds. (The Feynman Lectures On Physics Vol. I pp 32:8,9)? I ask this because I have not found anyone familiar with this. I understand none of the physics of the scattering by cloud droplets which he taught but I certainly can understand the results of this scattering which he describe. But as you stated about the lecture you attended, the focus of his teaching of the scattering of visible light, but he made statements that force the conclusion that it most apply to wavelengths up to 20µm, an average diameter of an ordinary cloud droplet from what I have read.
I know that you referred to reflection from surfaces and not to scattering by cloud droplets. But you seemed to actually understand the physics (QED) of the reflection (absorption/re-emission events), which I cannot pretend to do. Somewhere I have read his statement about reflection (absorption/re-emission events) or maybe this was in reference to scattering in general. I do not know.
You have motivated me to look at Vol III of The Feynman Lectures. Here I find the topic–Emission and absorption of photons–on pp 4:7,8 and the topic–the blackbody spectrum–on pp 4:8-12. I have not yet read either beyond the first paragraph of the first, but one sentence jumps out. “When the light is emitted, a photon is “created”. ” Since I have not found anyone who claims to have carefully studied what Feynman taught in Volume I, I expect they, like me, have never opened Volume III. With only this glimpse, I suspect the endless debates involving the greenhouse effect and the radiative warming and cooling of the earth’s surface would be ended by understanding what Feynman taught in Vol. III.
Be interesting if there are any of the debaters who would claim to fully grasp what Feynman taught in this Volume III. For as you wrote: “His theory of quantum electrodynamics (QED) that, as far as I know, still stands.”
Have a good day, Jerry
Yes, jerry l krause, I too have read his lecture volumes, some over and over, but not 100% but more like 80% skipping what I deemed already understood. But I don’t remember him getting into reflection and transmission in the light of QED there.
Try taking six quiet hours and listen to these four lectures:
http://vega.org.uk/video/subseries/8
Listening two or three more times over the next year and you should realize so much more, I have and I do without a doubt.
As a common sense – I see it that any amount of heating via IR of a water surface will cause it’s skin temp to rise. Otherwise you are saying that ALL the extra IR goes into evaporation. No, put a (perfectly insulated – sic) bowl of water into an oven and turn it to low. Do you suppose that over time the water will cool, via evaporation?
Everything else being equal (same amb air temp/humidity/wind strength/turbulent mixing down) then the skin HAS to warm in order to evaporate more. If it cools then it CANNOT have greater evaporation (energy required).
The effective process we are talking of here is a REDUCED cooling of the ocean bulk. All heat from the ocean bulk has to pass through its skin to escape to space. If the skin temp is elevated (even slightly) then the DeltaT in the top few cms of water is reduced. This REDUCES heat flow from the bulk (below) to the air above.
That is how IR (back-radiated) “warms” the Oceans. With the GHE this is happening 24/7/365 over ~70% of the Earth’s surface.
Of course the oceans are gaining heat (not loosing it as quickly as in the past). Hence we have the GHE – an imbalance that needs to be corrected before GHG’s no longer reduce cooling. And that is via a stabilisation of SST’s at a higher temp so that via SB greater emission will overcome the “restriction” of CO2.
http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
The notion that increasing skin temp will somehow reduce cooling of the sea surface and hence warm the ocean is incorrect.
In fact, the skin temperature increases on a diurnal basis, with the general heating of the sea per insolation, starting with sunrise and increasing until the sun passes the zenith.
Heat gained daily at the sea surface (and near subsurface) is returned to the atmosphere nightly via convective overturning of the near subsurface.
The ocean cools at night despite any increase in temperature at the air water interface.Oceans warm by isolation, not by LWIR.
Toneb,
Nice clear explanation. Too bad you are wasting your keystrokes.
Mike:
A a voice of reason/common sense!
Yes I’ve posted here before and know that full well.
Actually Scott said it better …..
” How absurd the notion that all IR photons have to induce evaporation. The notion assumes that the water molecules, having just gained energy from absorption of radiation, would then somehow choose to always evaporate instead of just warming up a fraction of a degree. This idea carries the assumption that energy loss is the net effect of energy gain.”
http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
In even shorter terms:
How can adding energy to anything make it cool quicker than otherwise??
Slayers a strange breed.
Toneb,
I heated up water using a LWIR CO2 light source. 100% experimental confirmation without the confounding agents that were present in Roy’s or Mike’s experiments. Yet, they don’t bat an eye and say that it’s “obviously different”. When asked how it’s different, nothing. When pressed, I got one to say it was due to “intensity”, but then he claimed I was harassing him…
Of course, their mechanism says nothing about intensity either….
-Scott
The skin temp falls
On the Thermal Boundary Layer of the Ocean.
Ewing G, McAlister ED.
Abstract
Measurement of the long-wave infrared radiation from the top 0.1 mm of the evaporating ocean demonstrates the existence of a cool surface layer characterized by departures of as much as 0.6 degrees C from the “surface temperature” found by conventional methods. Being very thin, the layer cools sufficiently rapidly to reestablish itself in less than 12 seconds after disruption by a breaking wave.
PMID: 17841610 [PubMed]
http://www.ncbi.nlm.nih.gov/pubmed/17841610
toneb,
You say “the DeltaT in the top few centimeters is reduced.”
####
No, it is not “centimeters”.
The interval of interest is the upper one mm, approximately. This is the interval that cools toward the surface, and this the interval of delta T. This condition is present both day and night and is due to the cooling at the air water interface, mostly by evaporation.
Increasing skin temperature is a diurnal event, as the sea generally warms with increasing insolation. However, this added heat is released at night. The near subsurface water cools by convective overturning.
The claim that IR heats the ocean by inhibiting heat loss is more AGW hypothetical speculation; IR absorbed at the surface cannot effect any warming of the sea by inhibiting the nighttime cooling processes: evaporation (mostly), radiation, and sensible heat loss at the surface and convection in the subsurface.
SST is determined solely by isolation; the thermodynamics of the sea surface precludes any contribution of LWIR to SST.
mpainter 5:36pm: “No, it is not “centimeters”. The interval of interest is the upper one mm, approximately….The claim that IR heats the ocean by inhibiting heat loss is more AGW hypothetical speculation;”
Yet the top post experiment shows proof the water temperature is measurably different a “few centimeters” below surface after modulating the LWIR incident on the water surface after sunset. In this case about 0.3F just before sunrise. Not hypothetical speculation any longer.
Right, no longer hypothetical, the sample without the canopy lost may more joules causing the -0.3°C compared to the covered. Is that not what you are saying in so many words? That is what the experiment clearly shows.
Ball4
Wayne has set you straight on the experiment’s results. The water shielded against the back radiation cooled the least and the water not shielded cooled the most. Fancy that, AGW doesn’t work.
Homework assignment:Get the temperature profile of the sea surface and study it.
mpainter,
The SST profile you like to refer to (available at http://ghrsst-pp.metoffice.com/pages/sst_definitions/ for those interested) is a TWA of all the processes involved. What would the SST profile look like if one subtracted the IR component?
-Scott
The SWIR is absorbed in the upper mm. This accounts for the shape of the daytime profile and explains why there is no convection at day, but only at night.
It can be concluded that SWIR is the predominant source for heat of evaporation in the daytime and sw insolation the source for the heat of evaporation at night, this energy being absorbed in the subsurface in the daytime and convected to the surface at night.
LWIR has no effect on the sea surface temperature profile, being absorbed at the air water interface or within a few microns of that.
Energy is being input into it, so there has to be an effect of LWIR. Your favorite plot is log scale, so the first few microns are important. Please explain how the absorbed IR changes the shape.
Oh, and “convection” in heat transfer lingo means a lot more than the basic buoyant convection taught to kids.
-Scott
mpainter wrote: “there is no convection at day, but only at night”.
Convection in the mixed layer of the ocean is caused mainly by wind, not by density gradients. There is always plenty of convection, with the possible exception of dead calm conditions.
Mike,
He is ignorant of the meaning of convection when discussing heat transfer. I’ve tried to educate him that it’s more than just the buoyant convection taught to kids, but when I linked to the Wiki page discussing convective heat transfer and an example of forced convection, he just mocked it.
Wasted keystrokes.
-Scott
wayne 6:36pm – Agree, except for your incorrect C units, no longer hypothetical as the top post demonstrates LWIR for the no shield is less than the LWIR with shield and the water temperature is thus modulated at depth of a “few centimeters” -0.3 to -0.4F (not your C) during the night time hours – grey curve being lower T than red curve at each sample time (except start) due less incident LWIR energy for no shield.
mpainter 7:30pm – I agree with wayne (except for units) no longer hypothetical so it is you being set “straight” by wayne – LWIR modulation does have an experimentally demonstrated effect on free to evaporate water temperatures at a “few centimeters” depth which proves no longer hypothetical by guess, experiment and analysis (you know the Feynman experimental method) that you are incorrect writing 5:36pm: “the thermodynamics of the sea surface precludes any contribution of LWIR to SST.” The same result as top post has also been demonstrated out on the open ocean by tests in early to mid-2000s.
NB: mpainter – see the link by Scott 7:53pm and, as you recommend, study up on SST a “few centimeters” depth, night time version: “measurements of SST made by different…in situ instruments.”
Ok Ball4, got me on the ‘C’, my slip, but it really doesn’t matter exactly how much does it? Just that it did record a small difference and the difference’s sign is so telling.
Ball4 says:
July 1, 2015 at 6:01 PM
If you really want to see CO2-based IR heating of water from above, just put a vial of water in the path of a CO2 laser. It will remove any doubts that IR can’t heat water.
-Scott
mpainter:
You are fixating on the micron that the IR photons affect. Thermal effects would penetrate further than that.
See:
http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/
Have you not heard of conductivity, turbulence.
Indeed conductivity is the prime “warmer” here as the reduction of deltaT is what reduces the heat flux from below from getting through the skin as effectively.
Look- it’s just basic thermodynamics and common sense.
You CANNOT cool something by adding energy to it.
“You CANNOT cool something by adding energy to it.”
You can if the process caused or enhanced by adding energy is a net cooling process which is a feature of evaporation since it takes up 5 times as much energy as is needed to cause it at 1 bar pressure.
However, you cannot cool it below the point at which the added energy is all used up.
So 1 unit of added energy causes evaporation and then that evaporation also takes up another 5 units of added energy with nothing left over.
“the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C (cp = 75.3 J K−1 mol−1)”
from here:
https://en.wikipedia.org/wiki/Enthalpy_of_vaporization
The point being that the energy required to heat water from 0C to 100C is enough to provoke ALL of the water to evaporate yet, in evaporating, the water takes over 5 times as much energy from the surrounding environment.
There can be no added energy in the form of IR left over to heat anything or even reduce the rate of cooling of anything.
This is simple established science that appears to have been forgotten or ignored.
Stephen 3:16am: ” There can be no added energy in the form of IR left over to heat anything or even reduce the rate of cooling of anything. This is simple established science that appears to have been forgotten or ignored.”
The top post experimental evidence demonstrates Stephen is here incorrect – proven wrong simply by using age old, long remembered & established science employing the full Feynman scientific method.
If Stephen’s imagination (meaning no counter experimental evidence provided) were correct, the red curve and the grey curve would overlay within experimental error. Only in Stephen’s imagination does evaporation speed up (or slow down) to exactly cancel the effects of modulating LWIR on the night time water temperature at a “few centimeters” depth. The resulting curves do not overlay thus a measurable temperature difference was found (0.3-0.4F outdoors over night) due LWIR modulation with water free to evaporate.
The curves differ because the overlying shield reduces free convection (and therefore the rate of evaporation) from the surface beneath so that the surface beneath the shield warms relative to regions with no shield.
It is nothing to do with DWIR.
Stephen 11:37am: “The curves differ because the overlying shield reduces free convection..”
This is simply a guess purely from Stephen’s imagination. To make Stephen’s guess informed science, Stephen needs to show the confirmatory experiment which demonstrates it “reduces free convection” per the Feynman scientific method as did Dr. Spencer using reasonable instrumentation & show the data (0.3 to 0.4F), including analysis based on experimentally proven science.
There is no reduction in free convection or any constraint on evaporation as shown in top post Stephen, these processes operate freely in Dr. Spencer’s back yard as over the ocean in the great outdoors. Dr. Spencer’s experiment has been confirmed at sea by others, LWIR modulation affects the sea water temperature in the same way & magnitude which thoroughly disproves Stephen’s wrong imagination using the scientific method.
Imagination busted.
Roy’s ‘experiment’ merely ASSUMES no reduction in free convection beneath the shield placed above the water surface.
Every cloud or any increased atmospheric opacity reduces convection beneath it which over water results in higher himidity, less evaporation and a warmer water surface.
Every cloud travels on top of its own column of higher humidity which is why the water surface warms when clouds pass over. Evaporation reduces and the water surface warms with no recourse to DWIR required.
Stephen 1:10pm – There is no such assumption; nothing shown to prevent evaporation or conduction/convection being free. Stephen simply imagines evaporation & conduction/convection will adjust to keep the temperature unchanged. Nothing is preventing that from happening naturally Stephen. It doesn’t. The temperature changes; curves do not overlay. Show us your experiment where temperature does not change.
Imagination busted.
Stephen Wilde:
“You can if the process caused or enhanced by adding energy is a net cooling process which is a feature of evaporation since it takes up 5 times as much energy as is needed to cause it at 1 bar pressure.”
No you cannot.
It is NOT a net cooling process.
As Roy has shown and as the study I linked above also shows.
ALL the IR photons do NOT go into evaporating water. The greater evaporation rate happens BECAUSE the skin surface is heated.
Why can you not see that if energy is added the it MUST warm. LH take up regardless.
It’s just basic thermodynamics and common sense.
AND proved in the real world.
Toneb,
At this point it’s probably a good assumption that they’ll never get it. I’m mostly just seeing how far they’ll go to deny the most basic of concepts and hard experimental evidence.
-Scott
How can a process that requires one unit of energy to initiate it but yet requires 5 more units of energy to complete the process not be a net cooling process ?
Heating and evaporation occur in parallel within the depth of the so called ocean skin which is always consequently cooler than the ocean bulk.
The net effect is zero because the enhanced evaporative process must use ALL the added energy whilstleaving the underlying thermodynamic effects unaltered.
Stephen,
As already asked, at what depth does the evaporation take place compared to the absorption of LWIR photons?
-Scott
— Scott says:
July 2, 2015 at 12:04 PM
Stephen,
As already asked, at what depth does the evaporation take place compared to the absorption of LWIR photons?
-Scott–
I was thinking one needs to know what most of wavelength which reach typical ocean surface would be. So in terms of percentages of the total LWIR, what are the wavelengths which are involved?
Toneb-
Long wave IR which is around 300 watts per square, would be add 300 watts per second, to a very thin film of water,
Sunlight can be 1000 watts per square meter, when 1000 watts of sunlight strike the surface of the water, very little is absorbed in first 1 mm of water- unlike Long wave IR.
You can make sunlight heat just 1 mm of water, by for instance spraying 1 mm of water on your car or any other surface in sunlight. And the sunlight will evaporate the water pretty quickly.
So 1 mm of water over 1 square meter of area, is 1 kg of water. To evaporate 1 kg of water requires:
“Latent heat of evaporation – 2,270 kJ/kg ”
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
So 2,270,000 joules or watt seconds. So 1000 watts require
2270 seconds- 37.8 minutes. Of course if surface is already warm, it cools it, using the heat to evaporate quicker.
So throw a gallon of water [3.78 kg of water] on hot driveway and might be evaporated in 10 mins.
So ocean water is evaporated by sunlight warming beneath the surface and the warmed water rises and evaporates.
Or takes quite a while for sunlight to warm a body of water.
So cool swimming pools requires days to heat up from sunlight. Pool covers blocking evaporation, warm pool faster, and having a darker material at bottom of pool will warm it faster, etc.
gbaikie
AND:
There is additional heating caused by LWIR in the process I describe above and shown both by Roy and the study I linked to.
” Reducing the size of the temperature gradient through the skin layer reduces the flux. Thus, if the absorption of the infrared emission from atmospheric greenhouse gases reduces the gradient through the skin layer, the flow of heat from the ocean beneath will be reduced, leaving more of the heat introduced into the bulk of the upper oceanic layer by the absorption of sunlight to remain there to increase water temperature. Experimental evidence for this mechanism can be seen in at-sea measurements of the ocean skin and bulk temperatures. – See more at: http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/#sthash.sZCHou6g.dpuf
Well, Peter Minnett (RSMAS) doesn’t appear to think the back radiation warms, but rather reduces the loss of the surface.
But were back radiation were to actually warm the surface it would cause evaporation.
DWIR from above the water surface creates a graduated transition between mostly warming at the top of the ocean skin layer to mostly cooling at the bottom of the ocean skin layer with approximate equality about halfway between.
DWIR from above the water surface therefore increases the temperature gradient through the vertical profile of the ocean skin layer but leaves the rate of energy flow from ocean bulk to atmosphere through the skin layer undisturbed
The increased thermal gradient through the skin layer deals only with IR from above and leaves the background gradient present to continue its work in moving energy from ocean to air at the pre-existing rate.
gbaikie:
“Well, Peter Minnett (RSMAS) doesn’t appear to think the back radiation warms, but rather reduces the loss of the surface.
But were back radiation were to actually warm the surface it would cause evaporation.”
No, what Minnett says is……
IR warms the skin temp (as well as increasing evap) AND reduces the heat flux up from the ocean depths.
Incident IR on the skin cannot increase evaporation withouit warming the skin a little first. There is not a perfect conversion of IR into evaporated H2O.
Toneb,
Here in a bit I’ll quantitatively show why it’s impossible to have perfect conversion to evaporation at the bottom of the thread. At this point, the thread is a psychological understanding thing for me…it’ll be interesting to see their reactions to this one.
-Scott
— Toneb says:
July 2, 2015 at 2:14 PM
gbaikie:
“Well, Peter Minnett (RSMAS) doesn’t appear to think the back radiation warms, but rather reduces the loss of the surface.
But were back radiation were to actually warm the surface it would cause evaporation.”
No, what Minnett says is……
IR warms the skin temp (as well as increasing evap) AND reduces the heat flux up from the ocean depths.
Incident IR on the skin cannot increase evaporation without warming the skin a little first. There is not a perfect conversion of IR into evaporated H2O.–
We know that reflective surface in space environment prevent the loss of radiant energy. So likewise, though greenhouse don’t reflect radiant energy, the random re-emission of part of longwave IR spectrum could act similarly to a surface which reflects radiant light- which acts insulation- preventing something from losing less energy.
To me it seems from Minnett explanation that this is what he means- that a slowing of radiant losses from the surface would in turn cause less convectional mixing and therefore less loss of heat from the water beneath the surface.
Of course there are lots of aspects of a ocean which result in the inhibition of convection losses, which no doubt Minnett is aware of. But it seems Minnett thinks what he talking about has an significant effect.
Toneb says, July 2, 2015 at 1:35 AM:
“Look- it’s just basic thermodynamics and common sense.
You CANNOT cool something by adding energy to it.”
And you CANNOT have a cool thing heat a warmer thing. IOW, the cool thing is not adding energy to the warmer thing. The warmer thing is adding energy to the cool thing. It is called “heat transfer”.
Kristian:
That is the argument used against AGW, and of course the answer is yes you can in the sense that there is an exchange of energy until a balance is reached. IR photons hitting the ocean skin do not know or care where those photons came from – they simply are added to the reservoir whilst the hotter object continues to cool (but at a slower rate). There is heat flux from below the skin to “warm” it as the skin receives back-radiated IR from the atmosphere that SLOWS its cooling. this then reducing the deltaT over a small distance. This what acts to reduce overall flux from the bulk ocean to space….. and retains heat.
Toneb says, July 2, 2015 at 2:22 PM:
“That is the argument used against AGW (…)”
No, it is not “the argument used against AGW”. It is a fact of nature. I’m merely pointing it out to you:
You CANNOT have a cool thing heat a warmer thing. Because ‘heating’ something in thermodynamics specifically refers to transferring energy TO that something, directly resulting in an increase in its internal energy, hence in isolation raising its temperature. Likewise, ‘cooling’ something in thermodynamics specifically refers to transferring energy FROM that something, directly resulting in a decrease in its internal energy, hence in isolation lowering its temperature.
The point is that these two opposite processes normally occur simultaneously and quite often occupying the exact same space. And this circumstance, then, is the reason why people (obviously like you) are confused into thinking that insulation and insulating layers around a heated object actually somehow function as a second energy source to the object, in addition to its actual energy source, the thing doing the actual ‘heating’. It’s not.
Listen. When it comes to the surface of the Earth, the Sun is its heater and the atmosphere its cooler. The surface warms by absorbing SW from its heat source (the Sun) and cools by emitting LW (plus conducting and evaporating away energy) to its heat sink (the atmosphere).
