Home/BlogMy previous post explaining a simple experiment to demonstrate that a cool object can make a warm object warmer still led me to give the experiment a try.
The purpose is to demonstrate that, energetically, the atmosphere’s greenhouse effect can make the surface of the Earth warmer than if the greenhouse effect didn’t exist even though the atmosphere is colder than the Earth’s surface. There is no violation of the 2nd Law of Thermodynamics, which states that the net flow of heat must be from higher to lower temperature, which does not preclude cooler object from emitting IR radiation in the direction of warmer objects.
If the atmosphere didn’t exist, the Earth’s surface would lose IR radiation directly to the cold depths of outer space, which is essentially at absolute zero temperature, and emits no energy back to the Earth; but instead the atmosphere, in effect, blocks some of that radiation, and emits some of its own IR radiation back towards the surface.
The net effect is that the surface and lower atmosphere cannot cool as rapidly to deep space, raising its average temperature.
The experiment shown below does not prove that greenhouse gases in the atmosphere perform such a function, only that it is not a violation of the 2nd Law of Thermodynamics for a cooler object emitting infrared radiation to keep a warm object warmer that it would otherwise be if the cooler object was not present.
The reason I am posting this is not to convince the rabid disbelievers, who are probably beyond hope. It is to reduce their influence on others. I’ve discussed all kinds of evidence here, the most convincing is just using a handheld infrared thermometer pointed upward at a “cold” clear sky, then measuring a warmer temperature when point it obliquely at the sky. That shows a cold object (the sky) can warm a surface (the thermopile in the handheld thermometer), even though the sky is colder.
Experimental Setup
The following setup (assisted by one of my daughters) includes a metal plate, painted flat black, and heated with a 250 W flood light. The heated plate has exposed to it a Styrofoam cooler containing ice. The hot plate is kept above the ice to minimize any air convection effects on the results.
Fig. 1. Experimental setup to demonstrate a cooler object can make a heated object even warmer still.
Since a part of the heat budget of the heated plate is its loss of infrared energy to the cold ice, it should be possible to measure an increase in the temperature of the hot plate if the view of the ice is blocked with a second sheet (painted with very high IR emissivity paint, Krylon white #1502) at room temperature.
I tried two setups: the one on the left in Fig. 1 has the temperature probe scotch taped to the hot plate, but it did not stay firmly in contact with the plate, and also the steel plate I used exhibited large temperature gradients. On the right side of Fig 1, a better setup had the temperature probe firmly attached to the back with aluminum tape, and the metal sheet is aluminum flashing, painted flat black. The results presented below are for the arrangement on the right.
Flir thermal imager measurements of the setup late in the experiment are shown in Fig. 2, where we see the ice is generally below 32 deg. C F (it came from the deep freeze), and the heated plate (in the second phase of the experiment, results below) has a temperature around 110 deg F.
Fig. 2. Flir thermal imagery of ice and hot plate during the second portion of the experiment.
I recorded temperatures every 5 secs with the plate alternately exposed to a view of the ice for 5 minutes, then with the ice covered for 5 minutes. This cycling was repeated five times. The results are shown in Fig. 3. What we see is just what I would expect, that the temperature of the hot plate increases with time when its view of the ice is blocked by the room-temperature sheet.
Fig. 3. Time series of temperature of the hot plate as it was alternately exposed to ice, then a room-temperature high-emissivity sheet, every 5 minutes.
Also shown in Fig. 3 are the results averaged over all five cycles, which smooths out some of the noise. The hot plate was so hot that just a small breeze of air from moving the room temperature sheet around the apparatus was found to cool the hot plate by a couple of degrees. The gradual warming trend was due to the ice slowly warming up over the ~1 hr period of the experiment.
Experiments like this often have sources of error when one tries to isolate a certain process. One concern would be whether the room-temperature sheet was slightly reflective to infrared radiation, which would cause the observed temperature effect by reflecting some of the hot-plate emitted IR radiation back on itself. I tested the sheet, which has very high infrared emissivity paint applied, by measuring its IR temperature with the Flir imager outside, both at right angles to the surface, and then at a ~45 deg. angle so the cold sky (today running in the mid-20s deg. F in the Flir measurements) could reflect off the sheet. There was no noticeable difference to a small fraction of a degree, so the paint appears to have an IR emissivity close to 1.0 and is indeed non-reflective in the infrared.
Another concern was the close proximity of the flood lamp to the ice and room temperature shield. If the shield was slightly more reflective to visible light than the ice, it could cause a little more heating of the hot plate. So I moved the lamp down and away from the setup so that there was little noticeable illumination of ice or shield (see the left side of Fig. 4). The hot plate cooled considerably since the lamp wasn’t nearly as close, but the effect of adding the shield was very clear with just one application, with temporary warming of the hot plate until the shield was again removed:
Fig. 4. Second setup where the flood light was place farther away from the plate it was heating, and the results of one cycle with the hot plate shielded from emitting IR toward the ice, then exposed to the ice.
It should be noted that heating a surface with incandescent bulbs will be mainly through infrared radiation, since their emission spectrum is more in the IR than at visible wavelengths. Halogen bulbs, which I did not use, have a somewhat closer spectrum to the solar spectrum.