The atmosphere DOES NOT WARM THE SURFACE OF THE EARTH BY LW. The surface of the Earth warms the atmosphere by LW (and by conduction and evaporation/condensation).
I seriously don’t understand how this is something one can get so utterly confused about. It is the most straightforward thing in the world.
Earth’s surface cools by shedding its heat into the atmosphere, the atmosphere being the surface’s ‘cold reservoir’. But it cools LESS EFFECTIVELY into the warm (massive) atmosphere than it does/would into cold (massless) space.
The simplest way of explaining the massive atmospheric warming (insulating) effect:
The solid, solar-heated planetary surface goes from disposing its heat directly into space, close to absolute zero, to disposing it rather into a massive atmosphere, a new, added, intermediate energy reservoir that will – by absorbing the heat transferred to it from the surface – naturally grow much warmer than space. Since the surface still has to shed (nearly) as much heat per unit time as before, only now into a much warmer reservoir than earlier, it will now necessarily have to be considerably warmer itself also, in order to still be able to ‘drive’ the required heat out. Basic heat transfer theory.
IOW: A solar-heated planetary surface cannot warm space, but it can (and does) warm a massive atmosphere, and in doing so, will – by necessity – be forced to become warmer itself as well.
It’s that simple.
Ignorant, Scott? Funny you should use that word. What is SWIR? and where does it come from? and why does it determine the daytime temperature profile?
And you, Mike M, it is established that convective cooling is at night, not daytime. Go read up.
mpainter,
Convection, in the sense used by engineers and scientists working in heat-transfer applications, happens all the time in the ocean. “Go read up.”
And I don’t care what else goes into the daytime temp profile, we’re just talking about LWIR and how it affects the profile. You say it doesn’t. But that absorbed energy has to go somewhere. Because effects happen in parallel, the presence/absence of SWIR doesn’t matter too much.
-Scott
Heating from IR and the consequent evaporation occur in parallel.
Since evaporation is a net cooling process none of the added IR is left over to heat anything or reduce the rate of cooling of anything.
They don’t seem to get it
The skin temp falls
On the Thermal Boundary Layer of the Ocean.
Ewing G, McAlister ED.
Abstract
Measurement of the long-wave infrared radiation from the top 0.1 mm of the evaporating ocean demonstrates the existence of a cool surface layer characterized by departures of as much as 0.6 degrees C from the “surface temperature” found by conventional methods. Being very thin, the layer cools sufficiently rapidly to reestablish itself in less than 12 seconds after disruption by a breaking wave.
PMID: 17841610 [PubMed]
http://www.ncbi.nlm.nih.gov/pubmed/17841610
They are too think to get it Stephen I have a facebook group you might want to look at – I have collected everything I can find on the subject – I would appreciate your input there
https://www.facebook.com/groups/1509057762686691/
Marc,
You’re right, we do “think” too much to get. And that’s because when someone thinks about this stuff and has a background in heat transfer, material and energy balances, and molecular-level interfaces, the answer is blindingly obvious.
You’re pointing to an abstract saying there’s a 100-um-thick layer that takes 12 s to reestablish. Tell me, what is the expected travel distance of a water molecule in 12 s?
-Scott
Thats your problem isnt it – you think that your the only one in the world with a background like that – well idiot I have that background also & trying to replicate what happens on the oceans surface & trying to replicate that with a laser & a bucket is just stupidity beyond belief
Marc Facer says:
July 2, 2015 at 9:07 PM
No I don’t think that. Tons of people have similar backgrounds. They just aren’t bored enough to put up with this. By the way, it’s not my problem, I’ve had that calculation done since long before Roy’s post. Mean travel distance of a water molecule in 12 s is 230 microns, also known as 2.3x farther than your 0.1 mm distance. Since we’re looking at averages, the average molecule will be traveling 4.6x farther than the necessary distance to get past your reforming layer.
Then use your background to come up with a quantitative model that can be tested. If intensity matters, then include that. If IR absorption causes some additional evaporation, quantify it. If you’re not doing that, then you either aren’t using your background or you don’t have what you claim.
Then use your background to actually show how it’s stupid, don’t just say it is. Derive what the maximum light intensity is where the effect goes from cooling to warming. The original claims were that LWIR could not heat water from the top. This has been definitely falsified with a CO2 laser. So your side either needs to withdraw the claim entirely or modify it to account for intensity effects. Just claiming it is wrong doesn’t make it so–that’s just proof by assertion.
Also, one thing that is “stupidity beyond belief” in my mind is claiming that adding energy causes the system to cool.
-Scott
Stephen,
You seem a bit more reasonable than these other guys. Yes, the two occur in parallel. So at what depths do the two occur?
-Scott
Within the cooler ocean skin layer.
Be more specific. If you can’t be quantitative, at least say which one is below the other.
-Scott
Obviously it is a graduated transition between mostly warming at the top of the ocean skin layer to mostly cooling at the bottom of the ocean skin layer with approximate equality about halfway between.
DWIR from above the water surface therefore increases the temperature gradient through the vertical profile of the ocean skin layer but leaves the rate of energy flow from ocean bulk to atmosphere through the skin layer undisturbed
The increased thermal gradient through the skin layer deals only with IR from above and leaves the background gradient present to continue its work in moving energy from ocean to air at the pre-existing rate.
So where is the best estimate of evaporation? Halfway through the “skin layer” it seems? How thick do you think that is relative to where the IR is absorbed? If someone would give me definitive answer on this then maybe this discussion would go somewhere. But since I seem to be the one providing actual numbers, I’m not holding my breath.
Impossible. The driving force for heat transfer is dT. So if dT changes then the thermal flux has to also change.
Driving force is difference in temperature, so the “moving energy” part has to change too.
-Scott
Scott says:
“The driving force for heat transfer is dT. So if dT changes then the thermal flux has to also change”
The dT involved in the transfer of energy from the ocean bulk to the atmosphere doesn’t change.
What happens is that a separate dT ionvolving evaporation to the atmosphere from the ocean skin is superimposed on that background dT to fuel the separate process independently.
The two processes are independent and each requires its own dT with no crossover between them.
Scott has had all of this expained to him repeatedly; a hard case.
One micron distance to the atmosphere, a thousand microns to one mm depth. Scott says the energy goes down because of CO2 lasers. A hard case.
mpainter,
And you’ve had it explained correctly to you repeatedly. And now have had three experiments show you wrong.
You say the penetration depth is one micron. I can buy micron-level depths (though deeper than ONE micron). Tell me–at what depth does evaporation occur?
-Scott
I see you’ve changed from 1 um here to 3 um elsewhere. Which is it? Why would you use 1 um here? Just trying to make the error you’ve imagined I made larger? You need to be consistent.
Also, for either the 1-um or 3-um value, what percentage of IR light does that amount to? 50%, 90%, 99%?
-Scott
mpainter,
Scott is right – arguing with you is wasted keystrokes.
To sum up:
1. Oceans accumulate heat during the day and release it at night. No amount of IR changes this basic fact, nor do all of the CO2 lasers in the world.
2. LWIR is absorbed at the interface; the 15 micron band (CO2) is absorbed within 3 microns.
3. IR energy absorbed in this minute interval is transient, being radiated or converted to latent energy within seconds. The upper three microns are evaporated in less than one half minute, given a rate of evaporation of one cm/day (typical rate on the tropics).
4. The sea surface temperature profile shows that heat is conducted from a warmer subskin to a cooler surface, night and day.
5. Convection from the subsurface occurs at night,not at day. Ocean swells are typically of a 2-3 meter amplitude and over one hundred meters length.Such geometry does not alter the surface thermodynamics.
6. The net effect of wind is to cool by increasing the rate of evaporation; the stronger the wind, the greater the rate of cooling.
the above is all from standard texts and references in physics and oceanographic studies. CO2 lasers cannot burn holes in this.
MPAINTER- you are exactly correct.
AGW enthusiast always want to change the rules, as they do with the data , to try to make it fit their agenda
Except I’m not an AGW enthusiast, I’m not changing the rules, I’m providing the data, and my “agenda” is probably closer to yours than the AGWers.
-Scott
Salvatore,
So true. For an example of what these types do with the facts, see Scott’s response below.
Except those aren’t facts, they are conclusions.
And for an example of what these guys do with experimental results, see how they dismiss the various experiments showing that IR can heat water from above out of hand with nothing other than proof by assertion.
-Scott
Scott, textbooks and reference tables. You can disagree with those if you choose, but it is all part of my education, which encompasses the natural sciences.
mpainter,
Here is my education.
BS Chem Engineering
BS Chemistry
Ph.D. Chemistry
Postdoc Atmospheric Chemistry
Currently, I build instruments and use them to measure new chemistries. My textbooks appear to be quite different from yours, so please provide specific references.
-Scott
mpainter says:
July 2, 2015 at 10:49 AM
True, but has no impact on this discussion because absorption of LWIR photons during the day increases the amount of daytime heat accumulation and decreases the amount of cooling during night…both of which yield a higher temperature in the end.
3 microns is an ok estimate of travel distance in water for this IR band. You should specify the amount (%) absorbed at that distance. However, the statement that this is the interface is strictly wrong…3 um is certainly part of the bulk H2O structure. Since we’re on the interface, what is the depth where evaporation takes place?
Wrong. The molecules have easily more than enough time to transmit energy before evaporating. Your presumed evaporation rate is 3.5 um in 30 s. The root(2Dt) distance traveled by water molecules in 30 s is 363 um–more than 100x farther. Also note that individual molecule diffusion is slower than the transfer of heat by other mechanisms. These calculations have been shown to you before (when you literally asked for a mechanism faster than evaporation) and you refuse to address them.
Net heat transfer via convective heat transfer, yes. But individual photon absorption events are still an increase in energy content and can slow the heat transfer being radiated upward, resulting in an increase in temperature over time.
Convective heat transfer always occurs anytime there is a temperature difference in the ocean. Your incorrect view of convection or your proofs by assertion will never change that.
Show your calculations for this. Waves on the meter scale can be generated by winds and if one of these lands on top of your precious subskin, the vast majority of the energy will end up below. An easy example case shows that your statement isn’t always true–in the case of 100% relative humidity, no amount of wind action will increase evaporation, so any mixing caused by waves would increase heat transfer to depths with no drawbacks.
Please provide citations. All of my above comments are fully justified by entry level chemistry, physics, and engineering texts too.
No one has yet to show how LWIR from a CO2 laser is different from LWIR from free atmospheric CO2. Your proofs by assertion do not change the results of this experiment, which you also literally asked for.
-Scott
Thanks, Scott. Now I don’t have to waste my keystrokes.
One quibble. You wrote: “3 microns is an ok estimate of travel distance in water for this IR band”. Surely the penetration must be at least on the order of one wavelength (15 to 17 microns).
Maybe mpainter is confused by the fact that the direction of convective heat transfer depends on the direction of the temperature gradient?
Or maybe mpainter is citing references, as, he says.
Mike,
I don’t think it has to be on that order. I don’t know if you access to journals, but this paper can give you the related molar absorptivities:
http://www.sciencedirect.com/science/article/pii/S0003269797921369
Very useful for calculating penetration depths for a range of %T. I’m still waiting for them to give me the depth where evaporation occurs before I really use those numbers.
-Scott
The IR from a laser would heat the water faster than it could evaporate which is inapplicable to naturally occurring IR from an atmosphere.
If our entire atmosphere were radiating downward at the intensity of a laser then there would be no water and no atmosphere because the accumulated surface kinetic energy would overcome the atmosphere’s hydrostatic balance and drive it off into space.
So how does that break the proposed mechanism? The mechanism doesn’t specify anything about intensity. Or temperature. Or RH. Wow, it doesn’t have any sort of quantitative construct.
At intensity level does it break? What is the critical threshold? Be quantitative.
-Scott
— July 2, 2015 at 1:28 PM
So how does that break the proposed mechanism? The mechanism doesn’t specify anything about intensity. Or temperature. Or RH. Wow, it doesn’t have any sort of quantitative construct.
At intensity level does it break? What is the critical threshold? Be quantitative.–
Isn’t the laser about 1 million time more intense per square mm, than back radiation could possibly be?
I think should generally try keep within couple order of magnitude as general plan.
You have been told a dozen times that intensity is dramatically more than what is experienced so shut up & listen for a change
But nowhere in your model is the intensity mentioned. If it’s not mentioned in the model, then it should be intensity-independent. Otherwise, your model obviously breaks down somewhere and you have no idea where.
-Scott
Scott wrote: “useful for calculating penetration depths”.
By assuming Beer’s Law? I think that implicitly assumes that light consists of infinitesimal particles. That assumption might work for some cases, but not for others. Once scales approach the wavelength, stuff gets weird: scattering cross sections up to four times the geometrical cross section, reflected light being affected by stuff well below the surface, etc. Sucn things depend on the delocalization of the photon.
So I think that if you measure the absorption of 15 micron wavelength photons by a 5 micron thick layer of water, some photons are absorbed before they reach the layer, some are absorbed in the layer, and some are absorbed after they leave the layer. But I was never good at that stuff. Too weird.
Mike,
Those are all very good points. They remind me of odd things you had to take into account in aerosol science, such as Mie scattering. Unfortunately, those topics are things in which I only have a surface-level knowledge in and don’t want to speculate wildly about. I do know that the linked paper says these molar absorptivities are useful down to 10 nm, but they may be able to correct for the “weird” stuff and change the measured optical density to a more useful true absorbance. Unfortunately, in my field (most commonly used light = 3-10 um) the optimal path length is in the 20-40 um range for most applications, and I’ve never personally made/used a cell below 6 um. In these cases, the only “weird” thing I have to account for is the presence of etalons in the measured spectra. Some of my new fabrication techniques should allow me to make far shallower cells (something I’m interested in anyway), so maybe I’ll dabble in it sometime to see what happens. Given that the linked paper used cells down to 1.2 um I didn’t notice any issues, it may be fine.
-Scott
Observation: If all of the energy that went into the arguments above were redirected into a collaborative effort to generate a climate model that accurately matches historical record and the next five years, then we all could focus on the truly important issues of this planet.
I think the basic model should start with planet completely cover by ocean which is uniformly 3000 meters deep. Which rotates in 24 hours.
Then add to model, land masses of a variety of types, but less than 30% of total surface area of the planet.
So have world covered in ocean, has 80% Nitrogen, 20% Oxygen.
And no land at the surface.
Let’s guess that at Earth distance from the sun the world could not warm to average temperature of 40 C. Average temperature being skin ocean temperature or a shaded white box floating 5 feet above the water. Now probably be a slight difference of either way it’s measured, but such small difference, will be ignored or will assume they are the same.
Rather then depending upon sunlight to warm, instead we going to have the ocean warmed from internal heat until such time and all water is 40 C. And later turn of such internal heating and have ocean just warmed by the sunlight.
So have a planet with uniform temperature [caused by internal heating- though sunlight could help warm it] of 40 C.
So planet rotating so have 24 hour day and 365 day year, and has axis tilted at 23 degree.
So first question is what happening in terms clouds and lapse rate.
It seems to me that one would have global a wet lapse rate.
Or saturated adiabatic lapse rate, wiki:
“When the air is saturated with water vapor (at its dew point), the moist adiabatic lapse rate (MALR) or saturated adiabatic lapse rate (SALR) applies. This lapse rate varies strongly with temperature. A typical value is around 5 C°/km (2.7 F°/1,000 ft) (1.5C°/1,000 ft).”
https://en.wikipedia.org/wiki/Lapse_rate#Saturated_adiabatic_lapse_rate
So laspe rate of 5 C per 1000 meter elevation, until one reaches elevation were air temperature is cooler.
And we have a super tropical troposphere. + 20 km of troposphere.
So at 5 K over 5 km elevation, one goes from 40 C to 15 C.
And from 5 km higher one follow Standard atmosphere tables:
http://usatoday30.usatoday.com/weather/wstdatmo.htm
which start with 15 C air temperature.
And Earth tropics has 4% water vapor, and this world would be around 5% water vapor.
Now first question is would this uniformly warm world be covered in clouds.
It seems one could argue it’s completely covered in clouds, or one could also argue that due to uniform temperature is has no or few clouds.
One could also say it has a lot of clouds and some clouds which would caused by the planet’s rotation.
Of course another aspect on could throw in, would be one can’t have uniform heating due to sunlight heating it.
If think starts with clear skies one could say sunlight warms the ocean, if instead it’s covered with clouds, the sunlight warms the clouds.
Or maybe other factors.
So question is would to tend to be completely covered with clouds, and also what kind of clouds [low, high, thick thin, mixture of all above, etc].
As as far as internal heating, the thermostat to attempt to keep the temperature of 40 C, is measured at skin or 5 foot in the box. And because there is sunlight most of added heating would done in polar regions. and less towards equator.
And finally, can you change how it’s internal heated, so that one gets a world almost completely cover with clouds. Or tweak the heating to get a world without many clouds. So in other words, you heat world so it’s uniform temperature. Then change that heating plan in such a way so get a lot of clouds, or very few clouds. And how would this be done?
— WizGeek,
But you can’t build a model unless you understand the physics.
And even the people who do understand the physics, exerting many orders of magnitude more effort than on display here, have not succeeded in building a model with useful predictive capability.–
Well if someone can answer the question [even broadly speaking] about what kind of clouds one would get on uniformly warmed water world at Earth distance sun with a globally uniform ocean skin temperature of 40 C. Then you would have basic model to which one add add complexity [land mass, CO2, orbital variation, volcanic variation, human effects, etc].
If you can’t answer this elementary question, AND you imagine you have good enough understanding Earth climate, that advocate spending tens of trillion dollars of public time and wealth- then you are without any doubt, delusional.
Or an uncaring psychotic insufferable evil prick.
WizGeek,
But you can’t build a model unless you understand the physics.
And even the people who do understand the physics, exerting many orders of magnitude more effort than on display here, have not succeeded in building a model with useful predictive capability.
Scott,
In my example of one cm/day evaporation rate, the upper three microns are gone in less than 30 seconds. Bye, bye LWIR. BYE, BYE back radiation. Hello insolation.
And by root(2Dt) the upper three microns are gone in 0.002 s (2.05 ms). Bye bye radiation, into the ocean below. LOL
-Scott
Well, Scott, care to use some terms that the rest of us understand, if you would so please to condescend?
You don’t listen to anything else I say, so why should I comply? And given that you’re the one making claims about evaporation being faster than heat transfer, how could you not know what this is? It usually represents a lower limit of heat transfer, though normally heat transfer, even conduction, is far faster than this.
But I will go ahead and explain even though you won’t probably listen in the hope that someone out there will.
root(2Dt) means the square root of the quantity (2Dt) and is the mean 1-D distance from the starting point that a molecule will have diffused in time (t), where D is the diffusion coefficient of the molecule of interest in the medium of interest.
-Scott
Scott,
Yes, I knew that you were referring to molecular diffusion but I wanted you to say so.
You claim that molecular diffusion transmits heat from the point of IR absorption to the ocean beneath.Thus you show that you do not comprehend the significance of the temperature Bradenton the warmer subskin to the cooler interface.
Correction:
Thus you show that you do not comprehend the significance of the temperature gradient from the warmer subsurface to the cooler interface.
Scott,
I wanted to thank you for taking time to post on this thread. Your interactions with mpainter are very useful to aid in outsiders to learn the correct physics.
mpainter does not reason so if your attempt is to get him to think it is a lost cause. He has a cult mind mentality from the Slayer group and it is a super strong religious belief system that is not based upon logic and reason or accepted science.
I just want you to know that other individuals can gain some benefit from the time you spend posting.
mpainter and geran are like twin minds. They make absolutist claims but never will provide any links to what makes them so certain (I am sure it is the PSI webpage).
mpainter says:
July 2, 2015 at 3:45 PM
Thanks for wasting my time then.
Oh wow, I totally missed that. Thanks for reminding me that individual water molecules can know the temperature of other water molecules almost 1000 microns away and choose not to diffuse in that direction. Here I thought that the absorption of the IR nearer the surface would raise the local temperature, decrease the temperature gradient present, correspondingly decrease the net heat flow from the 1-mm deep ocean, and result in a warmer temperature at depth than would otherwise be present without the absorption of a photon. And I thought pointing out that diffusion (molecular or thermal) is far faster than the evaporation and would therefore permit this mechanism to work to increase SST. Silly me, forgot that water molecules had ESP and are smart enough to migrate to the surface if struck by a LWIR photon and evaporate to cool the system instead of just randomly walking like the other molecules.
-Scott
Norman says:
July 2, 2015 at 9:14 PM
Norman,
Thanks for the kind words. I gave up on mpainter weeks ago, so it’s not like I’m trying to convince him. At first it started as just a bit of fun but also to teach people some real stuff. Now it’s morphed into a sort of psychological experiment for me–I want to see how long they keep rejecting things purely out of hand. The CO2 laser experiment definitely shows their claims are wrong at least some of the time, but they won’t both to even try to modify their claims to explain away the laser experiment…bizarre. mpainter even asked to irradiate with a dark gun, which a CO2 laser would qualify as. Of course, when he asked weeks ago to find a mechanism for heat to transfer faster than evaporation, I presented it and he ignored it. Now he’s trying to argue against it some, but it’s basically standard Slayers stuff that thinks no energy flows from cold to hot…never mind they use blankets when they go camping…clearly must be a mental placebo thing.
I’m glad to hear that you think what I type is worth reading. You’ll really enjoy when I go into the depths that evaporation occurs and its implications. I’ve repeatedly asked for what depth they think it occurs, but they ignore that too. Once the calculations are shown, there will be gnashing of teeth. Hopefully I can type that up and get it posted before bed…today has been long/hard and only compounded by all this posting.
Have a good day,
-Scott
At this point, will anyone step to the plate to discuss where the evaporation occurs relative to the absorbance of the IR photons? If not, I’ll go through the calculations here in a bit.
-Scott
Some posters here are effectively saying that there is no GHE.
Because you must be if you say that it does not “heat” the oceans. (slow heat flux to space).
Or are you saying it occurs over land only.
You know that the oceans comprise ~93% of the climate systems heat and that for the oceans to be radiating at the rate they are they MUST have a GHE acting on them. Without it they would soon freeze.
http://scienceofdoom.com/2010/10/23/does-back-radiation-heat-the-ocean-part-two/
–Some posters here are effectively saying that there is no GHE.–
Probably some. I am undecided.
I tend to think the radiant effect of greenhouse gas is probably about 1/3 of what it is said to be in the Greenhouse Effect Theory.
And it seems that there many things that cause a planet to be warming other than greenhouse gases.
I am probably a Lukewarmer, but all LuKewarmer seem to have different views.
For instance, it’s not that I think it’s unlikely earth will warm by more 2 C or more within a hundred year, I think it’s impossible- in terms of greenhouse gas causing the +2 C increased average temperature.
I don’t public policy at least at this point in time, can do anything effective in terms of controlling Earth temperature.
I likewise don’t think governments could terraform Mars.
I think terraforming Mars would easier for governments to do- cheaper, certainly. Not kill too many people, obviously. And teraforming Mars would be politically better and be more beneficial to world’s poor. Though not benefiting the so called “wallstreet 1%” or the political elite, the tyrant of the world, whatever you want to call them, or not the ones doing the least about the billions of people who living on $2 bucks a day or less. Mainly because terraforming Mars would be focused on doing something, rather than just corruption [though obviously if government is involved in anything there will be corruption, but it seems there would less, because there is something to do, rather just constantly lying about it.
But not saying terraforming Mars is something that should part of public policy, but rather just that it would be better- much better for everyone.
So governments have not done anything effective in terms of “global warming problem” despite incurring a global cost in the trillion of dollar range- and even worst then tossing trillion of dollar of wealth in the toilet, the is no accountability for these costs- so it’s very undemocratic, and elitist, which fuels more global insanity in term leadership.
But anyways it’s not effective in the least, it has no chance of controlling global temperature or weather, and it’s huge distraction to more important public interests.
So there nothing vaguely good about it, other than a feeding trough for those unworthy of any public money.
So summary I think greenhouse gas roughly speaking could cause about 1/3 of the 33 K or less of the warming they are purported to do. [And of course water vapor would cause most of this radiant effect].
And I do believe the Little Ice Age was actual event and was a global effect- as I also think the medieval warming period, was actually a period of global warming.
Or obviously the hockey stick was fraudulently made and presented as scientific work. {There is a lot of that in general, which unfortunately is not only in “climate science”.}
I think this paper will interest people here. This Stanford group made a device that cools BELOW ambient temperature even in sunlight. Very weird but it works.
Passive radiative cooling below ambient air temperature under direct sunlight
Aaswath P. Raman, Marc Abou Anoma, Linxiao Zhu, Eden Rephaeli & Shanhui Fan
AffiliationsContributionsCorresponding authors
Nature 515, 540–544 (27 November 2014) doi:10.1038/nature13883
The link is here:
http://www.nature.com/nature/journal/v515/n7528/full/nature13883.html
The point is that it cools by radiating directly to deep space via a n infrared hole in the atmosphere..
Well there no shortage of IR holes.
But claim is it’s passive. But I would guess it’s no more passive than hydro dam, coal plant, or solar panel.
But keeping something 5 degrees cooler than ambient air, while in direct sunlight, seems impressive. Not particularly useful in itself, though could be for something.
Now this reminds me of ideal blackbody that serves as basis
for Greenhouse Effect theory.
Which was that in direct sunlight of 1360 watts per square meter it could keep the surface cooled to 5.3 C. Which is more impressive, but it was magic.
Latex enabled on your site yet?
$ latex \displaystyle x \cdot 0 = 0 $
If an error or if nothing at all is displayed above, it seems not.
Gordon wrote (in response to my comment);
“@KevinK…”…they can merely delay the flow of energy through the system by acting as a sort of “hybrid” thermal-optical delay line”.
What energy? Are you referring to electromagnetic energy or thermal energy, they are different entities?”
Response from KevinK:
Yes indeed they (heat/IR) are different entities, but they easily “convert” back and forth. EMR gets absorbed and becomes heat. Heat gets radiated away and becomes EMR. Happens all the time continuously every minute of every day of every hour of every year, ALL THE TIME. The hybrid “thermal-optical” delay line delays the transit time of energy (alternating as thermal energy and then EMR energy) through the system. The energy simply “bounces” back and forth between the Earth’s surface and the atmosphere and gets delayed as it transits from the Sun to the Earth to the energy free void of space (note “space” is not “cold”, it has no matter and thus has no temperature).
“Also, what comprises your delay line? Are you talking about the 1% of the atmosphere deemed to be GHGs?”
Response from KevinK: Yes, the radiatively active gasses (CO2, etc.) form this delay line in the gaseous atmosphere of the Earth.
The much vaunted “Greenhouse Effect” in the gases in the atmosphere of the Earth merely delay the flow of energy through the Sun/Earth/Earth Atmosphere/Universe system and have no effect on the average temperature at the surface of the Earth.
Cheers, KevinK.
h
johnKl
Way up somewhere you had a post with a link to Claes Johnson.
http://claesjohnson.blogspot.com/2015/03/a-basic-model-of-greenhouse-effect-with.html?m=1
After reading some of this article I think the gentleman might be a brilliant mathematician but he is one terrible physics minded person.
In his 1,2,3 calculations he uses the Earth’s surface emissisivity as 0.7. That is really crap physics! The majority of the Earth is water and its emissisivity is almost 1.
http://www.engineeringtoolbox.com/radiation-heat-emissivity-d_432.html
So he uses 280 kelvin for Earth’s surface temperature but really does a super poor job of representing the REAL radiation that would be emitted from water of much closer to the 390 of the standard graph. The radiation budget graphs are bad too since they are assuming a black body. The 390 given should actually be a little lower but it is so much more than Claes who uses the crazy emissisivity of 0.7. The Earth’s surface will absorb 70% but the rest is reflected. You cannot use what is reflected as the amount that will be absorbed or emitted.
I would not use him as a source on anything related to climate with this type of assumptions he is making.
Hi Norman,
Claes Goran Johnson’s physics makes more sense than your post, however crappy you think the physics me maybe. Why? Well for one thing the Earth’s absorption of solar radiation is closer to .7 as Claes Johnson suggests. In fact NASA itself claims Earth’s albedo to be .3 or its absorption .7 as close Johnson claims. Try the following link:
Nasa (.gov) >earth observatory>view
As your toolbox link indicates water absorption exceeds .9.
To Continue:
However, Norman you seem to have forgotten that the Earth is covered by greatly by clouds and some of it by ice. NASA provides average cloud cover that is the percentage of the Earth’s surface covered by clouds to be .68 for an optical depth of .1 and .56 for an optical depth greater than 2.
Moreover Norman Claes figures appear reasonable. The Earth receives approximately 340 w/m^2 averaged over the Earth’s surface and absorbs .7 or ~240 w/m^2.
As to black body assumptions, you make a valid point but Claes sure isn’t the only one doing it. In addition, the 390 w/m^2 figure you provide is I believe a calculated one. I don’t know that that’s ever been measured directly from the Earth’s surface.
Actually, I’m more interested in the small 40 w/m^2 that flows from the Earth’s surface directly to space unimpeded by the atmosphere. Even if at some point in time the remaining 40 watts per meter squared were to be absorbed by massively increasing co2 the temperature differential would be small.
Thanks and have a great day!
Correction:
My statement should have stated that even if the 40 watts per metre squared will one day be absorbed bY additional greenhouse gasses the temperature difference would be small.
Ok, no one has been willing or able to give a quantitative value for where they think that evaporation takes place or even put it above or below where the IR absorption is taking place. I do thank Stephen Wilde for at least somewhat addressing it by stating that it is in the cooler ocean skin layer, which is correct, although not specific. A more specific qualitative answer would be right at the liquid/water edge of the cool layer. But that answer isn’t quantitative and therefore doesn’t allow proper evaluation of many of the claims being made here. Specifically, knowing the evaporation layer thickness would allow us to calculate the maximum percentage of water molecules that evaporate due directly to LWIR absorbance incidents (here, we’d assume that every absorbance occurrence in the evaporation layer results in a water molecule to evaporate…this is not at all true but is the most favorable assumption possible for the LWIR-causes-water-evaporate crowd).
To know maximum depth from which evaporation can take place, we need to calculate the Knudsen layer thickness (Google it if you want the equation). I calculated it for water and got 87 nm (1e-9 m) thick. It should be noted that this is an entirely new length regime compared to everything else discussed so far…it is far smaller than even the “minute” distances waved around by those saying LWIR can’t heat water. More importantly though is the realization that any IR absorbed past this layer cannot lead to direct evaporation of water and instead will raise the local temperature. Therefore, a good lower limit of the percentage of IR not leading to direct evaporation is the quantity %T, where %T is the percent transmittance. %T is easily calculated from Beer’s Law if you know the molar absorptivities of water at the wavelength of interest. Thankfully, the molar absorptivities are very easy to look up. What I consider as the best molar absorptivity data is that shown in Fig 1 of the paper in the link I gave Mike earlier:
http://www.sciencedirect.com/science/article/pii/S0003269797921369
However, that research did not go as long as 15-um light, so for that light I’m using the following source:
http://www1.lsbu.ac.uk/water/water_vibrational_spectrum.html
This second source seems to agree quite well with the first source in the overlapping regions. For 15-um light, I’m using 27 M-1 cm-1 for molar absorptivity, and for 4.3-um light, I’m using 0.9 M-1 cm-1 (note that the 4.3-um band has been mostly ignored in the above discussions, probably because it penetrates for deeper into the water but also because it does contribute less to total CO2 emissions). So what percentage of the photons in those two bands remain after the Knudsen layer? I calculate 97% for the 15-um light and 99.9% for the 4.3-um light. So, at the very most, only 3% of 15-um LWIR and 0.1% of 4.3-um LWIR directly lead to water evaporation. The actual numbers are almost certainly lower than these.
Out of curiosity, how does the assertion of saturation after just 3 (or 1 depending on when it’s being stated) um distance hold up? Depends upon your definition of saturation obviously. I’d normally use 99% absorbed for that value (I don’t know what is typically used in climate science) but will bump it down to 95% here. The values I get are 8.8 um and 263 um for 15- and 4.3-um light, respectively. Notice that not only are these significantly longer than the values used by the LWIR-can’t-heat-water crowd (though still well within the 1 mm distance posited by mpainter), but more importantly they’re ~100x and ~3000x deeper than the evaporation layer. Just for reference, I got 36% and 71% transmittance for 3- and 1-um depths, respectively, so I don’t know where those numbers came from.
So why is this important? Heat transfer. The main argument being thrown about is that the IR is absorbed right at the surface and (depending on who’s talking) either leads to evaporation or cannot cool below where it is absorbed because of the temperature gradient. However, we’ve now established that at most 97-99.9% of the LWIR makes it past the evaporation layer, depending on wavelength. So the direct-to-evaporation argument is quantitatively debunked. The temperature gradient argument still remains, but these results have important ramifications. The lower ocean surface temperature is posited (correctly) to be due to evaporation. But we’ve now established that the LWIR is almost entirely absorbed below the evaporation layer. They argue that it can’t possibly warm the ocean because of the temperature gradient, but notice that the dominant driver of the temperature gradient (evaporation) is farther away from the warmest layer of the ocean than where the IR is absorbed? Thus, the IR has to affect this gradient because it is “inside” of it. How does it affect the gradient? Each absorption event leads to a small amount of warming which in the end reduces the amount heat transferred from the depth to the surface because that energy can instead come from the absorbed IR.
Some people really don’t understand this concept so I’m going to explain it in a form that hopefully is the form most people here have been exposed to—a basic electrical circuit and Ohm’s Law (this is a part of a broader concept called the electrical circuit analogy). For the example, we’ll place the ocean surface at ground/0 V, place the warmest ocean layer at +1 V, and use a sign convention where flow goes from most positive to least positive values. The +1 V potential is like saying the warmest ocean layer is 1 C warmer than the surface (potential is analogous to temperature here). There is some amount of thermal resistance in the ocean between the warmest layer and the surface, and we’ll model this in the electrical circuit as a 1-ohm resistor. Heat flow in the ocean is analogous to the current flow in the circuit, which would be 1 A. This model I’ve described is for the ocean in the absence of incident IR. So how does the circuit change when IR is absorbed? An example would be to put a new voltage source between (in series) the 0 V and +1 V potentials with 0.5-ohm resistors on either side of it (note that total resistance does not change). In the absence of IR, the voltage here would be +0.5 V, but due to the IR absorbance (analogous to charge buildup) it has been raised to +0.51 V. The current flow exiting the +1 V source is now reduced and the flow entering the 0 V sink is increased. This is only a transient effect obviously because it leads to a decrease in charge at the +0.51 V point, which would then quickly go back to being +0.5 V. However, during the time immediately after the absorbance event the current leaving the +1 V source is decreased and at no time is increased relative to the no-IR situation. Consequently, the flow of heat from the warmest layer of the ocean to the surface is reduced with each photon absorption event, thereby causing it to have a higher temperature over time than it would have otherwise. You can call this warming or reduced cooling, I don’t care, the result is the same–a temperature in the presence of incident IR that is higher than in its absence.
Yes, I know the electrical circuit analogy isn’t perfect, but it gets the point across better than most. Sorry if this isn’t clear, it was written after midnight and my young boys wake up before 6 am. I would appreciate any sort of quantitative feedback, thanks.
-Scott
Scott,
Very convincing. Too bad it won’t convince anyone.
“To know maximum depth from which evaporation can take place, we need to calculate the Knudsen layer thickness (Google it if you want the equation). I calculated it for water and got 87 nm (1e-9 m) thick. It should be noted that this is an entirely new length regime compared to everything else discussed so far…it is far smaller than even the “minute” distances waved around by those saying LWIR can’t heat water.”
Well it should be noted I never said LWIR can’t heat water. And you didn’t say I did, but I will state it anyhow. And I would note one would have to heat water to cause water to evaporate.
So .. yawn.
Anyhow, I will bring up a couple of things.
Though I am generally interested in space and space rocks, I am not familiar with Knudsen layer thickness. But googling it, one gets that it’s considered useful for comets.
And factoid regarding comets, H20 is evaporated in vacuum at
about 100 K.
And in term limit of evaporation in depth of ocean on earth, it seems many gases other than H2O gas are entering and leaving the ocean surface. And not sure why the H2O gas can’t pass thru water as other gases do. So if H20 gas passes thru water in similar fashion as O2 or CO2
does, one could say that H20 molecule going thru water is not being evaporated.
But one could also say the that a gas molecule which evaporated doesn’t necessary need to go up and away from the water [or has to go in any direction].
Or basically none of molecules near the surface are going any where.
Each molecule hits other molecule in such a short period of time, individual molecules don’t travel any distance- despite having velocity far faster then a speeding bullet.
Instead all air molecules can be understood to move up and down in air packets.
So in space you don’t have gas which move in terms of air packets. One only get this when there is huge number of gas molecules in cubic cm volume.
So one could have a relatively huge packet of air- relative to 87 nm [though packet aren’t actually given a unit size], and the packet could be cut in half by the waterline.
So could say, rightly this is not a “normal air packet” or could it see as normal packet of air in a rain cloud- so “not normal” because it’s interacting with liquid. though quite normal because it’s common.
So the concept of packets is that one dealing with movement of kinetic energy rather than movement of individual molecules.
But things like thermals, or up and down drafts, is not individual moving, but galaxies of packets moving as a mass. So individual molecules don’t travel alone, they travel within a mass of molecules.
For one molecule to move say 1 foot higher above the waterline, one needs buoyancy and buoyancy is about moving galaxies of gas molecules.
Now if have a H2O molecule which became gas from liquid hours/days ago, and not part of the evaporation process you are focused on, and leaves the surface as part of a galaxy, it doesn’t matter when it evaporated, all that matters is that it left the ocean [or it will be replaced with another evaporated molecule going the “wrong way”.
Or another thing is general nature of water. Supposedly for instance gas/water molecules can stake across the surface at tens of meters a second [slow velocity for gas and not what gases do]. But need reference for that I suppose. Where? Not here:
http://www.chem1.com/acad/sci/aboutwater.html#STW
Anyways, can’t find it. So retract that. Oh here’s something while looking for that:
“Energy is required to increase the surface area (removing a molecule from a well hydrogen bonded interior bulk water to the lesser hydrogen bonded surface), so it is minimized and held under tension.”
http://www1.lsbu.ac.uk/water/physical_anomalies.html#Visc
So in summary, comets are cold, and don’t have much gravity.
And on earth the force of gravity is related to buoyancy will be related to evaporation. And related the max distance/depth involved.
And ice is much different than water. It’s got more organized molecular structure, and no surface tension.
Scott,
I’m not convinced that you are using the Knudsen layer correctly.
It is the region where evaporation takes place but that does not necessarily imply that all IR getting past it must heat the water below.
You have to consider the full depth of the cooler ocean skin layer and note that the evaporation occurring in the Knudsen layer higher up will be pulling kinetic energy upwards from the interface between the cool ocean skin and the ocean bulk below.
It is the pulling upwards of kinetic energy by the process of evaporation (faster than it can conduct upwards from below) that cools the ocean skin to below the temperature of the ocean bulk immediately below it in the first place.
To heat the ocean bulk any IR has to get completely past the full depth of the ocean skin and not just past the Knudsen layer.
Hence your assumption that 97% or more of IR gets into the ocean bulk is incorrect.
In reality no IR gets past the full depth of the cooler ocean skin because the evaporative action in the Knudsen layer is pulling it back upwards so your attempt at a quantitative assessment is flawed.
Stephen,
I am using the term “bulk” ocean as it would be used by a chemist working at the molecular level and thus anything beyond the Knudsen layer is by definition bulk. If you wish to use a different definition, such as one based on temperature, then that it fine. But it does not change the direct-to-evaporation percentage that I calculated.
You describe the evaporative cooling as “pulling” energy upwards, which I object to. For example, in the analogous fluid flow, there also is no such thing as “sucking”. When one drinks out of a straw, the atmospheric pressure is pushing the fluid into your body, which is at a reduced pressure. This is conclusively shown in the case of a critical orifice. Here, one gets to the point where lowering the downstream pressure has no effect on flow–the only way to increase the flow rate in that example is to raise the pressure of the inlet (or change your system to one with a bigger orifice).
But even using the “pull” terminology I can show that the overall ocean is warmed, as my electrical circuit diagram is still valid. When a photon is absorbed below the Knudsen layer, the result is an increase in energy below the Knudsen layer. Unless the evaporative layer can somehow “know” that that photon was absorbed and then “pull” harder, then one less photon’s worth of energy will be removed from the deeper ocean, placing it then at a warmer temperature than it once was. In that case, the “pulled” energy came from the photon instead of the bulk.
Here’s another example. Imagine placing a thin-film resistive heater between a block of ice and a rock. When the heater is off, that’s the equivalent of the situation above without IR. The rock and heater will be cooled by the ice, with the heater colder than the rock. Now, turn the heater on at e.g. 1 W. Guess what, the heater still ends up getting cold and being between the temperature of the ice and the rock. But, despite the heater being cooled (“overwhelmed”) by the ice, the final temperature of the rock (and the heater actually) would be warmer than they would in the absence of the 1 W. The same is true with the IR absorbance.