Conclusion
There is no violation of the 2nd Law of Thermodynamics in the experiment; a cool object can make a warm object even warmer still through infrared radiative effects. The phenomenon can only happen, though, if the cool object replaces something that is even colder, and thereby reduces the rate at which the warm object loses infrared energy to its surroundings. In this experiment, the room temperature plate takes the place of the ice which still emits at around 300 Watts per sq. meter; in the climate system, the atmosphere takes the place of deep space, which emits energy at close to 0 Watts per sq. meter.
The heated plate is placed above the ice so that there is essentially no intermingling of ice-chilled air (which will flow downward) with hot plate-heated air, which will flow upward. Ideally, the experiment would be carried out in a vacuum chamber, so that conduction effects by air would not be present.
There are changes that would make the experiment work better:
1) Dry ice in the place of water ice, to provide a colder target for the hot plate to lose IR energy to;
2) use an open chest freezer (running) covered with a single layer of plastic wrap, so the temperature of the ice won’t increase with time;
3) cover the hot plate holder with plastic wrap, which is about 90% transparent to IR, but will reduce the variations in heat losses due to air currents.
4) if the ice can cover a greater portion of the hemisphere that the plate is losing IR energy too, the effect will be magnified. An open chest freezer would accomplish this, too.
NOTE TO COMMENTERS: I intend to delete any comments which include personal insults.
NOTE TO READERS OF COMMENTS: Some commenters here throw around technical terms and make grand assertions and detailed arguments which I consider fallacious. I do not have time to counter them all every time they arise, although I have addressed virtually all of them in other posts over the years.
Roy Spencer an old fashioned scientist who actually did a real world experiment. I guess you have proved your point.
Hopefully John O’Sullivan and Joe Postma will look at your experimental set-up and do their own. See what results they get. If they think of something you have not considered that could create the results you obtained, they can change the set-up to eliminate these effects.
I think the PSI group is primary source of those with the “bad” physics. I think your test should make Postma apologize for the post he made about you here:
http://principia-scientific.org/observational-evidence-surfrad-sites-falsify-greenhouse-effect-hypothesis/
Whoops
Was having posting with a link. I copied it but did not delete all of the message.
From Postma’s blog:
A 3rd option. Dr Spencer exhibits that common fatal failing of man, foolish pride, i.e. never being able to admit youre wrong.
Joseph E Postma says:
2016/08/27 at 5:14 PM
Wellto me that categorizes as incompetent scientist.
Good point regardless.
yes, I’m just stubbornly refusing to accept the truth which Joe is trying to reveal to the masses. That must be it. 🙂
Because the cool object reduces the rate of radiative heat transfer.
Formula for radiative heat transfer:
q = ε σ (Th4 – Tc4) Ac
where:
q = heat transfer per unit time (W)
ε = emissivity of the object (one for a black body)
σ = 5.6703 10^-8 is the Stefan-Boltzmann Constant
Th = hot body absolute temperature (K)
Tc = cold surroundings absolute temperature (K)
Ac = area of the object (m2)
JDAM, see my response below to this equation…it’s a good jumping off point for discussion of what actually happens when the warm object (Th in your equation) is actively heated, such as the Earth is by the sun. In that case, Th can actually increase with a constant solar input if Tc is increased.
Got it sorry about the double post
“The phenomenon can only happen, though, if the cool object replaces something that is even colder, and thereby reduces the rate at which the warm object loses infrared energy to its surroundings.”
The hot plate is emitting IR. The ice is adsorbing IR. What is the causal connection? i.e.
Why would the hot plate not emit the same amount of IR whether or not the ice was present? Is this some part of basic physics I’ve been unaware of? Will a luminous source put out less light if it isn’t radiating onto a light absorbing surface?
Why would the cardboard cause less heating of the hot plate if the ice was not present?
because the ice (and the room-temperature sheet) emit IR back to the hot plate. It’s the NET (2-way) transfer of IR radiation that matters. EVERYTHING emits infrared radiation. To claim otherwise is just plain silliness.
I guess my thinking ability is faulty. The ice radiates some amount of IR back at the hot plate. The treated cardboard radiates more. Therefore the hot plate becomes warmer with the cardboard in place. So far, ok.
If there was no ice radiating some amount of IR back at the hot plate, why would that slow the rate of IR radiation from the hot plate? Because the IR radiated back from the ice heats the hot plate somewhat, increasing its rate of IR radiation even higher than it would be without the ice, thus increasing its net rate of energy loss?
As my quote from your writing indicates, you say that the cardboard would not cause the hot plate to become warmer without the ice present; isn’t that “the effect” to which you refer? I can’t see how that can be so, how the ice behind the cardboard makes any difference to the hot plate.
“If there was no ice radiating some amount of IR back at the hot plate, why would that slow the rate of IR radiation from the hot plate?”
Not sure what you are asking. If there was no ice, there would be *something”…in my experiment, it would be the table, at room temperature.
..The radiation from the ice cannot penetrate the room temperature sheet. It’s the difference between the [hot plate/ice] system versus the [hot plate/room temperature sheet] system. The latter will have a warmer hot plate because the net IR flow away from the hot plate is initially less, leading to an energy imbalance and warming…until the hot plate warms sufficiently to reach a new state of energy balance with its surroundings.