Essentially what people are doing is looking at the final temperature profile (which includes a multitude of inputs) and using that to claim that evaporation is the dominant input, which is correct. But where they error then is to effectively claim then that any other inputs counter to the dominant input then do not matter. Earlier I used a biking analogy, and I don’t know if you saw it. I almost always bike into the wind on my way to work. My momentum completely “overwhelms” the wind’s and I make it there. So since the dominant factor was my momentum, does that mean the wind didn’t matter? Of course the wind mattered, it took more energy (and time) for me to commute. All inputs combine to complete the picture, the dominant one doesn’t remove the effects of the others.
Also, the 97% number isn’t an “assumption”, it is a “result” or “conclusion”, and there is a big difference.
Have a good day!
-Scott
Scott,
The long post was well done. I think the big problem is with the use warming as in Global Warming. Every time Dr. Spencer or the physics trained people talk about it they are very clear with the “warmer than” clause.
I have no clue why the Slayers do not believe energy will flow from a cold source to a warmer one. Energy will flow in all directions available to it. They cannot grasp the idea of NET ENERGY. Net energy flow is from hot to cold but cold will still have an energy flow to warm, just at a much smaller rate. Any thermodynamics text book uses this fact in their equations.
They are unable to think and reason. Programmed cult mind. They have no desire to understand and learn.
Green house gases do not directly warm the surface. The energy flux from them to surface is less than that lost by surface. But the overall surface will be warmer with them then because of the constant incoming supply of solar input.
Greenhouses warm the surface by blocking upward convection so that heat can accumulate at the surface.
Descending adiabatically warming air (half the Earth’s atmosphere at any given moment) works exactly the same way.
Such descending and warming air dissipates clouds to let more sunlight reaches the surface and also reduces upward convection below the descending column so that the surface gets hotter than it otherwise would.
That is the correct (mass induced) greenhouse effect and it was once established science.
Stephen Wilde,
I believe Roy Spencer has addressed this claim in a past thread. If any air is moving down it must be balanced somewhere by air moving up. The upward moving air will do the opposite, it will move energy in bulk away from the surface (convection) and it will generally produce clouds which shut off a lot of solar input.
The mass greenhouse effect could not explain a 33C temp above what it would be without the GHE. It would warm some areas and cool others but in no way could it add total energy to the surface. That energy had to come from somewhere. It also does nothing to explain the satellite view of outgoing long wave IR. The surface emits an average of 390 watt/meter^2 but you don’t have that radiation in the satellite view. It can’t explain this empirical data at all.
That may be why this theory was once established but is no more because it cannot explain certain empirical data.
Norman,
I don’t recall Roy addressing that point.
The mass greenhouse effect takes 33C from the surface in uplift but returns it in descent.
It cannot cool the surface to less than the S-B temperature in the face of continuing insolation because conduction and convection is slower than radiation. To cool the surface below 255K conduction and convection would need to run faster than radiation.
Whilst that 33K is being extracted from the radiation budget during the first convective ascent the temperature as viewed from space drops to 222K because the energy removed from the surface by convection is no longer available for radiaton to space. THe actual surface temperature remains at 255K.
Then during the descent the surface is not warmed directly by the descending air but indirectly by the descending air reducing convection beneah the descending column as per the greenhouse analogy.
Once the first convective cycle completes you then have both continuing insolation PLUS returning KE limiting convection across half the surface so total energy IS being added to the surface.
Viewed from space the observed temperature rises from 222K to 255K but the actual surface temperature is 288K.
That explains the empirical data. The convective overturning cycle slows the release of radiative energy to space by locking it into convection for a while and thereby raises surface temperature to a lavel higher than the apparent temperature as viewed from space.
Ignore clouds at this stage because their effect needs to be dealt with separately from the basic mass induced effect.
Scott,
I understand what you are trying to say but your analogies omit the primary characteristic of the phase change of water from liquid to a gas which is that the phase change takes up the energy content of 5 more photons as a result of 1 photon causing or bringing forward the timing of an evaporative event.
That is what creates and maintains the cooler ocean skin and that 5 to 1 relationship is a consequence of 1 bar atmospheric pressure.
If 5 photons are taken into latent heat for every 1 photon that causes or brings forwartd the timing of an evaporative event then it doesn’t matter whether the additional energy required comes from above or below but it will most certainly come not from the warmer ocean below the cooler skin but from within the skin itself.
There will be no surplus energy left over to heat the ocean below or reduce the flow of conducted energy from the ocean below.
The enhancement of the evaporative process then stops as soon as all 6 photons of added IR have been taken into latent heat which leaves the background energy flow from the warmer lower ocean through the cooler skin to the atmosphere undisturbed.
That is as simple as I can make it for you and other readers and it was once well established science before the inadequate radiative theories came to the fore and omitted the thermal consequences of non-radiative energy transfers.
If you want quantification then there it is in the 5 to 1 relationship.
I will have to take you at your 5:1 word. I have not calculated the number myself.
Using that ratio and the at most 3% of 15-um light and 0.1% of 4.3-um light that can directly cause evaporation, I calculate these numbers (based on the original percentages):
15 um:
Energy absorbed by Knudsen layer = 3
Energy lost by Knudsen layer through evaporation = (15)
Energy absorbed below Knuden layer = 97
Net energy gain = 85
4.3 um:
Energy absorbed by Knudsen layer = 0.1
Energy lost by Knudsen layer through evaporation = (0.5)
Energy absorbed below Knuden layer = 99.9
Net energy gain = 99.5
Honestly, the rest of your description makes no sense at the molecular level. That is as simple as I can state it for you and other readers. My analogy with the ice even included a phase change, and you could easily replace the ice with liquid water and it would still hold. The temperature gradient is indeed established by evaporation and is much larger than the Knudsen layer, but the size of the gradient does not change where evaporation can happen. Any radiation making it past the Knudsen layer has to decrease the gradient and therefore reduce the heating from the lower ocean level.
-Scott
Energy absorbed by the ocean skin = 100% of DWIR.
Energy used by the Knudsen layer within the ocean skin to create additional evaporation = 16.66% of DWIR
Energy taken from the ocean skin by the phase change = 100% of DWIR.
Net energy gain = 0.00
The 5:1 ratio comes from here:
https://en.wikipedia.org/wiki/Enthalpy_of_vaporization
“the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C (cp = 75.3 J K−1 mol−1).”
Heating water to the boiling point of 100C is enough to vaporise ALL of the water yet the vaporisation process actually takes up 5 times as much energy as that required to achieve the initial heating.
Stephen (and others):
Please explain the following from …..
http://judithcurry.com/2014/05/21/mechanisms-for-warming-of-the-oceans/
which quotes …..
http://home.earthlink.net/~drdrapp/ocean.heating.v3.pdf
“The surface of the ocean is thus warmed slightly by IR absorption and this reduces the temperature differential from the bulk mixed layer to the surface. This reduces the rate of heat loss from the mixed layer of the ocean to the surface; thus the mixed layer does not cool as fast as it would without the IR forcing at the surface (i.e. it warms relative to the unforced state). It is important to understand that the IR absorbed at the surface does not flow down into the ocean. The energy flow is always upward.”
“ne thing we can assert however, is that warming of the oceans by an increase in down-welling infrared radiation to the surface is an efficient process, and the initial rate of heat loss by the mixed layer is only slightly less than the magnitude of the imposed forcing.”
You are asserting that 100% of DWIR is utilized for evaporation and 0% is available to be absorbed to reduce the deltaT in the ocean skin.
The above paper along both Roy’s experiment and the Ocean observations I linked to above show that your assertions (for they are only that with no experimental or empirical proofs).
Additionally it makes no common sense the the evaporation process (when caused by absorbed electromagnetic energy IRRESPECTIVE OF ITS WAVELENGTH), be 100% efficient.
Oh, I know this post will be “objected to” as I have conversed with you before (years ago) and do watch the bizarre ramblings of “slayers” across the ensemble of “contrarian” Blogs.
So some empirical proof PLEASE;
AS IN EXPERIMENTS AND OBSERVATIONS to deny the above and add something other than hand-waving. That goes for the others of similar disposition here.
Toneb.
The quote is simply wrong.
If IR raises the kinetic energy of a molecule within the ocean skin then it may raise the temperature of a molecule at the point within the skin at which it is absorbed but the effect is simply to provoke an immediate evaporative event or bring forward the timing of the next evaporative event and in the process cause conversion of 5 times as much kinetic energy into latent heat.
The quote fails to take into account that 5 to 1 ratio.
There can be nothing left over if the phase change mops up more energy than initially provoked it.
In reality, DWIR increases the temperature gradient through the ocean skin by enhancing evaporation even though the gradient between the top of the skin and the subskin may be reduced. It sharpens the temperature differential between the base of the skin and the warmer ocean below but pulls no additional energy upward from below the skin layer because it takes the shortfall from within the skin instead.
Toneb
That source appears not to consider the effect of the 5 to 1 ratio between energy required to provoke evaporation and the energy taken into latent heat by evaporation.
By virtue of that ratio 100% of any radiation absorbed within the skin layer must be converted to latent heat.
The empirical proof is the coolness of that layer itself. Evaporation creates it by cooling the surface faster than energy can conduct up from below.
To achieve that it must be the case that the evaporative process is converting ALL radiation absorbed directly into the skin layer into latent heat PLUS ALL energy conducted up from below.
There simply cannot be any surplus left over contrary to that article’s proposition.
Sorry,
This:
“In reality, DWIR increases the temperature gradient through the ocean skin by enhancing evaporation even though the gradient between the top of the skin and the subskin may be reduced.”
should be:
In reality, DWIR increases the temperature gradient through the ocean skin by enhancing evaporation even though the temperature differential between the top of the skin and the subskin may be reduced.
Indeed,
The oceans accumulate heat by day and release by night. A minute increase in LWIR does not change this.
Increases in SST are shown to be due to increased insolation through reduced cloud cover.
Toneb, July 4, 2015 at 1:43 AM, quotes:
“The surface of the ocean is thus warmed slightly by IR absorption (…)”
NO, IT IS NOT WARMED SLIGHTLY BY IR ABSORPTION!!!
What is it with you people!? How hard can this be? In what kind of warped reality are you all living!?
The ocean surface isn’t warmed by LW from the atmosphere at all. It cools by LW to the atmosphere. Just as it cools by conduction to the atmosphere.
However, it mainly cools by evaporation to the atmosphere.
The atmosphere is heavy (it has a mass subjected to gravity). It weighs down on the surface, exerting a pressure (a downward force) on it.
This pressure suppresses evaporation rates at a particular surface temperature level.
THIS is why the ocean surface is warm. It balances its solar input (which is doing the actual SW/LW warming) with its evaporative output under a certain atmospheric pressure.
The only reason we can keep an ocean on Earth in the first place is the MASS of the atmosphere resting on top of it. If we removed the atmosphere overnight, most of the oceans would quickly evaporate into space and the remains would be left frozen solid, until even these were all eventually (over a few diurnal cycles) gone, vaporised by sublimation from direct solar heating.
If we were to double the atmospheric mass, but initially kept the surface temperature the same, evaporation rates would plummet. Hence, the ocean would necessarily warm (through the accumulation of the solar energy input – more IN than OUT per unit time). If, on the other hand, we were to half the atmospheric mass, initially keeping the sfc temp unchanged, evaporation rates would soar and the surface temperature would inevitably drop (more OUT than IN per unit time).
Ocean temps and – significantly – the change in ocean temps through time are governed by the balance between solar input and evaporation rates. Trying to explain it rather through atmospheric “back radiation” is only silly …
Kristian is right as regards the effect of atmospheric mass weighing down on the ocean surface.
It is that weight which determines the relationship between the amount of energy required to initiate evaporation and the amount of energy taken up in the phase change.
At 1 bar atmospheric pressure one fifth of the energy required by the process of evaporation needs to be added in order to initiate the process.
At zero atmospheric pressure evaporation occurs immediately with no new energy needing to be added.
As one icreases atmospheric pressure more energy needs to be added to initiate evaporation at a given rate.
see here:
http://www.newclimatemodel.com/wp-content/uploads/2011/11/TheSettingAndMaintainingOfEarth.pdf
mpainter says:
July 4, 2015 at 4:40 AM
Correct, it doesn’t change the sign. It does however, change the quantity. It seems the only numbers you guys are capable of processing is a 5:1 ratio, which you apply only when you choose to.
-Scott
Stephen,
I’ve already shown that the upper limit of the evaporation induced by 15-um IR is small and that the amount induced by 4.3-um IR is completely negligible, even considering your 5:1 ratio. You just ignore that and press on.
Once the light makes it past the Knudsen layer, the energy added upon its absorption is no different from any other energy below. What you’re claiming is that the water molecules at the surface are counting IR photons passing them and then intentionally evaporating just the right amount to exactly offset the added energy in the layers below the Knudsen layer.
-Scott
No counting or intention required, just the basic laws of physics.
You haven’t ‘shown’ anything, you simply assumed that the Knudsen layer rather than the ocean skin is the region of absorption whereas the Knudsen layer is actually the region of evaporation which draws energy from the entire depth of the ocean skin.
Evaporation results in cooling and conduction towards the cooled region then inevitably takes place.
That conduction is fuelled by energy from IR being absorbed through the full depth of the skin layer until it is all used up with no effect on the subskin ocean.
As soon as evaporation has used up all the IR the underlying energy transfer from ocean subskin through the skin layer to the atmosphere continues undisturbed.
Stephen,
You clearly either don’t understand what I wrote or just didn’t read it.
I do not “assume” that absorption takes place in the Knudsen layer–I actually showed the opposite, that the majority of the absorption takes place below the Knudsen layer.
But there may be hope for you from this quote:
Ok, I can mostly agree with that. So tell me – what is happening with the energy transfer from depth BEFORE “evaporation has used up all the IR”…i.e., while evaporation is using the energy from the IR absorption?
-Scott
Stephen Wilde says, July 4, 2015 at 10:56 AM:
“That conduction is fuelled by energy from IR being absorbed through the full depth of the skin layer until it is all used up with no effect on the subskin ocean.
As soon as evaporation has used up all the IR the underlying energy transfer from ocean subskin through the skin layer to the atmosphere continues undisturbed.”
There is no energy for evaporation to “use up”, Stephen. Because there is no positive transfer of energy by LW from the atmosphere to the surface, unless it comes down as heat (which can only ever move spontaneously from warmer to cooler). The LW transfer is UP and only up, from sfc to atm. The ocean surface cools via three separate mechanisms: 1) Evaporation (by far the most important one); 2) LW (thermal) radiation; and 3) conduction. All of these energy transfers move OUT OF the surface. They constitute heat LOSSES. The ONLY thing heating the surface is the Sun. It transfers energy as heat TO the surface (and ocean bulk).
— Scott says:
July 4, 2015 at 10:24 AM
Stephen,
I’ve already shown that the upper limit of the evaporation induced by 15-um IR is small and that the amount induced by 4.3-um IR is completely negligible, even considering your 5:1 ratio. You just ignore that and press on.
Once the light makes it past the Knudsen layer, the energy added upon its absorption is no different from any other energy below. —
So back radiation is suppose to be light, but it is not limited to 15-um IR. Btw, what is the spectrum of this back radiation light, which this science is based upon?
Also we can assume that the ocean does not radiate energy above this thin Knudsen layer you mentioned, that any radiant light from the ocean come from below this thin Knudsen layer, correct?
And since the ocean does evaporate, we could assume some of light of the ocean heats water below the this thin Knudsen layer and also some radiant light [sunlight if nothing else]
warms water below this thin Knudsen layer. And therefore H20 gas which is made below this thin Knudsen layer will pass thru this thin Knudsen layer.
Also air above the layer is the atmosphere [which could have higher concentrate of H20 gas] would have this thin Knudsen layer as an barrier of sorts. Perhaps like very thin inversion layer. But this barrier is compromised by H20 gas molecules lighter than water molecule. Sort of holey balloon surface.
If you go more thin than this thin Knudsen layer, this holey balloon could progressively becoming mare compromised.
So at deepest point of this thin Knudsen layer the water might be together in some vague structure, and becomes progressively more vague in terms of a structure of a liquid.
Now at microscopic level the surface probably isn’t flat, nor calm. You could have comparatively large spheres of water, which might be rolled across the surface, one could flat sheets of water molecules being tatters of once uniform surface structure, “trying” to regain this structure, sometime succeeding in forming a coalition, other times rolling up and eventually becoming spheres which bounces across the surface, becoming bigger sphere and/or perhaps being launched above the water and leaving it.
Anyhow, point is at some point in the surface the radiant energy of ocean below meets the radiant energy coming from above the ocean.
Of course all matter doing the same thing- radiant energy meeting below it’s thinnest surface but the water of ocean is evaporating, other stuff may not have it’s material evaporating.
Kristian,
I do not subscribe to the concept of IR coming down from the higher atmosphere where it is cooler.
I do subscribe to IR emanating from each level of the atmosphere according to its temperature at that height.
Thus the IR impacting the water surface is coming from the temperature of the air molecules just above the water surface at the lowest point along the lapse rate slope.
That temperature is raised by adiabatic convective overturning which you do not accept which IMHO is a flaw in your otherwise good narrative.
This recent paper supports my understanding of adiabatic processes:
http://hockeyschtick.blogspot.co.uk/2015/07/new-paper-finds-increased-co2-or.html
–This recent paper supports my understanding of adiabatic processes:
http://hockeyschtick.blogspot.co.uk/2015/07/new-paper-finds-increased-co2-or.html —
Excerpt from summary:
“Moreover, based on the adiabatic model of heat transfer, the writers showed that additional releases of CO2 and CH4 lead to cooling (and not to warming as the proponents of the conventional theory of global warming state) of the Earth’s atmosphere. The additional methane releases possess a double cooling effect: First, they intensify convection in the lower layers of troposphere; Second, the methane together with associated water vapor intercept part of the infrared solar irradiation reaching the Earth. Thus, petroleum production and other anthropogenic activities resulting in accumulation of additional amounts of methane and carbon dioxide in the atmosphere have practically no effect on the Earth’s climate.”
I don’t think greenhouse gases cool earth.
I don’t think the interception by H20 gas or water droplets
in a clear sky of the sunlight cools earth. I think it obviously reduces direct sunlight and having less direct sunlight would effect the max temperature that land surface could heat up to but land surface and how hot they can become, isn’t a significant factor in increasing Earth average temperature- rather the hotter a surface gets that the more intense the IR emitted to space.
Though a hotter surface would convect more heat- one might get more up draft [thermals]. And a bit warming from addition mixing effect, but not very significant.
Now methane could be higher in atmosphere, so one could have similar effect of high elevation volcanic dust- a cooling effect. But I don’t think there is much methane in higher elevation. Methane converted into CO2 and higher it is the more it’s converted.
It’s a quantitative difference. Or said differently, dust can redden the sky, I doubt methane or CO2 would do this.
As for first reason, “they intensify convection in the lower layers of troposphere;”, I don’t think greenhouse gases do this.
Scott proves that energy moves from the interface to the ocean against the temperature gradient;from a cooler interface to a warmer subskin. See what 25 years of education and 5 years as a postdoctoral lab assistant can do. AGW can be likened to infestation of termites, munching away at our educational institutions.
Congratulations on a successful rebuttal.
-Scott
The point at issue boils down to whether DWIR increases or decreases the thermal gradient between the top of the ocean skin and the water below the base of the skin.
AGW theory suggests that with more DWIR the thermal gradient between those two specific locations becomes less steep and so,logically,energy will flow upward from ocean to atmosphere through the skin layer more slowly and the oceans will warm.
That proposition ignores the negative 5 to 1 ratio between energy required to cause evaporation within the ocean skin and the energy taken up when evaporation occurs.
If evaporation increases when the ocean skin receives more radiation of any wavelength then the coolness at the base of the ocean skin must increase rather than decrease. That is obvious because the coolness was caused by evaporation (via the 5 to 1 ratio)in the first place.
More of a causative agent must produce an enhancement of the effect.
AGW theory relies on no increase in evaporation or a ratio of 1 to 1 rather than 5 to 1. Both of those would indeed lead to a reduction of the thermal gradient between the top of the skin and the top of the warmer subskin.
However, that ratio of 5 to 1 results in stronger cooling of the ocean skin when evaporation increases so what we see may be a raised temperature at the top of the ocean skin but as a corollary the negative 5 to 1 ratio must also enhance the cooling below to a 5 times greater extent.
At the boundary between the base of the ocean skin and the top of the warmer subskin the thermal differential actually becomes greater. The cooling change at that point must be 5 times larger than the warming effect at the top of the ocean skin.
So, even though the temperature difference between the top of the skin and the top of the subskin might reduce the temperature differential at the bottom of the ocean skin increases more.
Donald Rapp, Mr Minett et al have all got it wrong because they omit the effect of that 5 to 1 ratio.
–The point at issue boils down to whether DWIR increases or decreases the thermal gradient between the top of the ocean skin and the water below the base of the skin.–
Assuming the DWIR is radiant energy. Or if DWIR is radiant energy it should manifest itself upon the top surface of the ocean. It should also effect every droplet of water on Earth.
As a thought, perhaps it’s complex mental coping mechanism to deny the existence of small droplets of water.
Little people aren’t important, something to step on the way to power and glory.
[[Simplify, simplify, the atheist once thought life was very simple also. They also thought life had billions of years to evolve. Also if given enough time the oppression of Socialism also works. Or the problem with Stalin was he was tiny, weenie bit off- close, but not quite what you want. Though there are those who still maintain that the man was perfect.
While others worship the still living god of Castro.]]
Now if DWIR radiant energy do exist and can warm the top surface of the ocean, perhaps it heating a lower part of the ocean via convection of gases. Fish breath, and so there is oxygen, as well as nitrogen [and CO2] in the ocean. Plus there is an idea that water vapor can only go up.
And a water gas molecule has no idea about what is up. It’s careening around not going anywhere. Sort of like this thread.
–AGW theory suggests that with more DWIR the thermal gradient between those two specific locations becomes less steep and so,logically,energy will flow upward from ocean to atmosphere through the skin layer more slowly and the oceans will warm.–
This energy flow being convection. And what goes up, goes down.
But the more important aspect of the ocean is the uniformly
if it’s heat. Ocean is mixed by wave motion. Ocean is warmed uniformly by diffused sunlight. It’s salt water [getting back to experiment] and salt water inhibits convection of heat.
But time is not important to the faithful, time is only important as it serves as an excuse. Time is tenure, and tenure is excuse to do nothing.
The results don’t make a lot of sense in terms of downwelling LWIR effect.
The temperature difference is too spiky. In two places the delta changes by 0.4F in single sample period. In many places it changes by 0.3F in a single sample period while the average delta of the whole night is less than that.
The temperature changes seem physically impossible so I suspect your instrumentation is inadequate.
You may be able to improve it by having two of each, shielded and unshielded, and two temperature probes in each. It would help to synchronize the exact time each sample is taken and see that each of the redundant tests are giving the same result.
The most worrisome thing is it appears to my eye that a shift to the right of the unshielded result by a single sample period of 5 minutes would negate virtually all the temperature difference. This indicates that the entire difference in temperature took place in a step change during one or just a few sample periods.
The step change appears to have happened in the time immediately before and after the clouds moved in which were probably accompanied by some wind changes.
Nice try but no cigar.
FYI just to improve the simulation of the ocean the interior of the test chambers below the water line should be black, not white.
“Nice try but no cigar.”
I am not sure what Roy W. Spencer experiment is suppose to prove.
I would think it could prove the salt water reduces convection losses. Or since I don’t think the location in the tropics, and therefore lacks a huge amount of water vapor, yet water did not cool much.
It seems also that shade may have had a small effect. The effect of the clouds seems not likely- maybe, but not very definitive.
The idea of inhibiting evaporation in next experiment, seems to me one could more dramatic result.
In terms of open container, the most significant variable could the presence light breeze, and sealed container a small amount of wind would have a smaller effect. Also think that shield could amplify wind effects- so that tends to make me believe it probably was fairly calm conditions during the tests.
I am not sure there is a way to monitor low wind effects- ie, if someone walked near a instrument, would it detect the wind effect created.
Toneb:
You demand mpirical proof.
###
Very simple; direct a dark heat gun at a bucket of water. Make sure the water is at ambient air temperature. Do this yourself and you will see that LWIR does not heat water.
This demonstration is ignored by those like Donald Rapp, Scott, Norman and yourself.
mpainter,
I did this. CO2 laser is a dark gun. It heated the water without an issue. You rejected the results.
Your lack of understanding used to be laughable, but at this point it is just sad.
-Scott
Scott,
Someone needs to explain to you why a laser is not something that mimics natural processes.
I see nothing in your mechanism about “natural processes”. Maybe you should modify/improve your mechanism?
-Scott
The difference is between a dark heat gun emitting diffuse IR at the same rate as the atmosphere as against a CO2 laser which gives an intense focused beam capable of heating the water surface faster than evaporation can occur.
Scott needs to apply more thought.
I’d like to see him heat his bathwater with a hair dryer 🙂
Stephen,
Would you please be quantitative on how fast evaporation occurs? I’ve gotten numbers from mpainter in the past and showed that molecular diffusion alone is far faster than evaporation, and overall bulk heat transfer (via conduction or advection) is far faster than heat transfer by molecular diffusion.
Until you provide a number and compare it to heat-transfer mechanisms, it’s just proof by assertion.
-Scott
I think Scott is Doug Cotton who is the only person I know who ignores convection in favour of his own personal interpretation of molecular diffusion.
Stephen,
Would you answer my question about how fast evaporation is? You claim that the laser heats the water faster than evaporation occurs (which I agree with), but you haven’t quantified either speed for the laser case or for the ambient-condition case. The driving force for evaporation is difference in chemical potential, which is heavily influenced by temperature, whereas the driving force for heat transfer is dT itself. So both are dramatically sped up by temperature…it’s not like the evaporation rate stays constant as temperature changes.
I do not ignore convection and used it as an argument for ways to get LWIR to depths. mpainter attacked my claims because his view of convection is limited to kid-level understanding of buoyant convection. Check out how engineers treat convection, you really just need to read the first paragraph and compare it to my 11:34 am comment to see that convection is not ignored:
https://en.wikipedia.org/wiki/Convective_heat_transfer
As to my own “personal interpretation” of molecular diffusion, root(2Dt) as the mean travel distance came about due to the work of Albert Einstein. You should take it up with him.
-Scott
Stephen Wilde says:
July 4, 2015 at 11:00 AM
Okay, I did that Stephen. Here is the description, along with results:
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194595
Summary: the water heated up without issue.
-Scott
So at this point I think it’s safe to say that the disguise is gone and the claims being made are now clearly exposed to be those of the Slayers’ misunderstanding of the 2nd law of thermodynamics.
I’ve been thinking of some examples that show how silly their misunderstanding is.
Here’s one: room-temp LED lighting. According to their understanding, I should not be able to see an light from an LED until the LED reaches 37 C because otherwise my eye would be absorbing a photon from a source that is colder than the eye, which is purportedly in violation to the 2nd law. So I’d see nothing from the LED until it was raised to ~37 C and the BAM, I can see the full intensity.
-Scott
In your dreams.
In any event I have not been advancing Slayer arguments.
It’s quite possible you (specifically) haven’t been. I’d have to go through all your comments specifically to see. Many of the objections I’ve seen to how the IR absorption affects the conduction from beneath are just Slayer arguments modified from radiation to conduction.
-Scott
Looks like Kristian is using Slayer approaches to argue with you Stephen (see July 4, 2015 at 12:18 PM). He says there is no IR-absorbed energy from below to fuel the evaporation b/c the IR source is from a lower temp. I’d have to go through all the comments on this thread to see who’s specifically pushing the Slayer approach vs who isn’t, but I wouldn’t be surprised if almost all of the people I’ve argued with are using the Slayer misunderstanding as their reasoning.
-Scott
See my comment below @ 12:49 pm
–Looks like Kristian is using Slayer approaches to argue with you Stephen (see July 4, 2015 at 12:18 PM). He says there is no IR-absorbed energy from below to fuel the evaporation b/c the IR source is from a lower temp. I’d have to go through all the comments on this thread to see who’s specifically pushing the Slayer approach vs who isn’t, but I wouldn’t be surprised if almost all of the people I’ve argued with are using the Slayer misunderstanding as their reasoning.–
I seems there becomes a lot of slayers if you divide the issue between people who believe that back radiation adds joules of heat to the surface vs prevents joules of heat from leaving the surface.
My view of the slayers view is that that greenhouse gases have near zero effect. So if atmosphere had zero CO2 a slayer would say this would have no effect upon global temperature, or doubling CO2 would have no effect upon temperature.
Personally I think it’s possible the doubling CO2 could have no effect upon temperature, but also seems possible that doubling could increase global temperature by as much as 1 C.
But I don’t think that the temperature increase over last century or so, supports the idea that doubling CO2 causes as much a 1 C increase in global temperatures. And questionable in terms of supporting it causing .5 C increase or more per doubling of CO2.
It seems to me the cause of there not being evidence of .5 C increase of in global temperature from doubling CO2 levels is because, the measurements have not been accurate enough and we don’t have good grasp of “natural variables”.
So I think it possible that doubling CO2 might cause as much as 1 C increase, but it could be less.
I think there is no evidence that CO2 does anything to lower global temperature. Or any kind of evidence that CO2 is in anyway involved with cooling the atmosphere.
So, unlike UHI effects which do [without any question] warm the surface in terms of increasing average temperature. And UHI effect unrelated to greenhouse gases.
But also UHI effect also can’t measured in terms of having a global warming effect.
So obviously warms in terms of local/regional effect, and “in theory” should have some small global warming effect, but it’s so small, that likewise it’s immeasurable in terms of on the global scale. And there are possible other human causes of increasing global average temperature, other than urban changes or CO2 emission.
I am skeptical of slayer’s view that CO2 can not cause global warming. Though if view instead is that CO2 gas increase can not cause a significant amount of warming [which could mean, not more than 1 C from a doubling of CO2] then I would tend to agree with that.
I also think that a doubling of CO2 is not likely any time soon- within 50 years. And unless there is some significant improvement in climate science [not a given] were to get a doubling of CO2 at such time we could also be unable to measure the increase in temperatures.
Or it could continue to be so insignificant that at best it’s hard to measure, and continuing to have no significant effect upon people or life on this planet.
Notwithstanding the foolish scare stories in the news, the only real and measurable effect of CO2 has been greening of the world’s deserts, and an measurable increase in crop production.
Or increase in CO2, has been one of reasons that “overpopulation” has not had a food shortages.
Though a far more significant cause of the abundance of food supply is related to improvement in technology.
I would say this interesting and somewhat related:
http://onlinelibrary.wiley.com/doi/10.1002/2015JA021220/abstract
And linked from:
http://judithcurry.com/2015/07/04/week-in-review-science-edition-12/#more-19200
Scott says, July 4, 2015 at 12:41 PM:
“Looks like Kristian is using Slayer approaches to argue with you Stephen (see July 4, 2015 at 12:18 PM). He says there is no IR-absorbed energy from below to fuel the evaporation b/c the IR source is from a lower temp. I’d have to go through all the comments on this thread to see who’s specifically pushing the Slayer approach vs who isn’t, but I wouldn’t be surprised if almost all of the people I’ve argued with are using the Slayer misunderstanding as their reasoning.”
Says the man who employs troll tactics to avoid real discussion, by not responding directly, but rather moving to somewhere else on the same thread and then outright lie (“He says there is no IR-absorbed energy from below”) to misrepresent the opponent and then round it off with a nice dose of dismissive ad hominem (“Slayer approaches”).
So you, Scott, believes that IR from a cool atmosphere to a warmer surface will actually separately heat (directly raise the temperature of) the surface, so that this added energy doing the extra heating needs to be evaporated away? And by pointing out this stupidity, I’m all of a sudden a Slayer?
The ocean surface isn’t warmed by LW from the atmosphere, Scott. A child would understand this. It cools by LW to the atmosphere. Just as it cools by evaporation and conduction to the atmosphere.
How hard is this to grasp? It is the most basic thing ever!
Says the man who employs troll tactics to avoid real discussion, by not responding directly, but rather moving to somewhere else on the same thread and then outright lie (“He says there is no IR-absorbed energy from below”) to misrepresent the opponent and then round it off with a nice dose of dismissive ad hominem (“Slayer approaches”).
Well that’s a pretty strong comment. Where to begin… How about with the claim of an ad hom. I said you use “Slayer approaches”. How is that an attack against your person? I described your approach with what I believed to be an accurate description. I said nothing about you, just your approaches. This is no different than you saying I “employs troll tactics”. And I find either to be far better than your “warped reality” statements that you’ve used twice now. Pot, meet kettle…
As far as the “troll tactics”, I didn’t “not respond correctly”, I just flat-out didn’t respond at all. My comment about you was to show Stephen that Slayer arguments were being advanced by some commenters. But if that is a “troll tactic”, then maybe you should point it out to mpainter who routinely did it on this thread even when in direct response. Oh wait, his conclusions agree with yours…
The “outright lie” about no IR-absorbed energy. I’ll just quote the relevant 12:18 pm passage:
Absorbance of a photon is an increase in energy by definition. So if there is not added energy, then either you’re not absorbing IR or you’re violating the First Law of Thermodynamics by destroying the energy contained in that photon. Perhaps I was wrong and you think the process does happen and no energy gain happens.
Yes and no. I believe that IR is absorbed by the cooler water surface. At the point of absorption, the local temperature has to increase transiently b/c by definition that absorbance event causes an increase in energy. I don’t necessarily “believe that this added energy needs to be evaporated away”, that’s Stephen’s thing. I’m saying that the added energy decreases the temperature gradient, thus decreasing the loss of energy by the bulk ocean beneath. Just look at the equivalent electrical circuit I described earlier. Actually, considering that circuit, the evaporation rate would be expected to increase somewhat.
No. Is English not your native language? Saying you use a “Slayer approach” is not the same thing as saying you’re a Slayer. Probability that you’re a Slayer certainly is higher if you’re promoting those arguments though.
I agree that the ocean cools by LWIR. But it cools more slowly when another IR emitter is above it, no matter its temperature. Otherwise, everything would radiatively cool at the same rate as if it were surrounded by absolute zero. Because it cools more slowly, it’s temperature is higher than it would be otherwise. I don’t care if people call this “warming” or “less cooling”, the result is the same–water temperature will slowly increase over time relative to what it would be otherwise.
-Scott
Scott says, July 5, 2015 at 9:36 PM:
“Where to begin… How about with the claim of an ad hom. I said you use “Slayer approaches”. How is that an attack against your person? I described your approach with what I believed to be an accurate description. I said nothing about you, just your approaches.”
The “dismissive ad hom” lies implicit in the term “Slayer”, which is meant to connote “nutcase”, “crackpot” and/or “anti-scientific” ideas. It is a commonly used warmist rhetorical tool to dismiss any argument against one’s own, employed in the cases when one doesn’t want to (or can’t) go through the process of piecing together an actual, proper counterargument. You do not actually address the argument itself, what is really being said. You just dismiss it offhand. You know this, Scott. And still you used the term to define my “approach”.
If someone proclaimed that you, whatever you actually wrote or said, promoted fascist or racist ideas, would you be fine with this? After all, they didn’t call you specifically a fascist or a racist.
Please leave out the name-calling, Scott. If it’s meant for me personally or my ideas or my “approach” to a particular issue, just drop it. It leads to no good. It pisses off your opponent. Which is just what a troll tactic is meant to accomplish … And I’m sure you know this.
“I just flat-out didn’t respond at all. My comment about you was to show Stephen that Slayer arguments were being advanced by some commenters.”
See above.
“The “outright lie” about no IR-absorbed energy. I’ll just quote the relevant 12:18 pm passage:
There is no energy for evaporation to “use up”, Stephen. Because there is no positive transfer of energy by LW from the atmosphere to the surface, unless it comes down as heat (which can only ever move spontaneously from warmer to cooler). The LW transfer is UP and only up …”
I know what I said, Scott. No need to remind me. You clearly stated that “He says there is no IR-absorbed energy from below to fuel the evaporation b/c the IR source is from a lower temp.”
I said no such thing. I didn’t mention IR absorption at all. And I distinctly referred to DWLWIR (which is from ABOVE, from the cool atmosphere to the warm surface): “… there is no positive transfer of energy by LW from the atmosphere to the surface, unless it comes down as heat (which can only ever move spontaneously from warmer to cooler). The LW transfer is UP and only up, from sfc to atm.”
Just admit that you lied, Scott, and be done with it. Stop with the hole-digging.
“I believe that IR is absorbed by the cooler water surface. At the point of absorption, the local temperature has to increase transiently b/c by definition that absorbance event causes an increase in energy.”
Yeah, this is how I see that you don’t really understand what is actually going on in a heat transfer process, what a heat transfer really is. For instance, do you know what the “thermodynamic limit” is?
“I’m saying that the added energy decreases the temperature gradient, thus decreasing the loss of energy by the bulk ocean beneath.”
Exactly. You’re saying that there’s an extra transfer of energy to the surface next to the solar input, only this second, separate transfer is from a source that’s actually cooler than the surface. But it doesn’t matter, it still “heats” the surface some more.
See, this is what I’m talking about. Your view of how the world works is a warped version of reality.
“Saying you use a “Slayer approach” is not the same thing as saying you’re a Slayer. Probability that you’re a Slayer certainly is higher if you’re promoting those arguments though.”
See above.
“I agree that the ocean cools by LWIR. But it cools more slowly when another IR emitter is above it, no matter its temperature.”
Yes, Scott. THEN WHY DO YOU SAY IT HEATS/WARMS THE SURFACE!? It doesn’t. It never does. The SUN heats/warms the surface. The LW cools it. Only more or less.
Why is this so hard?
“Otherwise, everything would radiatively cool at the same rate as if it were surrounded by absolute zero. Because it cools more slowly, it’s temperature is higher than it would be otherwise. I don’t care if people call this “warming” or “less cooling”, the result is the same–water temperature will slowly increase over time relative to what it would be otherwise.”
You don’t care. Of course you don’t. Well, you should care.
Please read my first two replies (to Roy) on this thread:
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-193991
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194002
Kristian says:
July 7, 2015 at 12:56 PM
Funny how you know what I know. That said, I see you’ve not bothered to quote where I called out your hypocrisy. “Warmist”, “troll tactic”, “warped version of reality”…all in this post alone. Hypocrite.
If I was promoting fascist/racist ideas, I should be called on it. It’s just stating a fact.
Again you presume to know what I know. And again you show your hypocrisy using the term “troll tactic”.
See above.
I won’t admit to something I didn’t do. I still wonder if you’re a native English speaker given your lack of English comprehension. Look at your bolded words. “IR-absorbed” is a compound adjective (clearly, since it is hyphenated) describing the noun “energy”, and the location of the energy is given by the prepositional phrase “from below”. You specifically said “There is no energy for evaporation to ‘use up’, Stephen. So no lie at all, just your lack of understanding, because you clearly say there is no energy there to use up, so by definition there is no IR-absorbed energy. If you’d read my long post from July 3, 2015 at 1:16 AM, which is what the discussion was revolving around, then this would be abundantly clear—the energy is from IR absorbance below the Knudsen layer that can go to fuel the evaporation. Sounds to me like you’re just jumping into an argument and saying someone’s lying because you didn’t know the context and can’t properly interpret English. That should probably be called a “troll tactic”, but that’d be an ad hom and only you can do that.
Any non-absolute-zero body will have radiance that can be absorbed by a warmer body. That absorption event increases the energy of the absorbing body by definition. Is the result a colder body making the warmer body have a higher temperature than it started with? Of course not. Is the result a colder body making the warmer body have a higher temperature than it would in the absence of the colder body? Certainly possible.
Also, I’ve read your first two responses to Roy. Nothing special. If you agree that downwelling LWIR radiation from any body will cause the ocean to be warmer than it would be in the absence of said downwelling radiation, I have no argument with you.
-Scott
–Yeah, this is how I see that you don’t really understand what is actually going on in a heat transfer process, what a heat transfer really is. For instance, do you know what the “thermodynamic limit” is?–
Made me look.
Of course wiki has something to say about.
But what I found amusing, was this bit:
**Cases where there is no thermodynamic limit**
….
…
“Any system that is not H-stable; this case is also called catastrophic.”
**Therefore** it is “obvious” that “global warming science” isn’t a case where there is a thermodynamic limit.
Anywho, I now have a vague idea if what the term means, though the trip wasn’t too helpful- as it’s mostly common sense.
So then, I listened to:
https://www.youtube.com/watch?v=WMVNVhOKwek
[part 22 thru 24] Which was less amusing.
But I guess I do see why it’s could be relevant question, now. And though it’s interesting thing to consider, not too sure it’s particularly relevant. Or I think there are simpler things to consider.
And what the general topic makes me wonder about is what is thermodynamic limit of Earth in terms of it’s temperature.
And what is causing the cooling of Earth.
What caused the Little Ice age. Change in solar and volcanic activity doesn’t seem like whole picture. Or it reminded me of this question I have had.
And as for humor of wiki, the alarmist are convinced that they isn’t thermodynamic limit of Earth.
Of course there could be two types thermodynamic limit- one could idealized planet at earth distance from the Sun- idealized in terms confined to 1 atm or say less than 3 atm
masses, but everything else is maximized to reach highest average temperature [or maybe highest tropical temperature]. And then thermodynamic limit with about the earth will have and the Sun we have. Or it would appear that Holocene Maximum was much warmer than today- one can assume that this was roughly the Earth we have and the sun we have today.