Suppose the ice is removed. Treated cardboard, or something equivalent, is placed six feet from the hot plate or any further distance. Equilibrium is established. Now place the original treated cardboard closer, in-between the six foot barrier and the hot plate (as before: 1 to 2 feet from the hot plate?). Doesn’t the effect prevail? Doesn’t the hot plate warm more? (unless one assumes the original conditions involve no loss of energy from the system)
No. If the surroundings are all at room temperature it doesn’t matter what you put in front of the hot plate, it’s temperature won’t change.
Actually, if the measurements were sensitive enough, a plain sheet of cardboard by itself should make a difference.
If there is nothing nearby (say nothing within about 6 ft), then the hot plate will see entirely ~ 20C background of the room. The distant objects won’t warm noticeably.
If you then put a cardboard sheet nearby (say 1 ft away), it should start to warm — somewhere between the temperature of the room and the hot plate. With a small thermal mass, it should warm a fair amount fairly quickly (until it reaches some equilibrium). As such, the warm cardboard will have more back-radiation than the cool surroundings. The top plate should warm a little.
yes, you are right…I stand corrected. So, it seems important for the shield to be at the same distance away as the ice, and for it to cover the same area. So, maybe the experiment would benefit from two identical sheets, but one is chilled, and the other room temperature, and they are swapped out. Still be nice to make it work with something that is actively cooled, say with an open chest-style freezer. -Roy
AndyH says: August 28, 2016 at 4:04 PM
(If there was no ice radiating some amount of IR back at the hot plate, why would that slow the rate of IR radiation from the hot plate?)
Roy W. Spencer, Ph. D. says: August 28, 2016 at 4:17 PM
“Not sure what you are asking. If there was no ice, there would be *somethingin my experiment, it would be the table, at room temperature.”
Indeed Dr. Spencer,
You are clearly falsifying the concept that ‘all things radiate (thermal EMR) proportional to ‘only’ their own temperature^4“.
You clearly demonstrate that all ‘effective’ thermal EMR power transfer ‘always’ is limited by the magnitude of the opposing EM radiance. There is absolutely nothing in your demonstration that even suggests a power transfer in the direction of a higher potential.
Such a suggestion continues to be only the desperate claims of some AlGoreasta CAGW nonsense.
OK Dr. Spencer Let us try some science for which none know the answer.
Two 100 cubic foot insulated steel bottles 55 kg empty with 45 J/gm specific heat. Large line with one 6 mm diameter choke orifice in the middle. One bottle 100 atmospheres N2 at 1000 Kelvin bottle and all. other bottle 0.1 atmosphere (1.5 psi) nitrogen N2 at 80 Kelvin, bottle and all. Open all valves, blow all blocks. Please show the temperature and pressure profiles vs time for each bottle.
Any Meteorologists would have to consult with some apprentice engineers for six months. A Master engineer would have some journeyman engineer look that up in a table and be correct within 10 minuets.
I forgot! Then we go measure to find out just who was corrector!
The scientific method!! Not some fantasy!
You say the hot plate is about 110 F and graph shows range of 240 to 260 F. Though the “second set-up” of flood lights placed further away graph has range of 120 to 105 F.
Seems a bit confusing. Why did select a temperature which higher boiling or warms as much as surface of the Moon?
Also I will note that if cool air is falling, other air must replace
it. And would guess that having hot plate indented creates a pocket
of warm air- which air convection could “in a sens”e could have a more amplified effect upon.
Or trapped heat of that air is allowing the plate to warm up more than it would otherwise [air could freely go up].
And/or in regards to this you said:
“The hot plate was so hot that just a small breeze of air from moving the room temperature sheet around the apparatus was found to cool the hot plate by a couple of degrees”
Or small breeze destablized the pocket of air- making a large effect on temperature of hot plate. And delayed effect because the hot air pocket of air had to heat up again- which allow plate to heat up, again.
Also you said rising trend in temperature was due to warming ice-
I would say it was also due to heating box around plate and the rest of the room in general.
And the thrashing about begins….
Just questions and comments.
How about I make prediction.
Use water and add dry ice the colder smoke should warm more than cardboard.
Btw after 5 min of cardboard under heated plate, how warm did cardboard get?
Dr Spencer,
I don’t know about thrashing about, but if you turn your heat source off, the heated plate cools. Any passive radiator beneath it cannot heat it, or stop it from cooling.
Just like the Earth’s surface at night.
Cheers.
But the heated surface will cool MORE QUICKLY when it is facing the cold ice than when it is facing the warmer cardboard. Just like the earth would cool more quickly without GHGs than with them.
Cheers.
Tim,
And in the arid tropical deserts, that’s precisely what happens. In the absence of the most important GHG, H2O, the surface cools more quickly. Conversely, it attains higher daytime temperatures. Simple observation, supported by physics.
I’m not sure what your point is. Are you disagreeing with yourself, or something I have said?
Cheers.
The point is that the GHE works as advertised, helping the sunny side warm up, and hindering the night side from cooling down as fast.
Or perhaps more intuitively …
IF the GHGs keep the earth from cooling down as quickly over night, then the next day the sun can warm the earth more than it could have otherwise.
Tim,
No, the arid deserts get hotter with less GHG, not more. They also cool down faster with less GHG, not more.
Insulation can be used to reduce the rate of cooling, but also to reduce the rate of heating – say around a freezer.
Do you disagree?
Cheers.
Tim,
You wrote –
“IF the GHGs keep the earth from cooling down as quickly over night, then the next day the sun can warm the earth more than it could have otherwise.”
No. A rock in the desert may reach 70 C during the day, yet still be below 0C at night. If, due to insulating during the night, it only gets down to say 5 C, it will not get any hotter than a rock which fell to below 0 C during the night. A rock heated to 30 C at dawn will be no hotter than the freezing rock, when maximum temperature is reached during the day. Both will be at the same temperature.
Temperatures cannot be added, any more than W/m2.
No GHG heating. Cooling more slowly is not heating. No GHE global warming. Remove the heat source, cooling results.
Cheers.
Tim Folkerts says:
August 28, 2016 at 7:52 PM
“The point is that the GHE works as advertised, helping the sunny side warm up, and hindering the night side from cooling down as fast.”
Dr. Spencer’s experiment does not prove the GHE works as advertised (although I can’t be sure how you define the GHE). He says in the introduction “The experiment shown below does not prove that greenhouse gases in the atmosphere perform such a function, only that it is not a violation of the 2nd Law ….” Since he took care to suppress convection, no conclusion can be drawn about what happens during the day.
Mike Flynn
YOU STATE: “Temperatures cannot be added, any more than W/m2.”
I might disagree with your second point. You can test this with some lamps. Turn one lamp on a surface and wait until equilibrium and turn on a second that can hit the surface equal to the first and see if you get an even warmer temperature. If you do this and get a higher temperature reading with two lamps (which you will) this will demonstrate your statement as false.
Anthony Watts once posted this experiment. -Roy
“A rock in the desert may reach 70 C during the day … ”
Replace “rock” with “swimming pool” or “lake” in your argument and you hopefully will see that it falls apart.
Imagine two large, deep, identical swimming pools. At the beginning of the day, one is filled with 0 C water; the other with 5 C water. I guarantee that at the end of one sunny day, both will have warmed, but the one that started warmer will also end warmer.
Tim Folkerts
Maybe Mike Flynn does not understand thermal inertia. Based upon the mass and heat capacity of an item in a radiant flux it will determine the rate it heats. If you have a variable flux like solar understanding this is critical. I am not completely sure what desert he is talking about. Most deserts (unless really high above sea level) in the summer will not go to 0 C at night. I am not sure if a rock is frozen and put in the sun it will reach the same temperature as a rock that is heated. I might try it out to see.
I think your pool example was a good one.
Mike Flynn says:
“No. A rock in the desert may reach 70 C during the day, yet still be below 0C at night.”
It might or might be close to this.
But air temperature has to be warm enough for a rock to reach 70 C- else convection air losses from cooler air will prevent this temperature.
So having day high air temperature of 30 to 40 C and dry air- low humidity, is still holding a lot water vapor were it to cool near 0 C. And 1000 meters up will be cooler than 0 C. Or said differently 0 C air can’t hold much water vapor as compared to +30 C air- and +30 C air at 10% humidity would be quite dry air.
“Practically all of the Earth’s surface contributes water vapor to the atmosphere open water and ice (which cover three-quarters of the planet), the soil and vegetation. And once water vapor is in the air, wind carries it everywhere even across arid deserts. The world’s lowest recorded relative humidity value occurred at Coober Pedy in the South Australia desert when the temperature was 93 degrees and the dew point was minus 21 degrees producing a relative humidity of 1 percent.”
http://articles.chicagotribune.com/2011-12-16/news/ct-wea-1216-asktom-20111216_1_relative-humidity-zero-dew-point
Or “Coober Pedy” might have had below 0 C in following night and
70 C rock during that day.
–Tim Folkerts says:
August 29, 2016 at 5:18 AM
A rock in the desert may reach 70 C during the day
Replace rock with swimming pool or lake in your argument and you hopefully will see that it falls apart.
Imagine two large, deep, identical swimming pools. At the beginning of the day, one is filled with 0 C water; the other with 5 C water. I guarantee that at the end of one sunny day, both will have warmed, but the one that started warmer will also end warmer.–
Water and rock are vastly different.
A 0 C rock could warm to 70 C in one day, and 5 C swimming pool
can’t.
With Al Gore’s gas fire swimming pool heater, it would be unable to do this in 24 hours with heat dial set to 11.
As far as your 0 and 5 swimming pool, it depends it it’s saltwater
or fresh, and what you measuring. Average pool water or surface temperature. Fresh water density is lowest above 0 C, therefore the 0 C pool water density will increase when heated by sunlight.
Or roughly speaking the 0 C water should mix better than 5 C water. And there could other environmental factors involved.
But in any case one is only going increase pool average temperature by few degrees in one day- unless it’s a small toy swimming pool. If no wind and just measuring the surface the 5 C
swimming pool has advantage of lacking as much mixing, and top 1/4″ may reach 30C or more in one day- but if low humidity or windy then evaporation will limit it’s top surface temperature
Back to rock, it has poor heat conductivity so shaded side of rock starting at 0 C could remain near this temperature or warm nearer air temperature whereas sunlit side of rock could warmed to 70 C at it’s surface.