But find other limit more interesting, so assume one wants Earth with higher average temperature, how and what could be it’s highest average temperature- if given say 10,000 years to warm up. Of course another less interesting question is how cold could earth be made- assuming you liked skiing.
But it seems one thing which warm earth would be to flatten all the mountains. Say makes all continents with elevation of 200 meters [if pushed it all into the ocean it would not increase land mass area by much]. Now many could think that it would make it warmer by removing the white snow on the mountains- I don’t think that is main issue of why it would make it warmer.
Anyhow, relevant to your point, I assume you talking about adding to heated bodies to together and this takes time reach an equilibrium? And/or heated gas isn’t at equilibrium as just been heated by hot heating elements, or maybe also that heat gun is heating water vapor in the air.
Is that generally the point?
Scott says, July 7, 2015 at 9:22 PM:
“Funny how you know what I know. That said, I see you’ve not bothered to quote where I called out your hypocrisy. “Warmist”, “troll tactic”, “warped version of reality” … all in this post alone. Hypocrite.”
Scott. I’ve explained exactly WHY I call “troll tactics” on your part. I’ve explained exactly WHY your version of reality is clearly “warped”. And, frankly, I can’t see how referring to you as a “warmist” would qualify as an ad hom. You do think more CO2 in the atmosphere makes the world warmer. That DWLWIR from the cool atmosphere makes the already warmer surface even warmer. Or don’t you? It’s not like I call you “anti-science” for believing so. If you called me “natural warmist”, “solarist”, “oceanist” or “atmospheric massist”, what would it matter? It would point to my scientific position on this issue. I can see why calling people “alarmists” or “warmunista” would be objectionable. A bit like calling someone a “denier”. But “warmist”?
If you had cared to explain exactly why and how my “approach” is a “Slayer approach”, then we could’ve discussed it. But you didn’t respond to me directly. You didn’t “say it to my face”. You went somewhere else (a pure coincidence I even saw it) and called my argumentation a “Slayer approach”, without qualification. (Well, you purported to give an explanation by directly misrepresenting what I had actually said; foul play!) That’s more like slander.
“If I was promoting fascist/racist ideas, I should be called on it. It’s just stating a fact.”
Of course. IF you did. If you didn’t, someone could still claim that you did. And I’m sure you would find that totally fine also.
“Again you presume to know what I know. And again you show your hypocrisy using the term “troll tactic”.”
See above. When you use troll tactics (I explained which ones they were) and don’t actually address directly any argument of mine, I call it out. That’s not ad hom, Scott. Ad hom is used to dismiss an opponent’s argument by calling him (or his argument) names instead of addressing the actual argument. You did that. I called it out.
“I won’t admit to something I didn’t do.”
Hahaha! Big surprise!
“You specifically said “There is no energy for evaporation to ‘use up’, Stephen. So no lie at all, just your lack of understanding, because you clearly say there is no energy there to use up, so by definition there is no IR-absorbed energy.”
*Facepalm*
It is common practice, when responding to another person’s comment, to make sure you know what you are in fact responding to. You clearly didn’t, Scott. I was replying to Stephen discussing how the alleged added energy to the surface from downward IR from the atmosphere was “used up” by evaporation, preventing it from having any effect on the subskin.
OF COURSE there is IR absorbed by the ocean surface, warming it. But it’s from the SUN, not from the atmosphere. I wasn’t talking about any IR from below, I wasn’t talking about any IR being absorbed. I was talking about the conceptual DWLWIR ‘flux’ not having to be “used up” by evaporation at all, which seems to be Stephen’s contention.
I responded to STEPHEN’S comment specifically, Scott. Not to any of yours.
“If you’d read my long post from July 3, 2015 at 1:16 AM, which is what the discussion was revolving around, then this would be abundantly clear—the energy is from IR absorbance below the Knudsen layer that can go to fuel the evaporation.”
What “energy from IR-absorbance”!? The only IR flux added to the ocean, warming it, comes from the Sun, Scott. There is no energy added to the surface by way of LW from the atmosphere. The ocean loses energy via LW to the atmosphere. Which was my point. And I wanted to point it out to Stephen, not to you; you appear to be a lost case anyway …
“Sounds to me like you’re just jumping into an argument and saying someone’s lying because you didn’t know the context and can’t properly interpret English. That should probably be called a “troll tactic”, but that’d be an ad hom and only you can do that.”
I’m not “jumping into” anything, Scott. I don’t care about your ramblings. I simply wanted to point out to Stephen that there is no IR energy from the atmosphere to the ocean surface for evaporation to “use up”. Because the surface loses energy by LW to the atmosphere, it doesn’t gain it. And you felt the need to call that a “Slayer approach”. Good for you. But then you should also expect me to maybe call you out on it. And when you also choose to blatantly misrepresent my reply to Stephen, to sort of justify your ad hom, then it also shouldn’t really come as a surprise that I might object, should it?
“Any non-absolute-zero body will have radiance that can be absorbed by a warmer body. That absorption event increases the energy of the absorbing body by definition.”
No, this my point. There is no thermodynamic (macroscopic) flux of energy transferred from a cold body to a warmer one. There is always the potential, but the potential is never realised until the cold body faces an even colder body, or colder surroundings. The only thing we ever observe is the actual transfer of energy, which in a heat transfer is always unidirectional and always (spontaneously) from hot to cold.
This is why I asked you about your knowledge of the “thermodynamic limit”. Do you understand the difference between conceptually looking at individual ‘photons’ and looking at probabilistically averaged macroscopic radiant fluxes?
Kristian says:
July 9, 2015 at 2:45 AM
Kristian, I’ve explained exactly WHY I said you use “Slayer arguments”. Because you do! You said ”There is no energy for evaporation to ‘use up’, Stephen”. The only way that is true is if there is no IR absorption by the water from the atmosphere. Your argument is that there is no DWLWIR from the atmosphere, which is the same argument as used by Doug Cotton, et al. Associate with them or not, I don’t care, it’s the same argument. Basically what I’ve quoted up above can be summed up as “I can call people names but others can’t call me names”.
Wow you’re a hyprocrite. Worse than that, you’re claiming I’m doing what I’m not when you’re the one actually doing it. Go back and look at Stephen’s comment and trace it back to its roots—it all stemmed from my explanation of the Knudsen layer. Your lack of understanding of this is clear from the bolded text in the above quote. No one is talking about IR from below!
Yes, and I responded to Stephen’s comment specifically too. Yet I get blasted for it, what a hypocrite.
Net energy change equaling loss does not mean that the atmosphere’s IR photons are not changing the energy state of the water.
You’re saying there isn’t downwelling IR from the atmosphere because of temperature differences. That’s a Slayer argument, like it or not. I’m glad you “called me out on it”, because it’s the truth. As for “blatantly misrepresent”, I stated that you said there is “no IR-absorbed energy” (and if you trace the context of the thread it’s clear we’re talking about from directly above) and right here you say “there is no IR energy from the atmosphere”. See above for my explanation on net vs individual events.
Okay, I want you to clarify this. Are the photons from the colder body never being radiated toward the hotter body? Or are the photons sent that way and just not absorbed?
Your understanding of the thermodynamic limit is flawed. All it is the observed average result, which is analogous to the central limit theorem in statistics. All of the underlying occurrences are still happening.
Have a great day,
-Scott
Stephen/Kristian:
“The point at issue boils down to whether DWIR increases or decreases the thermal gradient between the top of
the ocean skin and the water below the base of the skin.”
It does, as I have shown in the above links:
Give me YOUR experimental proof that it doesn’t.
LH ratio has nothing to do with it.
It is NOT a 100% efficient process in conversion of energy.
You cannot have a 100% efficient conversion of one form of energy Electromagnetic energy taken up to evaporate water.
The molecules that are next to the evaporated ones MUST be excited some amount before they are left behind.
That is added heat. The Hydrogen bond MUST cause the molecules left behind to be agitated. The evaporated ones CANNOT cleanly convert the impacted IR energy to break the bond whilst unaffecting adjacent molecules. Vis some molecules receive the required amount of energy to break bonds with their adjoining molecules. Some get more
than they need and some not enough. Either way there are unevaporated molecules that have been heated.
IOW: It is not a quantum process whereby magically the IR photons just happen to hook up with individual H2O molecules and deliver the exact amount of energy and no more and no less than needed such there is no impact on on others nearby.The evaporated molecules are not an isolated system.
From: https://www.bluffton.edu/~bergerd/NSC_111/thermo6.html
“It is not possible to construct an engine which
does nothing but convert heat into useful work.
……….or more generally,
It is not possible to convert one form of
energy to another with 100% efficiency.
The only exception is that 100% of any useful energy can be dissipated as waste heat.”
Stop hand-waving please.
I’m waiting for experimental/real world proof of your assertion.
I have provided several…. naturally becasue scince knows it to be the case.
Oh, and Kristian … “What is it with you people!? How hard can this be? In what kind of warped reality are you all living!?”
Now, as you have with the above resorted to suggesting that I “am living in a warped reality” – I take that as ad-hominem and as such turn it around onto yourself with the self evident truth of the real “reality” that you cannot see …. you may (eventually) find that empirical science is the “reality” I, Scott and the rest of the non-ideologically challenged world inhabits …. and NOT your mind. No, it is your warped reality that has to bend the world to fit and not the experts dating back to Fourier/Plank/Stefan/Boltzmann/Maxwell
etc. along with rconfirming experimental and real world data.
The Hubris of that statement is breathtaking and QED why you cannot see that, actually science is correct and not yourself.
Hand-waving and doing so loudly and often is not science my friends.
Shut up or put up ….. and that means real world experimental data – as Roy has done and I have linked to.
Toneb
I gave you real world empirical data and you ignored it and I have no doubt that you will continue to ignore it
Because
It refutes
Your position
___utterly___
Hand-waving. You post results from said experiment.
-utterly-
Toneb says “You post results from experiment” “utterly”
###
Results: the water doesn’t warm, and I am astonished that this obvious and elementary fact of radiative physics cannot be grasped by the AGW enthusiasts. Utterly.
Do it yourself experiment:
Can be done with a hair dryer and a bucket of water, or any suitable container. The cooler used by Roy in his experiment would be ideal.
More hand-waving. Some science please….
See my comment above at 1:23 pm
Toneb,
I ran the experiment. My experiment is described here:
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194595
Yes, the water warms without issue. Makes you wonder why they haven’t been specific about their experimental details…
-Scott
Toneb says, July 4, 2015 at 12:43 PM:
“”The point at issue boils down to whether DWIR increases or decreases the thermal gradient between the top of the ocean skin and the water below the base of the skin.”
It does, as I have shown in the above links:”
No, it doesn’t. And you most certainly haven’t ‘shown’ it. If it’s Minnett’s interpretation of his data you’re referring to, then … no. If you’re interested, I’d be delighted to discuss Minnett’s claimed ‘findings’ with you.
This is what you apparently don’t understand, Toneb. The mathematically derived potential DWLWIR ‘flux’ from atmo to sfc doesn’t do anything in terms of thermodynamic effects (like a change in temp or induced evaporation). Because it is not itself a separate thermodynamic entity, a separate transfer of energy. It is but the one (and smaller) conceptual part of the actual (observed) LW transfer between the two systems: the radiant heat. The radiant heat between the sfc and the atmo flows UP and only up. From warmer to cooler. The hypothetical UWLWIR and DWLWIR ‘fluxes’ making up the integrated ‘net’ flux (the upward radiant heat), cannot and don’t generate independent (distinct) temperature effects. How did you ever come to believe such a ridiculous thing?
“No, it is your warped reality that has to bend the world to fit and not the experts dating back to Fourier/Plank/Stefan/Boltzmann/Maxwell
etc. along with rconfirming experimental and real world data.
The Hubris of that statement is breathtaking and QED why you cannot see that, actually science is correct and not yourself.”
*Yawn*
Please read my first two comments on this thread (in response to Roy directly). They provide some context:
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-193991
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194002
This is what I’m talking about. Until you’re able to show even a crude understanding of basic heat transfer theory and the laws of thermodynamics, I would say it’s rather inexpedient for you to be pontificating about the subject at hand. You’re only making a fool of yourself. You will remain wrapped up in utter confusion. In a warped reality where energy heats in all directions. The universe doesn’t work like that, Toneb. Go read a book …
“It is not possible to construct an engine which
does nothing but convert heat into useful work.”
Huh, so Sterling engines are impossible?
His points 1 to 7 were stated more carefully.
He must have meant to write:
“It is not possible to construct an engine which
does nothing but convert heat into useful work if the source and sink are at the same temperature.”
There, fixed. I have seen a Sterling run off of ice cubes, isn’t it then converting heat into useful work.
Careful what you search for on the web, much is stated incomplete or plain wrong, ven from .edu pages.
… that is, the ice cubes being the colder sink.
I could already feel someone pouncing all over that incomplete statement. 😉
Scott imagines the slay the slayers with his ever ready CO2 ray gun.
“LOOK! Another one!”
Zap zzz cracklepop
Congratulations on a successful rebuttal.
-Scott
Scott
The heat gun demonstration refutes your position utterly.
All you have is your CO2 laser which you brandish about while muttering darkly about “slayers”.
mpainter says:
July 4, 2015 at 1:29 PM
I’d be interested to know the conditions that this was experiment was performed and the quantitative results. Model of the heat gun, power settings, before/after temperatures, relative humidity, amount of water used, status of the control/blank, etc. My lab has enough equipment that I should be able to run this experiment myself this week when I resume work. I’d like to know how my experimental setup compares. Could you please provide this and/or send a link to a description of a similar experiment that you or someone else has performed and uploaded to the internet? If I run this experiment and find no temperature increase, I would need to modify my model.
No, I also have Roy’s experiment, Mike’s experiment, my case of 100% RH where evaporation is not occurring, my quantitative demonstration of how small a percentage of the absorbed photons directly cause evaporation (and the result that most of the IR absorbance happens far below the evaporation point and is therefore starts “ahead” of it in heat transfer), my electrical circuit equivalent model (which I really hope Stephen looks at b/c I think he may be the only person on your side willing to change his mind), my basic root(2Dt) calculations showing how heat transfer is orders of magnitude faster than evaporation, my willingness to learn, and my nice demeanor. I’m probably forgetting some other things too.
Have a nice day!
-Scott
mpainter,
I may be able to run a heat-gun experiment as early as tomorrow. So I’d really appreciate it if you could give me your detailed experimental setup and quantitative results so I can compare. My experimental design could be more optimized if I had some estimate of what the results will be.
Thanks and have a great day,
-Scott
Wiki, heat gun:
“A heat gun is a device used to emit a stream of hot air, usually at temperatures between 100 °C and 550 °C (200-1000 °F), with some hotter models running around 760 °C (1400 °F), which can be held by hand.”
https://en.wikipedia.org/wiki/Heat_gun
Or there’s wide variety and/or a gun can different levels of heat settings, but generally they like a hairdrier- though can obviously have much hotter air- since people don’t want to light their hair on fire.
Now, I would think what is desirable hotter air and less blowing of air.
Though also any surface which is hot- say over 100 C, will radiate Shortwave IR.
And Shortwave IR is transparent to water. So not sure why anyone would want to use heat gun, if want something which acts like Longwave IR.
I think the CO2 laser would be better. Though problem with laser is it’s beam is too narrow. If you could widen the laser beam to say 2″ diameter, I think that would be better in terms of seeing what Long wave IR does to water.
But it might be complicated to do this right, and why I would suggest having water over ice.
So submerge ice so about 1 inch below water and see what kind of hole the CO2 laser can make. And varying distance try say 1/2″ under water and say 2″ under water.
Other idea, is that CO2 laser should cut thru saran wrap, very quickly. And one could replace ice with saran wrap- that should be easier.
gbaikie,
Thank you for your input. Unfortunately, my CO2 laser results were rejected out of hand with no explanation and later led to mockery directed towards me. Upon pressing them, I basically got the answer that it was too intense and with the laser the heating was faster than the evaporation. However, their models didn’t include anything about intensity, so the laser experiment does require them to modify/improve their model, which they haven’t done. As for the heat transfer being faster than evaporation, I agree with that in the case of the laser but also say that is what is happening in all the LWIR-absorption cases…an idea which is also met with scorn.
Unfortunately, I’ve never seen the results of a heat-gun experiment where someone tried to heat water from above. When asked by myself and Toneb, they have yet to provide results or the experimental setup. I have some good ideas for the experiment. However, as you pointed out, I think SWIR could be used as an escape route even they they were the ones that suggested this experiment.
Have a great day,
-Scott
— Scott says:
July 6, 2015 at 7:42 AM
gbaikie,
Thank you for your input. Unfortunately, my CO2 laser results were rejected out of hand with no explanation and later led to mockery directed towards me. Upon pressing them, I basically got the answer that it was too intense and with the laser the heating was faster than the evaporation. However, their models didn’t include anything about intensity, so the laser experiment does require them to modify/improve their model, which they haven’t done.–
I don’t know which models you are referring to, but is the lack some kind of a variable for intensely, the major thing wrong with them?
–As for the heat transfer being faster than evaporation, I agree with that in the case of the laser but also say that is what is happening in all the LWIR-absorption cases…an idea which is also met with scorn.–
Evaporation is very powerful way to dissipate heat, water has a high specific heat and is poor conductor of heat.
The amount of heat conducted is greatly increase by the difference of heat, and basically everything other than metals is poor conductor of heat- though diamonds are one of the best conductors of heat.
Water can convect large amounts of Heat, but convection is related to gravity, and it’s matter of warmer liquid being lighter than cooler liquid- so hot water doesn’t convect downward and hot water doesn’t conduct much heat [up or down].
There is interesting video regarding lava and ice, I linked to before relating to another topic. So… here:
https://www.youtube.com/watch?v=yvSmPqqZB3Q
So, maybe not directly related but I think it’s interesting, nevertheless.
mpainter,
I have now run a heat-gun experiment. Water temperature increased considerably. Yes, water can be heated from above.
http://www.drroyspencer.com/2015/06/can-infrared-radiation-warm-a-water-body-part-ii/#comment-194595
-Scott
Toneb said:
“The molecules that are next to the evaporated ones MUST be excited some amount before they are left behind. ”
That can only be true if the phase change of liquid to gas takes up less energy than is available.
If molecule 1 receives 1 unit of IR and evaporates,taking up 5 units more in the process do please explain how adjoining molecules can nonetheless be MORE excited.
–If molecule 1 receives 1 unit of IR and evaporates,taking up 5 units more in the process do please explain how adjoining molecules can nonetheless be MORE excited.–
They are cooled- a lot.
JohnKl,
From somewhere way up above I saw your response to my post.
Here is an older article of direct measurements of radiation given off by grass (maybe more in the book but this should do).
The IR given off by grass give somewhere around the 390 watts/meter^2.
https://books.google.com/books?id=GhkGo9ZztAcC&pg=PA157&lpg=PA157&dq=direct+measurement+of+Earth%27s+Long+wave+IR&source=bl&ots=dM7KP4Ye-g&sig=5kMUin4LqL-5xau4H9Bqqcu-GaQ&hl=en&sa=X&ei=uRabVfaLEIHYoASQ-KaQAg&ved=0CB0Q6AEwADgU#v=onepage&q=direct%20measurement%20of%20Earth's%20Long%20wave%20IR&f=false
Ok, so I put together a quick experiment at work today trying to heat water from above with a heat gun. Here is the summary:
Experimental:
Variable-temperature heat gun used (model # available on request). Blowing speeds set to “low” with the only other option being “high”. The temperature was adjusted so that the steady-state temperature read by a mercury thermometer held ~1 cm in front of the nozel was 75-95 C (temperature varied considerably). This temperature was chosen because of gbaikie’s recommendation to keep it under 100 C.
Water container was a polystyrene petri dish with ~125 mm diameter. Thickness was ~1 mm, and depth was ~1 cm. Water mass used was 83.48 g (only measured after the experiment, so no estimate of evaporative losses is possible this time). Note that this gives an approximate depth of 1.7 mm, which is far deeper than the penetration depth of LWIR.
Heating was done by first warming up the heat gun by operating for 20 s at the desired settings while pointing away from the experimental setup, then the gun was pointed down at the water surface at a distance of ~1-3 cm for 60 s. The gun was moved some side to side, but care was taken to avoid getting close to the edge of the petri dish to avoid significant direct warming of the polystyrene.
Room temperature was ~21.1 C. Relative humidity was ~35% according to a sensor in an adjoining room ~20 m away. That room is typically warmer, so the experimental RH may have been a bit higher.
Results:
Initial water temperature: 21.3 C (very close to room temperature)
Final water temperature: 24.1 C
Net temperature change: +2.8 C
I also tried running the heat gun at maximum for a 60-s test and found the water temperature to change from 21 C to 31 C. Didn’t measure the water mass that time but it was similar.
Clearly, I think we can conclude that water can be heated from above. The evidence speaks for itself.
Sorry for the lack of experimental details. Some of them (such as model of heat gun) can be provided upon request b/c I can look them up. 1) My wife called and I had to leave work early. 2) The naysayers keep rejecting the experiments proving them wrong even when they request said experiments. I’m not going to put massive efforts into setting up a rigorously quantitative experiment until I hear their complaints on an experiment like this.
Have a good day!
-Scott
Try it outdoors with wind and an open sky on a water surface fully insulated from conduction from any container.
The room prevents convection and wind and allows humidity to increase which slows evaporation.
Warm air flowing from the gun would have warmed the sides of the petri dish in any event so the water acquired heat by conduction from the dish.
Maybe the gun also warmed the interior of the room?
Nothing but qualitative handwaving Stephen. I did exactly what you said I couldn’t do. And if these issues were critically important, they should have been specifically stated from the start.
Try running some of the numbers on your claimed mechanisms of failure and you’ll see that they are unlikely. I include numbers in my description for a reason. If you don’t run the numbers, then I will throughout the course of the day.
-Scott
Nor do I tnink that the setup of a small amount of water in a petri dish whacked with an unnaturally powerful IR flow of 75C adequately replicates conditions above an ocean surface.
Yet you suggested using a hair dryer on bath water?
Run some numbers Stephen. Alternatively, we can resume our that ended with my post at July 4, 2015 at 11:39 AM, where you were very close to understanding my argument.
Unfortunately, I’m about to leave for work so my communication will be spotty for most for most of the rest of the day.
-Scott
Stephen Wilde
I think Scott did answer the critics who believe it is impossible for IR to warm ocean water. It does not matter if it directly mimics the oceans (which would have degree of variablility of evaoporation rate based upon water temp, air temp, relative humididy of the air etc) he did prove that IR is capable of heating water from above. A significant point.
Only by doing it in a room thereby eliminating wind and convection and in a petrie dish (containing a tiny amount of water) which could not possibly have been prevented from receiving IR directly from the heat source and conducting it to the water.
He has manipulated every relevant parameter to get the result he desires.
Utterly pointless.
Stephen Wilde,
Yes experiments can be set up to achieve desired results. He may next take temperature of the petri dish to determine if your point is correct. The thing I like is he is actually testing things which is far superior to endless talk.
On another thread a person named Curt Wilson did elaborate tests on lamps to demonstrate backradiation can warm an object above the temperature it would be without such returning energy.
I may try my own experiments. I have IR heat lamp at work. I can weigh the water up before the test and after and monitor temperatures. Philosophy is the pure thought and reason (which is what most these threads are, even your own theory) but science is a branch that works best with experiments and tests. You are free to run your own tests to see what happens. Better to run your own and use your tests to prove Scott wrong and not just idle untested criticsm.
Stephen Wilde says:
July 7, 2015 at 10:27 AM
Holy cow, I somehow eliminated convection! Wow! Never mind that the heat gun has a fan in it that generates a significant wind that is easily visible on the surface of the water. Never mind that the relative humidity of that heated air was far lower than what is present at the ocean surface and therefore evaporation would be sped up. Never mind that I did exactly what was asked for within the precision specified. Never mind that I’ve specified the depth of the water (1.7 mm) and shown that the IR would be well over 99% absorbed at that depth of water.
Can you start being quantitative now?
-Scott
Okay, he’s had enough time to try to be quantitative, as I’ve been ask for quantitation for days. Regarding this:
What is the outside temperature of the petri dish that is required to conduct that much heat to the water? Do that math. Oh wait, you don’t seem to do that, so I’ll calculate it for you. For a 1-mm thick polystyrene at 125-mm diameter and 1.7-mm deep, the outside temperature of the petri dish would need to be 187 C warmer than the average water temp, so ~209 C. That’s well above the output temperature of the heat gun and also well above the melting temperature of the polystyrene used in that petri dish (~100 C as confirmed in my lab’s oven).
There is a reason that being quantitative is important.
-Scott
–A IR lamp does make much of the IR which is suppose to be
backradiation.–
Grr.
Should be:
A IR lamp does NOT make much of the IR which is suppose to be
backradiation.
And my post is in moderation, I suppose because of all the friggin links [perhaps to those ads and/or just too long of a post- or something]
In terms of suggestions regarding making something which could be like backradition
[[From earlier post, we know saran warm absorbs about 30% or more long wave IR [8–15 µm].
And if saran absorbs then it also emit this IR.
Or use something else which also absorbs a higher percentage of the 8–15 µm ]]
Could wrap one side of some panel with saran wrap. Make panel somewhere in range of 1 or 2 foot square. If on smaller size, have closer to water, if larger one can be at a further distance.
The flatter and more smoothness of a surface will cause more “directed” light.
[One could also try some conclave shapes- maybe something with large golf ball dimples.]
Now with panel need means of warming it- you could use a IR lamp to do this, or any other incandescent light bulb which warms the back of it.
To be closer to backradiation something should be closer to 30 C rather a warmer temperature.
Or 30 C is 303 k. At 303 K a perfect blackbody radiates
470 watts per square meter.
Now a blackbody at 303 will radiate some shortwave IR.
And I assume that backradiation does not have any shortwave IR.
So this is a detail one can worry about or not.
But something like a largish size warmed surface area like this would radiate less shortwave, as compared to just an IR lamp.
Of course the panel area will inhibit evaporation losses and convection losses, and this is more details one can worry about or not.
“Holy cow, I somehow eliminated convection! Wow! Never mind that the heat gun has a fan in it that generates a significant wind that is easily visible on the surface of the water. Never mind that the relative humidity of that heated air was far lower than what is present at the ocean surface and therefore evaporation would be sped up. ”
It’s basically all convection- an unusual kind of convectional heating. Probably closest thing in nature would be thermal updraft into a rain cloud.
Or warm wind blowing over heated land surface, and passing over a cooler land surface or water. And warmer wind going across water is probably only time land areas could warm ocean surface [a very minor phenomenon in terms of “global warming”- mostly the ocean warms the land areas rather than land areas warming ocean areas]
As far comparing experiment with the the ocean, most ocean in tropics is 24 C. And “I also tried running the heat gun at maximum for a 60-s test and found the water temperature to change from 21 C to 31 C.”
I bet it would as work well in terms of warming to 35 C- which about the warmest ocean surface temperature, though I think in coastal regions, shallow seas, etc, I think open water can get as high as 40 C.
Anyhow the warm tropics is critical in terms of Earth’s average temperature. In three ways.
First the tropics is very large part of entire surface area of entire Earth. So if averaging the surface temperature, Earth average temperature is 15 C, because the tropics has much higher average temperature than 15 C. Or continental US is has average temperature of about 12 C. And if you include Alaska, the US is still around 12 C- and Alaska is no where near average temperature of 12 C. Or countries like Canada have average temperature of about 0 C. So for Canada, Russian, Europe their world is not 15 C- only when one includes the vast region of tropics does it make their world 15 C.
Second, the tropics warms the rest of the world- more Europe, than Canada or Russia- Canada and Russian are still warmed by the tropics.
Third, tropical ocean can not be warmed by much warmer. So the difference of tropics that raises the average global temperature can not be “improved” to allow a higher average global temperature. Or the tropics has never changed much over million of years in terms of it’s average temperature- what changed is amount the rest of world is warmed by tropics, plus average temperature of tropics might cool by 1 or 2 degrees. Or the cold period of ice ages or glacial periods, is mostly related to with the Temperate Zones being colder- and mostly the Northern Temperate zone which has the majority of world’s land area.
Or in distant past of +50 million years ago when you had average temperatures of 25 C [and no polar ice caps] the average temperature of tropics may have been 1 to 2 warmer than warmest it ever been in the current icebox climate, but big difference is temperate zone are more tropical in their temperatures- they never get to freezing even in winter [though poles would freeze during the winter- they would have to, but it’s seasonal [or the is sea ice is gone shortly after springtime or sometimes in winter it would not freeze- depending on the weather. Though in higher elevation it would be a different story]
I said:
“And I assume that backradiation does not have any shortwave IR.”
It should be noted that as far as I know, backradiation is not characterize in terms of it’s spectrum.
So all got is greenhouse gases emit it, and we know what they emit- which is long wave IR.
Were one to measure it, and it had SW, that SW would be going from the sun or some other source [moonlight? or other].
So with blue skies it could be scattered light [why sky is blue], it could also be diffused sunlight, the sky reflecting sunlight, the sky reflected a hot surface- or 60 to 70 C surface emits SW, which could be reflected by sky.
Or whatever. But if going to talk about 300 watts of radiation which is warming as opposed to stopping radiant energy from being emitted, one should have some decent characteristic measurement of it {what your frequency, Kenneth?}.
Scott,
Great job for doing an actual experiment! That is the best way to settle any debate.
Norman,
I ran an experiment pretty early in the thread with the CO2 laser. It answered the question in the title of the OP unambiguously. They rejected the experiment. Looks like they’re rejecting this one too.
-Scott
Air at 75K-95K (hot enough to scald skin) was blasted into a petrie dish containing a small puddle of water (less than 100 grams)
The hot air obviously flowed all over the rim of the dish which then conducted heat to the water in the dish.
Hardly surprising.
Stephen,
What temperature would the petri dish have to be to conduct that much heat to the water? I’ve shown my calculated value. How about you show yours to establish if it’s even a possibility?
-Scott
water in plastic coffee lid, 200 ml
11:59 and 45 sec: 120 F
12:06: and 45 sec: 102 F
12:10: 96 F ** 12:37: 75.5 F [at least .5 F -/-]
12:11: 95 F
12:12: 93 F ** 12:38: 75 F
12:13: 92 F
12:14: 90 F ** 12:39: 75 F
12:15: 89 F ** 12:40: 74.5 F
12:16: 88 F ** 12:41: 74 F
12:17: 87 F ** 12:42: 73.5 F
12:18: 85.5 F ** 12:43: 73.5 F
12:19: 85.5 F ** 12:44: 73.5 F
12:20: 84 F ** 12:45: 73 F
12:21: 83 F
12:22: 82.5 F
12:23: 81 F
12:24: 81 F
12:25: 80.5 F
Table wood near it 70.5
12:27: 78.5 F
12:28: 78.5 F
12:29: 78 F
12:30: 78 F
12:31: 77.5 F
12:32: 77 F
12:33: 76 F
12:34: 76 F
12:35: 75.5 F
12:36: 75.5
At around 12:46, measured water/mixed it, +175 ml and 74 F so maybe 20 ml or less
evaporated in about 45 mins
12:50: 73 F
12:51: 73.5 F
12:52: 73 F -note error in measurement and convecton increase may due to .5 increaee
12:54: 72 F – wood table: 71 F
12:55: 72.5 F
12:56: 72.5 F
12:57: 72 F
12:58: 72 F
12:59: 71.5 F
1:00: 71.5 F wood table 71 F
Leave it for 5 mins
Meanwhile using as baseline, going to put container of hot water [boiled though will
cool on top of this and check temperature in same fashion- hand held IR themometer- cheap
one.
1:04 pm: 71.5 to 72 F and wood table 71
105: 71.5 F
So near equilibrium.
1:07: 71 F and wood 71.5 F
1:08: 71 F and wood 71 F
1:10: 71 F and so is wood with -/+ .5 F [at least].
So added two red bricks to elevate hot water.
hot water going to use two ice cream 1 gal pail one inverted
and other with hot water.
And check to temperature before hot water pail added: 1:16 pm: 71 F
1:20: 71 F and wood is 72 F
1:21: 70.5 and wood is 72.5 F [not sure why- got two brick and inverted
pail which dried with paper towel. Maybe angle of taking measurement,
it pretty steep angle.
1:24: Same measurements
1:25: About same measurement
So put boiling water in pail with a lid above the inverted over the
coffee lid which has about 175 ml of water
1:30: water: 70.5 and wood table: 72 F
Hot water over it 1:35: watter: 72 F and wood: 73.5
1:37: 72 F and wood 73.5
1:38: same
1:39: same
1:40: 72 F water, 73 F table. Hot water is 172 F
1:42: Same, hot water has 125 F nearer bottom and 172 F near top
so terms radiant heat it’s the 125 F temperature is more relevant.
1:45: 72.5 water, 73.3 wood, notice hot water at bottom has wide difference
as low as 110 F- but this plastic of pail not water, as would need to remove lid
and would work anyhow. Top of lid is 145 F, top water is 165 F [measuring plastic
in all cases]
1:49: water 72 F and wood 73 F- up lid is 145 F, bottom part is now around 120 F.
Lid pretty constant sides and lower part part of hot water varies- probably due to convection
of air outside of pail
152: water 72 F table 72.5
1:55: water 72, table 73 F hot water pail lid: 134 F, and coldest measured temperature at
bottom part of hot water is is 102 F. Whereas top of this water is 151 F.
1:58: 71.5 and wood table: 72.5, hot water lid: 128 F and lowest low level water: 112 F
higher level: 148 F
2:00: 71, wood 72.5, lid: 122 F. bottom lowest: 108 F and highest hot water 148 F
Or Using bottled water.
And it does appear to warm the 175 ml, and going to pour it out and mix and measure it
now.
And seems little has evaporate and mixed it’s about still about 71 F.
Remove buckets and wait couple mins. And temperature of ceiling is currently 77 F
So left the two bricks, btw.
2:08: 71 F and wood is 72.5 F
2:10: 70.5 [also reading of 71 F and wood 72.5 F]
So not changing much.
So hot water above cooler water did not add a measurable amount heat nor measurable amount of evaporation.
Could lessen the distance, inverted pail was about +6 inches, could try 2″ distance. though that would interfere more with convection. It could made to be warmer- if merely mixed the hot water one should a higher temperature. Could also use more hot water- I used about 1/4 of gallon. So half gallon and more mixing [and somehow have a lid would give more time with higher temperature and one could get something actually measurable
Another thing about it, was buckets, had about 17 cm diameter bottom with a 20 cm diameter top. In terms of radiant energy it would be better if this reversed, having wider diameter bottom.
Oh, also rather the bother of mixing water, it would be simpler if used salt water, as this one should have less gradient of heat of the water [warmer at the bottom].
Or all water should be water as salty as sea water- since
it’s on topic of global climate/temperature.
I was thinking {LOL} that it would be nice if one could buy
mid ocean sea water. Coastal water is not like this water.
Of course problem is why would there be any market for it.
If it was health food type craze, it would be in the stores,
But don’t see how it could considered a health benefit- it seems more plausible than mineral water- and could be cheaper:).
You could say a 1000 miles from civilizations, or something.
Water 500 miles from Tahiti or something. I wonder what would be considered the most pristine sea water?
gbaike, you get my kudos on this thread. Don’t know if you have a string of degrees and letters by your name like Scott, matters not to me, but you have come up with cloest to the correct experiment needed.
Feel real sorry for Scott, string of degrees yet evidently, by all of his own words throughout this thread, knows nothing of scale, know nothing of equivalence, can’t consider rates of different parts of a system, possesses no physical intuition. This entire thread is going up for the students wanting to be and think like a good scientist for an example of what NOT to think like. Priceless. Hmm. CO2 lasers that can probably cut steel as his example of proof… oh my! And the heat gun, just as bad… conduction! He shouild have known.
The only improvements you could do making your experiment more solid is to double insulate as Dr. Spencer did in the photo at the top. You need a very large, like 3’x3′, flat and shallow copper tray a few inches deep to be supported high enough over the test salt water (much like the canopy or shield shown at the top of the thread) so absolutely zero conduction can pass downward from the hot pan above and the test water below by conduction. Rule out any conduction but convection around the hot tray should do the trick. Use temp measurements in the air below to make sure that is true, no conduction down to the test water and that the room temperature stays constant. We are not testing conduction here, rule that differential out of consideration.
After the test water has equalized to the stable ambient room temperature start the experiment and keep the hot water in the pan above at, say 80°C, radiating IR downward to the test water below. Constantly remove some hot water if it falls below 80 and adding more boiling water to raise the temperature back up above 80°C just as much as it fell below to average to 80°C. Now measure the increase in temperature of the test water below that is at room temperature at time intervals that is supposedly being warmed by that infrared. Bet it will also show zero change. If so, proof made, if IR from something even hotter than the air above the surface in our atmosphere cannot literally raise the temperature then air colder than the water surface also cannot be made to cool more slowly. I think that is QED if performed and true.
Dr. Spencer has all of the setup already but the pan and boiling water and moving it into a room that can be kept the same temperature. Maybe implore him to run it when he gets some spare time. May not be what he expects but that is science, right? If it does warm you have the distance, areas and temperatures to calculate an very approximate rate.
I was testing a variant.
Used stainless with copper bottom frying which is at least 2″
wider the bucket. And used the same bucket to store more thermal heat and kept lid on it and and 1/2 full- 1/2 gallon of hot water. And put boiling water in pail, but also add boiling water into frying pan. And remove water from frying pan cooled and add more boiling water into frying, but not change the pail with lid.
Also used sea water salt for coffee lid and in bucket of water.
I noticed that sea water didn’t help very much in terms of temperature gradient with hot water.
So bucket would do same thing- hot water on top and cooler water at bottom.
So had bucket in frying pan, and when add boiling water into frying, it would heat bucket of water [or mixing it].
Anyhow was fussy around it, just see how it worked.
And I could keep bottom of frying pan around 140 to 150 F.
And with this arrange, I could measured this surface.
And also put coffee lid on some foam plastic insulation about 1″ thick. And this allowed me to slide it under the frying pan bottom [which sits on same bricks as before.
Put the water in coffee lid was still too hot- 80 F. But thought practice/test sliding under pan.
It’s a bit harder to measure with the 2″ space.
But what seem to happened when fry pan was 150 F, is the wter in coffee lid would appear to go up couple degrees, in less than 1 min. It seems to be warming. Put then slide out, and temperature would return to 80 F. And repeated this oh a few times and it do the same thing.
So I figured it was not mixing from the movement.
But My plan wasn’t to do water at 80 F, I was just tested it while waiting for it to cool.
Then time passed and waited for the 80 F to cool- and added more boiling water to frying pan.
Coffee lid cooled to 78 F, and when bottom of frying was about 150 F, slide it back under. and when back and forth because of odd reading. It seemed to be about 5 F hotter when under frying pan, but immediately cooled when pulled out. And it took a long time to cool to 78, but was reading 82 F under pan, and then 78 not under the pan.
So plan was get the coffee lid to room temperature- about 74 F.
So didn’t really start the test.
My guess is it’s evaporating and my instrument is measuring that. Or just very thin layer of water warming, the movement mixes it.
Anyways unlike the other experiment, having bigger surface area and making far closer [2″ vs + 6″] and being able to measure the warmed surface directly and have around 150 F,
resulted in some result which seemed to be easily measurable.
gbaikie,
Thanks for trying several experiments. I have trouble comprehending exactly what you’re doing each time, but it appears that in that test you had no problem measuring a water temperature increase. Is that correct? If the water is cooling quickly it is likely not well mixed. Have you left the water underneath the pan for several minutes (5+)?
-Scott
–I have trouble comprehending exactly what you’re doing each time, but it appears that in that test you had no problem measuring a water temperature increase. Is that correct? —
The first attempt no. But when tried bringing heated surface
closer to the water, there was a measurable change.
–If the water is cooling quickly it is likely not well mixed. Have you left the water underneath the pan for several minutes (5+)?–
The second attempt, I posted what one could say were some initial results. In prior attempt, it was indoor and was daytime with what might be said is reasonable ventilation. The second attempt was at night though also indoor and less ventilation, though this difference probably not
significant.
The bigger variable was bringing heated surface
[around 120 F or more] closer the the water.
Also later that night I continue it, and increase the distance by a bit more then 1 inch [about 3 1/4″] it was about 2″.
And made some notes:
[[[
frying pan: 8 1/2″ [22.5 cm]
black plastic coffee lid: 6 1/2″ [16 cm] and holds +225 ml. Use 200 ml
[[Have 4 bricks rather than 2]] ***
1:07 am: coffee lid water 73 F
1:12: 71 F
1:20: 71 F put under bottom of pan at about 150 F –
top lid to bottom pan distance 3 1/4 inches
1:22: 75 F pan: 141 F
1:23: 45 F pans 138 F
1:24 remove: 72 F put back
1:25: 74 F pan 132 F, remove: [about 10 secs] 71.5 F then put back
1:26 73 F pan: 130.5 F
1:27: 72.5 F pan: 127 F, remove: first reading 71 [5 sec, then 70.5]
left out for min 69 F
1:30: 72 F pan: 122 F
1:31: 72.5 pan 120 F
Done. As temperature too low on pan
Leave out for 2 mins:, starts at 69.5 F, remains 69 F
Ceiling 67 F, table 68.5 F
Try later time
1:44: 70 F, pan still cool and at 111 F- put under it:
1.45: 72 F, then remove: 71 F [in about 10 seconds] return:
1:46: back to 72 F
Interesting.