Tim,
” Or perhaps more intuitively
IF the GHGs keep the earth from cooling down as quickly over night, then the next day the sun can warm the earth more than it could have otherwise.”
It seems this is what happens everyday. One easily notices that overcast nights do not cool as quickly, nor warm as quickly. This would be for the same reason the cardboard works. It is both an insulator and reflector. (even if the r value is small)
So it is not whether GHG have an effect, obviously they do, but how much difference the small percentage change in CO2 actually changes that effect. I suggest the difference is so small that it is not noticeable.
This is simply because there is 1000 times more water vapor in the atmosphere than CO2.
…but if you turn your heat source off, the heated plate cools.”
I wonder how fast.
My guess is it’s about 5 min to cool from 120 C to 70 C.
So turn it on, and heat up a bit, then expose to ice, heat it up for few minutes more, then turn off heat lights until it’s 70 C.
Timing it.
Then turn on again, heat it to 120 C with cardboard. Turn it off again. Time it again, until heated plate is 70 C
Just a minor thrash. Applying the Stefan-Boltzmann law to a 0 deg C black body with emissivity 0.97, that of ice, gives around 300+ W/m^2. Where does that IR heat come from, since absorbed IR will be endothermic due to the latent heat of fusion of ice and will not be emitted but go into break up of lattice bonds. Seems like I should be able to bring in a large snow pile to my living room and with the 300 watts/m^2 cut my heating bill this winter.
Or maybe I am going a little senile in my old age.
Roy,
Here’s a very simple variation to try on this experiment — a variation I used in a similar experiment.
Other than to make it seem more like the earth, using the flood lamps to warm the top plate is not necessary. Any heater would work just as well. Buy a small heating pad at your local drug store and position it behind the top sheet under the styrofoam insulation (just make sure there is no automatic thermostat built in). The electrical heating will be more controllable and more uniform. With a couple quick measurements of I & V, you can even calculate the power input.
yes, I thought of doing this. Couldn’t think of how to easily make it non-thermostatically controlled…I’m not an electronics buff. I decided on using a radiant energy source, more like the sun, because someone would surely object “but the Earth isn’t heated with a heating pad!”
You could always cut off whatever controller the heating pad comes with and plug it straight into the wall so it is always full power. Additionally, you could rummage through the back of the university’s labs for an old VARIAC if you want to be able to adjust the power.
Hi Roy,
“Couldnt think of how to easily make it non-thermostatically controlledIm not an electronics buff.”
If you cannot problemsolve how to by-pass the thermostat how can you problemsolve (explain) an experimental result?
Have a good day, Jerry
Jerry, why do cats have green wings?
Ball4 says: August 29, 2016 at 2:03 PM
“Jerry, why do cats have green wings?
Trick,
Jerry is worse than you at science. But he can explain that the green wings go well with the orange toenail polish!!
What I would like to know is how cooling of the hot plate would proceed if the lights were turned off at a particular temperature and the cooling rate graphed say for the ice and then reheat the plate and record and graph the cooling with the shield in place and then repeat for different distances of the shield from the plate. If the cooler IR from the cooler sheet does add heat the plate, the cooling rates will decrease as the shield gets closer to the hot plate. I have trouble when the system is not an isolated system with external energy input while cooling. It would be closer to the cooler atmosphere retarding the Earth’s temperature decrease on a cloudless night.
Would you know whether watts/m^2 power output is an extensive property or not? If it is, then it depends on configuration of the matter and in most cases will not necessarily depend on the temperature as in a black body or a suitable gray body. Temperature is an intensive property and does not depend upon the amount or configuration of the matter. Energy/second is extensive so watts are extensive. So it would be necessary to be careful in which direction one calculates, from temperature to power output or from power output to temperature. I kind of remember some warning about that in Gerlich and Tscheuschner (2007).
Bravo, Dr. Roy. Next step you should move the lights around until the plate at a steady 288 and keep the ice at 273, which is about cloud temp, and then see how much CO2 you have to to the enclosure to make the plate warm up by 0.8 degrees, like the atmosphere since 1850. /sarc, Gang I already know why it won’t work, thanks for your comments anyway.
Indeed.
I’m sure there are a multitude of reasons to avoid actually using CO2 in an experiment to demonstrate the greenhous effect. The primary one, of course, is that the presence of CO2 in the atmosphere does not raise surface temperatures.
This is why global warming enthusiasts avoid acknowledging that there is a distinct lack of surface heating when the Sun is absent. At night, for example.
Cargo Cult Scientism. Ignore inconvenient truths such as four and a half billion years of global cooling.
Maybe I’m just thrashing about.
Cheers.
MF
“Maybe Im just thrashing about.”
Replies:
Like a dying fish!
Is the pope Catholic?
Is the sky blue?
etc etc
Thrashing about? Ummm…yeah. Since CO2 absorbs strongly in a portion of the IR spectrum where H2O does not, near the peak of the Earth’s surface radiation spectrum, by the way, increasing the concentration of CO2 in the atmosphere will increase the average temperature of the atmosphere and, due to the transmission of energy from the atmosphere, the surface; that is physical fact. The question, in regards to climate warming, is whether the increase will be mitigated by some temperature dependent processes in the climate and by how much.
Yes, Slipstick, radiative heat from above causes water vapour and air to convectively rise in the atmosphere, where it forms clouds due to the air temperature linearly decreasing to -55C up to 12 km altitude. The clouds then reflect away a larger amount of incoming radiative heat from the sun. If it were not so, the earth would have overheated millions of years ago…
While what you say is true, it is unclear whether the effect at current conditions is actually increasing cloud cover sufficiently to compensate for the rise in temperature due to CO2 concentrations forced above “natural” levels. Given that the average temperature of the climate is rising, it appears that is not the case.
Slipstick,
First you would have to prove your premise. AFAIK, here is no definitive evidence that CO2 above present concentration would have any further effect on temperature.
Other than, of course, that no other reasonable model explains why 2015 was the hottest year in the modern climate record.
Because because because because because……
What explanation is needed? It’s warmer. That’s like trying to have an explanation of why some horse set a need track record – the first at this track.
So, just to be nice, I’ll give you a layman’s version of a actual answer.
Since records have only been kept for a geologically short period of time, the records only record a small sampling of temperatures. We will expect to continue to set high and low records regularly for the next 200 years or 2000 years, as the records slowly come to contain the variances we know have and will occur in the earth’s temperature.
Slipstick,
Night is colder than day. Winter is colder than summer. Now is colder than when the surface was molten.
Physical fact.
No heating effect from placing CO2 between a heat source and target has ever been experimentally demonstrated. Fact.
I assume you deny some, or all, of what I have said. I don’t know why. Maybe you could provide some facts to rebut my position?
Cheers.
Actually, we are currently conducting a planet-wide experiment in artificially increasing the CO2 concentration. So far, the results indicate that your “no effect” supposition is incorrect.
Mike Flynn: You say:
“No heating effect from placing CO2 between a heat source and target has ever been experimentally demonstrated. Fact.”
Utterly false! The US Air Force has spent decades measuring how radiation from a heat source is absorbed in the atmosphere before it gets to the target sensor. This work has been essential in making sure that heat-seeking missiles can work in all types of atmospheric conditions.
Mike Flynn says:
“No heating effect from placing CO2 between a heat source and target has ever been experimentally demonstrated.”
“Observational determination of surface radiative forcing by CO2 from 2000 to 2010,” D. R. Feldman et al, Nature 519, 339343 (19 March 2015)
http://www.nature.com/nature/journal/v519/n7543/full/nature14240.html
Lipstick, water absorbs in the same bands as CO2. Clouds absorb in the same wavelengths as CO2. So what is the contribution of CO2 to the GHE?
mpainter says:
“water absorbs in the same bands as CO2.”
Not so.
See Pierrehumbert’s Physics Today article, Figure 2, lower right corner, where he compares CO2’s bands to water vapor’s, from 600/cm to 670/cm.
https://geosci.uchicago.edu/~rtp1/papers/PhysTodayRT2011.pdf
Then read the last paragraph of his sidebar, “Saturation Fallacies.”
slipstick, not lipstick
“Ignore inconvenient truths such as four and a half billion years of global cooling.”
I suppose I could just as easily say you are ignoring the billions of years of global warming as the Sun transforms into a red giant and fries the Earth. It would be just as relevant to the current discussion as your comment.
This is a fair and square experiment that met its realistic objectives.
The question remains: Does radiation from the cold plate reach the surface of the hot plate? This is very difficult to prove in any laboratory experiment.
If radiation from the cold plate reaches the surface of the hot plate, then we’ll have a perpetual engine and temperatures of both plates run out of control, which we know this won’t happen. Something, sort of interaction between forward radiation from hot to cold and backradiation from cold to hot, potentially exists.
Nabil,
Not so difficult. Make hole in hot plate. Use photon detector behind hole, with crossed slits ensuring that only photons from cold plate are detected. Photons travel in straight lines. In a vacuum, no photons from the hot plate will reach your detector. It’s been done down to individual photon detection and energy measurement level.
Or just make a hole, and shine a torch through it from the cold plate position. Same principle.
However, describing the interaction between the lower energy photons and the “warmer” matter on which they impinge is not so easy. Bugger!
Cheers.
Nabil,
Happy you agree the experiment worked and confirmed its hypothesis.
“Does radiation from the cold plate reach the surface of the hot plate?”
What happens to it in between? Does it vanish? This is unphysical.
The plate clearly responded to the temperature of the plate below. How do you understand that, if not by transfer of photons?
One can buy toy perpetual motion machines-but there always is a hidden battery inside. Here there is power input as well, to the hot plate. So perpetual motion is not relevant. Is it?
Nabil Swedan says: August 28, 2016 at 8:40 PM
“This is a fair and square experiment that met its realistic objectives.”
Yes indeed objective “Demonstrate that thermal flux from any surface is NOT but proportional to only some power of its own surface, but always must be limited by any opposing field strength (radiance) at each frequency and in every direction. Gustav Kirchhoff.
“The question remains: Does radiation from the cold plate reach the surface of the hot plate? This is very difficult to prove in any laboratory experiment.”