Now done.
So still has this effect even when pan quite cool- not as much, oh wait still under
1:49: still 72 F and pan at 107 F.
And remove at 152- was 72 then remove and 71 F and wait a bit:
153: 70 F
154: 70 F
Ok, enough, do another time.
Note 60 C [140 F] is 333 K and is 637 watts per square meter
And 41.6 [107 F] is 314.8 K and is 556.8 watts per square meter.
Watts per square meter assuming perfect blackbody- and bottom of frying pan
is somewhat black in color:)
]]]
*** So changed the bricks to be two high vs 1 high- and had ends bricks facing rather than bricks lengthwise. A brick
being about 3 1/2 by 8″ and 2 1/2″ high.
Anyhow, to answer your question the last time, I did leave under for about 5 min {but not on purpose- I thought I finished, but left it under there and then “went, oh wait”.
It’s a bit of hassle to heat all water, and it was cooler at that point- but even at 110 to around 107 F it still was warming, and it still cooled immediately when slide if out from under the heated bottom of frying pan.
I tended to not leave under for long, because I knew the movement would mix the water. So if every minute it’s making it more “consistently mixed” by movement.
So there are many problems. One thing would like to do is increase the distance further, but it may need a larger heated surface then frying, or smaller diameter “coffee lid”.
So why closer “works” is I believe because getting more “directed radiant” energy.
Or everything heated does not emit directed light, instead
the emitted light goes in random direction.
Sunlight is directed light at Earth distance, because the portion of light from the sun to earth has to be coming from essential one direction. Or the outer edge of sun and middle of sun is a small angle- so sunlight is mostly parallel light [or directed light. And one make light go parallel with parabolic reflector. Or as wiki says:
“The parabolic reflector transforms an incoming plane wave traveling along the axis into a spherical wave converging toward the focus. Conversely, a spherical wave generated by a point source placed in the focus is reflected into a plane wave propagating as a collimated beam along the axis.”
Or continuing wiki, and says what I am talking about:
“Since the principles of reflection are reversible, parabolic reflectors can also be used to project energy of a source at its focus outward in a parallel beam, used in devices such as spotlights and car headlights.”
https://en.wikipedia.org/wiki/Parabolic_reflector
So heat/light radiate into spherical space, and if a surface, then it’s it’s a sphere cut in half- a hemisphere.
And most of energy emitted by say wall would not come directly towards you- or it have will radiate most energy at 35 degree or less. Or as we know, the tropics at 23 degree of the hemisphere is about 40% of area of the hemisphere. Or radiant energy from a wall is like the polar region of earth- about 5% of surface area. Assuming you are couple feet from the wall.
And in addition that effect, you also have the inverse square law, or if make directed light it is also diminished over distance.
Of course btw, lasers are somewhat different then “mere” directed light- but are related/dependent to the general idea.
Anyways back to the point, my problem with my experiment is lacking the ability to precisely measure. Though anyone will have this same problem, to a significant degree, but simply said hand held IR thermometer has limits in it’s usefulness.
And there are number of things we trying demonstrate via an experiment. And by increasing the distance, it’s one of limiting the conventional effect, and inhibtion of evaporation. So I would the above experiment is more relevant, but I would not think of it as conclusive of much.
I find it more “interesting” than proving anything
wayne says:
July 8, 2015 at 9:40 PM
It’s statements like these that make me suspect some of you may be paid by Greenpeace et al. to make skeptics look like idiots. Scale–who calculated the depth of the evaporation layer and showed it to be orders of magnitude less than the depth of the region of IR absorbance? Equivalence–who put together a quantitative equivalent electrical circuit to show how the system is working? Rates–who showed that the molecular diffusion and therefore heat transfer are orders of magnitude faster than evaporation? Not that it matters, considering the energy from the IR is absorbed below the evaporation layer, so it’s “ahead” of the evaporative-cooling heat transfer. Physical intuition–who is the only one left arguing with someone who thinks that thermal radiation doesn’t occur in all directions?
Wayne, please read the title of the original post and then tell me how a CO2 laser does not qualify. It’s possible it could be excluded from other discussions like we’re having here below, but when I’ve asked why, all I’ve gotten is qualitative handwaving. First it was “it’s different!” Then it was “intensity!”. Then it turned into “heats faster than evaporation!” Hey, guess what?…that last argument is my argument for why the temperature in the presence of IR is higher than when it is absent. When asked for quantitative arguments I got “5:1 ratio!” Yet, no explanation as to why the 5:1 doesn’t exist for the laser.
Two people literally asked for the heat-gun experiment. You claim “conduction!”. Sure, that could be an issue and should be raised. When it was raised as expected, I calculated the temperature the polystyrene would need to be in order to conduct enough heat to warm the water that amount–and it was > 200 C. That temperature is > 100 C hotter than my excitation temperature and also well above the melting temperature of the polystyrene used. If you calculate a different number, then show it…but don’t just handwave conduction when it’s been shown that it can’t account for it. The material type and all the dimensions are present for you to do the calculation. As for me “should have known”, guess what?…I did know! That’s one of the reasons I chose polystyrene–its melting temperature is low enough to put a hard upper limit on the amount of conduction occurring. Of course, since I used a heat source with a temperature lower than the melting point it really doesn’t matter.
So that’s the trick for non-quantitative people? “Absolutely zero conduction”. You tell him to “rule out conduction”, but I already have in my example. But it’s not “absolutely zero conduction” (which is impossible), so it’s not good enough. I guess that’s a way to reject every experiment out of hand, because “absolutely zero conduction” doesn’t exist.
-Scott
A shame that it is necessary to spend so much time and effort re-inventing the wheel by having to show from first principles stuff that was common knowledge in primary schools 50 years ago.
Well, 50 years ago, we did not have theory of plate tectonics-
though there was general ideas about, I knew an old prospector that occasional raved about the idea, and other ideas. And I believe generally it was pointed out how the continent looked like they fit together, but wasn’t a theory as much as it was a few ideas.
And didn’t know the big rocks fall from the sky- and we still have not completely grasp the idea. It’s not really part of climate science, for instance.
And of course 50 years ago, or 30 years ago, schools were not trying to scare the kiddies with global warming. What waste of time that must be.
But 40 year they had the pseudo science of Chariots of the Gods. And over population was something to worry about. And had the idea that socialism might work.
It was said that Nature abhors a vacuum: Wiki,
“In physics, horror vacui, or plenism, is commonly stated as “Nature abhors a vacuum.” It is a postulate attributed to Aristotle, who articulated a belief, later criticized by the atomism of Epicurus and Lucretius, that nature contains no vacuums because the denser surrounding material continuum would immediately fill the rarity of an incipient void.”
Of course it’s not Nature [the universe is a vast vacuum]
but it’s humans which can’t tolerate it.
And it explains why we still have this really stupid theory called, the Greenhouse Effect theory.
No one in the future is going to believe anyone thought this theory was taken seriously.
Need a new theory.
Have planet with 1 atm of Nitrogen and Oxygen. And it’s completely covered with one ocean [3000 meter deep].
And it’s earth distance from star like our Sun.
So it’s had ocean for billions of years and will continue
to have it for billion of year. And over the billions of
years the ocean has boiled a couple of time, from biggest impactors. But it’s been at least 1 billion years since something big enough hit it.
What’s it’s average temperature?
Stephen Wilde says:
July 9, 2015 at 3:35 AM
Exactly right Stephen. I’ve spent a ton of time doing calculations based on first principles that just keep getting ignored.
-Scott
Scott (or rather Doug),
Your contentions just need adjusting to recognise the adiabatic process of convective overturning. Once you accept that, your convoluted concept of molecular diffusion becomes unnecessary.
I saw nothing 50 years ago remotely comparable to your molecular diffusion ideas. Conduction and convection in a gravity field were quite sufficient when combined with the concept of hydrostatic balance (and still are).
Stephen,
Einstein’s “convoluted concept of molecular diffusion” isn’t even invoked in my equivalent electrical circuit. You were very close to understanding what was going on there when you admitted that the radiation was below the Knudsen layer. Once the radiation is absorbed below the Knudsen layer, how is it any different from other energy below?
-Scott
“Need a new theory.
Have planet with 1 atm of Nitrogen and Oxygen. And it’s completely covered with one ocean [3000 meter deep].
And it’s earth distance from star like our Sun.”
We just need the old theory that the entire greenhouse effect of 33K is a product of atmospheric mass held within a gravity field in hydrostatic balance.
For Earth with 1 bar pressure at the surface the average surface temperature will be around 288K and the radiative characteristics of the atmosphere make no difference except that the air circulation needs to adjust a bit to keep the surface temperature stable at 288K
That 1 bar surface pressure also sets the amount of energy needing to be absorbed by the oceans so that they evaporate enough to not disturb the hydrostatic equilibrium.
The water cycle makes it easier for convective overturning to keep things stable because it enables faster loss of energy to space from within the atmosphere so that convection need not work so hard to get energy back to the surface before radiation to space.
This was all implicit in the knowledge of 50 years ago but it was not known to the astrophysicists who moved in on climate science 20 to 30 years ago without knowing any meteorology which has always been a highly specialised discipline.
–We just need the old theory that the entire greenhouse effect of 33K is a product of atmospheric mass held within a gravity field in hydrostatic balance.–
I disagree.
In number of ways. First the 33 K is bogus.
So mass of atmosphere does not warm by 33 K.
Roughly speaking if one obsessed with the number 33 K,
then I would say ocean adds up to 15 K, mass of atmosphere up to 10 K
Clouds add about 5 K, radiant effects of GH gas about 5 K [both could be more]. And evaporation/latent heat about 3 K.
Of course, 5 + 5 + 3 could be roughly put into a box of what is called the greenhouse effect according to Greenhouse Effect theory {not specified or anything one might call scientific, but sort of assumed as “everyone knows” [but are clueless].
So clouds cause warming, Clouds are not gases, but droplets of water. But greenhouse theory is assumed to count clouds as radiant effect of “greenhouse gases”.
So the warming effect of clouds is very easy to measure- unlike radiant effect of gases in an atmosphere.
Yet you are say that clouds do not cause warming AND you using this 33 K number in the same breathe.
So idea of global water planet is to replace the mickey mouse concept of ideal blackbody which one adds atmosphere and clouds [and maybe snow] on top of.
So global water planet would have clouds and not snow.
So figuring out the clouds would be the hardest part. But that is not a new/different problem. Rather it “should” be a simpler problem.
So your idea is cloud don’t warm, which almost as bad as saying clouds are radiant gas, but warmist and everyone else recognizes that cloud are a warming effect. In fact some think clouds are 1/2 of the greenhouse effect- and have wrote a paper or two about it.
Both you and the religious believers appear to be ignoring
latent heat. And those studying Mars don’t ignore the latent heat of CO2 freezing out at the mars poles. And latent heat water also warms our poles- and does other warming besides this dinky area of the poles.
And many realize the ocean is warming Europe. Though you not counting the warming effect of the ocean [nor are the believers].
–That 1 bar surface pressure also sets the amount of energy needing to be absorbed by the oceans so that they evaporate enough to not disturb the hydrostatic equilibrium.–
Well, if had twice 1 bar or 1/2 1 bar, it’s not going to make much difference in this regard.
Or you can find 1/2 bars on Earth. If temperature at 15 C
at sea level, then 1/2 bar is at around 5000 meters:
http://usatoday30.usatoday.com/weather/wstdatmo.htm
And at 18,000 ft there not a huge difference in terms of evaporation.
Of course without any atmosphere [and Mars has some atmosphere
as far this is concerned]. Water evaporate at 100 K, or would evaporate a significant amount at 150 K. Or presence of water
in vacuum requires a surface temperature of below 150 K, even in sunlight.
But one could just have a planetary ball of just water, it makes it’s atmosphere- out of H20 gas, and so at say Earth distance sun and had enough water mass to make an Earth gravity. Then it’s temperature would not be 150 C or colder, instead it average temperature one be higher than Earth’s 15 C average temperature.
But it could not help having rocks in it, in our solar system, because there lots of rocks. There is massive amount of water, but more rocks [though far more hydrogen than rock, and far more Oxygen than rock, but oxygen in chemically bond to rocks and hydrogen].
–The water cycle makes it easier for convective overturning to keep things stable because it enables faster loss of energy to space from within the atmosphere so that convection need not work so hard to get energy back to the surface before radiation to space.
This was all implicit in the knowledge of 50 years ago but it was not known to the astrophysicists who moved in on climate science 20 to 30 years ago without knowing any meteorology which has always been a highly specialised discipline.–
Well, agree about astrophysicists thing a bit. But let’s put
the astrophysicists into a larger bin, and say everything to do with space exploration- everything other than earth.
Venus was a problem for the “astrophysicists and all them” of course we have the nut James Hansen- who failed to understand Venus.
But as for, water cycle makes …. faster loss to space.
I don’t agree.
I think one can blame it more on rocks than water.
For instance, we know eruption of volcanos has some cooling effect- that is more rock related than water related.
But more generally I think if increase in land area, tends to cool and increases of water area tends to warm. So from that basis one could blame rock more than water.
Or you choosing to blame water, for which rock is probably more to blame:)
Or water is active, it eats rock, and I can see how it might piss people off [particularly those that appreciate
the mostly quiet rocks].
But it seems there is consensus [not that this means much] that rock caused our icebox climate. Rock is why we have Antarctica at the south pole- which is regarded as main cause of Icebox Climate. As well as a lot of other activity of rock. Of course the huge stack of ice, is making it worse. And the ice is crushing the relatively passive minding it’s own business rock. But without the rocks getting in the way, earth would be about 10 C warmer in terms of average temperature.
Since most of the people here are either ignoring quantitative arguments or don’t understand them, I thought I’d dig up this old statement from a comment on Roy’s first post on this subject. There are no numbers, though it’d probably be worth running the numbers at some point. The statement of interest is the second paragraph, pasted below:
http://www.drroyspencer.com/2014/04/can-infrared-radiation-warm-a-water-body/#comment-111331
How is this observation explained?
-Scott
from above link:
“Of course LWIR can heat water. Put cold water in a thermos without a cover. It warms faster than with cover. The difference is water is absorbing IR from the air without the cover.”
LWIR is not air, it’s radiant energy [apparently].
So it’s not air radiating it’s molecules of air hitting at about 500 m/s per second. Or could fill a room with argon or helium [not gases which are greenhouse gas] and one gets the same result.
Plus if had thermos 1/2 full it will warm faster- so convection heat of air will warm the interior of thermos walls.
So have thermos with 2 liter capacity filled with 1 liter, as compared to 1 liter capacity with 1 liter.
So 1/2 full one will have in addition of conduction of heat of walls to water.
But in any case thermo without cover will not warm very quickly.
Or I put few ice cubes in thermos and shut lid, and ice last a few hours, no doubt without out cover it would melt much quicker. Similarly with hot liquid and full thermos
one does not get much loss of heat- it’s mostly evaporation losses which limited by small opening.
So one get a lot evaporation losses at +40 C. So comparing
a full thermos at say 30 C, with open cover vs close would much less cooling over time, as say 50 C water.
Roy
Of course your shield slows cooling of the water a little.
But your experiment does not emulate a planet’s troposphere wherein water vapor makes the equilibrium temperature gradient (aka lapse rate) less steep and thus makes the thermal profile lower at the surface end.
Moist regions thus do in fact exhibit lower mean maximum and minimum temperatures than do drier regions at similar latitude and altitude, as per my study in the Appendix of my peer-reviewed paper “Planetary Core and Surface Temperatures” linked from the “Evidence” page at http://climate-change-theory.com – which site has had nearly 10,000 visitors this year.
You still have a lot to learn about thermodynamics Roy.
The Sun’s radiation is nowhere near sufficient to raise the mean surface temperature to observed values in the first place. The only place you’ll learn about what really happens is the above website and linked papers.
As I said about two years ago, Roy, radiation into a planet’s surface is not the primary determinant of the mean temperature of that surface.
Roy, as I have explained in the above comment, we simply cannot explain a planet’s surface temperature with radiation calculations, and nor can we quantify the net transfer of thermal energy by non-radiative processes into or out of the surface. But the good news is that we don’t have to do so in order to determine what we would expect a planet’s surface temperature to be.
Unless we understand the actual process whereby a planet’s surface temperature is determined, we have no hope of understanding why, for example, increasing the concentration of water vapor actually leads to a lower surface temperature, as my study showed, not a temperature that the IPCC thinks is increased by about 30° due to back radiation from (mostly) water vapor.
Now, once you recognize that water vapor must be cooling the surface (because it makes the temperature gradient less steep) then of course you are left with the need to explain just why the mean surface temperature is about 33K hotter than the planet’s radiating temperature. And to do that you need to explain the possibility of other energy input that is not coming from radiation. That is what I have been the first in the world to do, based solidly on the Second Law of Thermodynamics.
Consider studying what I have written one day, because, when you understand it, it will blow your mind, just as it was a “Eurika” moment for me about three years ago.
http://climate-change-theory.com
Doug
What do you think:
**Is a mini ICE AGE on the way? Scientists warn the sun will ‘go to sleep’ in 2030 and could cause temperatures to plummet**
http://www.dailymail.co.uk/sciencetech/article-3156594/Is-mini-ICE-AGE-way-Scientists-warn-sun-sleep-2020-cause-temperatures-plummet.html
“‘Over the cycle, the waves fluctuate between the northern and southern hemispheres of the Sun. Combining both waves together and comparing to real data for the current solar cycle, we found that our predictions showed an accuracy of 97%,’ said Zharkova.”
I think it might cool a bit, but even if it does, it could recover and still end up around same temperature now or maybe warm as much as 1/2 C by 2100.
But we could also see all the gain of last century basically disappear, and sea level rise halt, CO2 level drop, and by 2040 China will be past it’s “peak coal” So human CO2 could be lower per year by 2040, say 10 or 20% lower than present. But I don’t people will worried about global warming by year 2040. And think possible alarmism
could end before 2020. Of course we get something else to be worried about- probably more demented than worrying about earth getting warm.
Please go to this comment.
Dr. Roy, Thanks for reporting on this interesting experiment, but I question the photo showing “…coolers, with one partially shielded with AL flashing painted with high-emissivity white paint.”
It seems to me, an ancient 91-year old radar engineer, that the cooler on the right has the radiation reflector supported above it about a couple of feet higher than the edge of the cooler to allow air circulation. The reflector is supported by four plastic(?) rods attached to the corners of the cooler, but it appears to sag from front to back in a sort of broad catenary(?) curve.
I wonder if this reflector, as intended, may be preventing the escape of heat radiation toward the sky and cold outer space, but also is allowing the entry of heat radiation from your surrounding earthylawn(? ) by being reflected downward into the cooler. This may be limited due to the reflectors curvature, but some lawn radiation will doubtless enter the cooler to be bounced around within it.
If you repeat this experiment, instead of allowing a catenary curve with your reflector, I would suggest a larger V-shaped reflector (sharply bent greater than 45 degrees at its center) to reduce the possible entry of heat radiation from the area surrounding that cooler.
Maybe my use of the word “broad catenary” is stretching the definition a bit horizontally — “catenary” refers to the drape of a chain, supported at each end, that is caused by the earth’s gravity; your reflector is hardly a chain. But the concept still applies somewhat, although maybe the curvature of your reflector is due to a retention of the shape of the roll of the AL flashing from which it was cut. But my suggestion is still valid. Next time use a sharp V shaped reflector, bent considerably greater than 45 degrees.
I would suggest using another styrofoam cooler, could keep same height or say 6″ lower.
Spray paint or put aluminum foil in the interior.
To Roy and others:
To be clear about this: if back radiation from carbon dioxide is related to “1.5°C climate sensitivity” then you are intrinsically agreeing with the IPCC, K-T and NASA implication in their energy diagrams that the flux of back radiation can be added to the solar flux and the total used in Stefan-Boltzmann calculations to get 15°C. We need 390W/m^2 to “explain” that mean temperature. So we have …
Solar radiation: 168W/m^2
Deduct non-radiative heat loss: – 102W/m^2
Sub-total: 66W/m^2
(So far so good)
Now add back radiation: 324W/m^2
Bingo: 390W/m^2
This is not just a claim that back radiation is slowing cooling, Roy. It’s an outright fraudulent hoax that politicians and the public are being fooled by that backradiation helps the Sun to raise the temperature of the surface each morning. What is really happening is explained with correct physics here Roy
Given that the principle means of cooling the earths surface is via convection and conduction (proved in experiments in England in 1905) Im confused as to why the talk is all about radiation which is not the main cooling mechanism until the upper atmosphere. Cloud cover reduces convection therefore retains heat. CO2 is less of a ‘greenhouse gas’ than Oxygen, which causes decomposition which releases heat (and CO2). Best we ban oxygen!
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