Such thermal EM flux W/m^2 has never been observed nor measured anywhere not in any laboratory, not anywhere in this atmosphere.
“If radiation from the cold plate reaches the surface of the hot plate, then well have a perpetual engine and temperatures of both plates run out of control, which we know this wont happen. Something, sort of interaction between forward radiation from hot to cold and backradiation from cold to hot, potentially exists.”
Almost correct! Radiance (Poynting) vectors (E x H)/normalized, exist in space from any potential source. Only the vector summation of all Poynting vectors at that point, results in a singular Poynting flux vector (E x B) for each frequency. This is always true as long as that flux is measured over some multiples of coherence interval. For Thermal EMR that interval remains indistinguishable from zero. One way thermal flux only, in direction of lower radiance.
2nd Law, no perpetual motion. Such detracts not at all from experimental objective!
own surface, own surface temperature
Nicely done, Dr. Spencer.
Nabil, what is this perpetual motion misconception ? Maybe don’t think of it as radiation travelling from the cold to the hot plate. Think of it as the summation of all those Schrodinger wave equations that sum together with the result of something we call heat energy moving from the hot plate to the cold plate.
Radiation is energy, regardless of its origin whether from the hot plate or the cold plate. If radiation from the cold plate reaches the hot plate it must add to that coming from the light. Consequently, the energy striking the hot plate should increase progressively with time, and the temperature of the hot plate should increase indefinitely. This does not happen and steady state is typically reached.
Nabil,
Certainly it appears that CO2 enthusiasts believe the Earth is a little warmer each day, due to “heat trapping” or some similarly obscure and undefinable mechanism.
As you point out, not physically possible.
As even Kevin Trenberth said – “It’s a travesty”, referring to missing heat. Maybe he was confusing modelled heat with reality. Defining the ratio of a circle’s diameter to its radius as 3 doesn’t make computer calculations correct, regardless of the number of significant figures. Looking for calculated missing area is as pointless as looking for non existent missing heat.
A good joke!
Cheers.
Still thrashing!
Your comments are a source of consistent humor.
Cheers!
Nabil Swedan, the temperature increase of the hot plate will result in an increase in its radiative emission. This effect, repeated, will eventually stop the increase that you propose will continue indefinitely. It may be likened to the finite sum of a convergent geometric progression.
Nabil, I am genuinely concerned that you are making things up to confuse people about what you know is not true.
Using your logic, why doesn’t the plate warm indefinitely when exposed to the flood light? Why would some additional energy input from the room temperature shield be expected to do this, when the flood lamp didn’t?
The answer, of course, is the heated plate always has a way to lose energy…especially conduction to the air, and its view of the room temperature surroundings.
Even if the hot plate was surrounded by a spherical shell that was transparent to visible light and opaque to IR light, the hot plate would stop warming when the spherical shell warmed sufficiently so that the whole room-shell-plate system reached an energy balance.
Trust me it is genuine. To prove it, long before the experiment, I made the question. I do not recall that this question has ever been addressed in the literature and genuinely feel that it is a good subject of research for physicists.
I agree with you that there is convection to the surrounding air and that my case of course theoretical. It only assumes the hot plate as the system and cold plate as the only surroundings, since we are interested in flow of radiation.
“I do not recall that this question has ever been addressed in the literature…assumes the hot plate as the system and cold plate as the only surroundings..”
Under that assumption, Nabil should indeed find his question addressed up front in a good beginning text on atmospheric radiation. For example, Bohren & Clothiaux 2006 p. 16 Fig. 1.6. As Christopher writes, the infinite series implied converges to simple terms used often in this field.
I am inclined to think that Nabil is genuinely mistaken in his belief rather than that he is trying to mislead. I think he is sentimentally attached or fixed to a view that he has expressed in the just previous article http://www.drroyspencer.com/2016/08/simple-experimental-demonstration-that-cool-objects-can-make-warm-objects-warmer-still/#comment-221512
as follows
“The question is that does this valid statement supported by observation can only be explained by the greenhouse gas effect and rdaiative forcing? The answer is no. There are other ways to explain it through thermodynamics.”
To expand on this, he links to
http://www.seipub.org/des/MostDownloaded.aspx
DOI: 10.14355/des.2015.03.001
Change is very difficult for an intellectual who holds a fixed view with sentimental attachment. It can be apparently or nearly impossible to reason with and persuade such a person to reconsider his view, because he has an emotional investment in it.
I think Nabil is aware that warmism is a very harmful scam. I think he has the feeling that the best way to deal with the scam is to undercut its basic postulate. Against that feeling, I think in this case, we have to accept the basic postulate that man-made carbon dioxide emissions can cause some degree of global warming. Our task us to stay the destructive actions of those who exaggerate the likely man-made effect. We have to combat the exaggeration and political over-response rather than annihilate the basic idea.
At present right here we are responding to Nabil’s story that back-radiation will not reach the ground. He writes above:
“The question remains: Does radiation from the cold plate reach the surface of the hot plate? This is very difficult to prove in any laboratory experiment.”
Perhaps Nabil may consider the Helmholtz reciprocity principle, one of the mainstays of Planck’s immortal The Theory of Heat Radiation. It is quoted in Wikipedia as follows:
“As cited by Kirchhoff in 1860, the principle is translated as follows:
“A ray of light proceeding from point 1 arrives at point 2 after suffering any number of refractions, reflections, &c. At point 1 let any two perpendicular planes a1, b1 be taken in the direction of the ray; and let the vibrations of the ray be divided into two parts, one in each of these planes. Take similar planes a2, b2 in the ray at point 2; then the following proposition may be demonstrated. If when the quantity of light ”i” polarized in the plane a1 proceeds from 1 in the direction of the given ray, that part ”k” thereof of light polarized in a2 arrives at 2, then, conversely, if the quantity of light ”i” polarized in a2 proceeds from 2, the same quantity of light ”k” polarized in a1 [Kirchhoff’s published text here corrected by Wikipedia editor to agree with Helmholtz’s 1867 text] will arrive at 1.”
In ultra-simplified form, this may be expressed ‘if I can see you, then you can see me’.
This principle has been tested thoroughly for well over a century, including experimental tests. I think it is reliable, and I commend it to Nabil.
“At present right here we are responding to Nabils story that back-radiation will not reach the ground.”
I do not recall saying this statement in the last two posts at this blog. Pleases refresh my memory.
Anthropogenic global warming (AGW) is a fact based on thermodynamics, period. This is my point supported by mathematics, physics, and observations.
AGW is unconvincing based on the radiative model. Think of it: When Donald Trump before the world says that he does not believe in AGW based on the current science. This is a serious matter, and if I were in your place I would at least rethink about the radiative model. After all backradiation is calculated and has never been measured. Pleas see my reply to Wim Rost farther down in this thread. This is the greenhouse gas effect, never been measured. I challenge anyone who can deny AGW based in thermodynamics.
OK, Nabil, how do you explain that an IR thermometer, on a clear day, will measure a higher temperature pointed at the sky at an oblique angle than it measures pointing straight up? The IR thermometer doesn’t actually measure IR radiation per se, it measures small temperature changes in response to the changing NET IR radiation at the surface of a thermopile. That warming of the sensor when pointed at an oblique angle toward the sky, *IS* evidence of “back radiation”. How else can you explain it? -Roy
Roy says, in response to Nabil Swedan says:
August 29, 2016 at 2:58 PM
“OK, Nabil, how do you explain that an IR thermometer, on a clear day, will measure a higher temperature pointed at the sky at an oblique angle than it measures pointing straight up? The IR thermometer doesnt actually measure IR radiation per se, it measures small temperature changes in response to the changing NET IR radiation at the surface of a thermopile. That warming of the sensor when pointed at an oblique angle toward the sky, *IS* evidence of back radiation. How else can you explain it? -Roy”
Roy: “How else can you explain it?”
WR: If there is no answer to this question, it doesn’t mean that the statement thereby is proven. You know I am not technical, therefore a 100% guess: what, if it is not back radiation that is measured, but an effect of what is elsewhere mentioned here, an effect of scatter radiation by gas molecules? Just to mention something. Perhaps there could be many more possibility’s other than back radiation. And: “What if the calibration equation [for back radiation] is wrong?” I (WR) understand everyone uses just one formula. Where is the ‘check, check and double check’?
When the evidence for back radiation is so strong, why do we need to point an IR thermometer to the sky and interprete the result ourselves?
WR: I am just posing questions. For me the whole climate story is a big puzzle and I try to find out the truth behind it. Like many of us. I am not ‘a believer, not in one and not in another ‘opinion’. I just think there have to be good answers on simple questions. Also when reality is complicated, as is the case when it is about ‘climate’.
Thank you, Nabil, for your reply. My words “At present right here we are responding to Nabil’s story that back-radiation will not reach the ground” were intended as a summary paraphrase of what I have understood you to be contending here. I reached my understanding of your view partly from the following words that I quoted from you: “The question remains: Does radiation from the cold plate reach the surface of the hot plate? This is very difficult to prove in any laboratory experiment. And these words quoted from you: The question is that does this valid statement supported by observation can only be explained by the greenhouse gas effect and rdaiative forcing? The answer is no. There are other ways to explain it through thermodynamics.
For better understanding of your view, I would like to ask you:
Do you think back-radiation is absorbed by the ground?
Do you think back-radiation is added to by added carbon dioxide?
Do you accept the Helmholtz reciprocity principle?
I have lost track and cannot be sure what you refer to by your words: “Pleas see my reply to Wim Rost farther down in this thread.”
Christopher Game says: August 29, 2016 at 2:36 PM
At present right here we are responding to Nabils story that back-radiation will not reach the ground. He writes above:
The question remains: Does radiation from the cold plate reach the surface of the hot plate? This is very difficult to prove in any laboratory experiment.
“Perhaps Nabil may consider the Helmholtz reciprocity principle, one of the mainstays of Plancks immortal The Theory of Heat Radiation. It is quoted in Wikipedia as follows:”
That is a deliberate lie written into Wikipedia by William STOAT Connolley in order to confuse all.
At the same frequency there is absolutely no reciprocity whatsoever! If there were there would be no beam steering radar at all. Please learn something, anything of EMR.
“In ultra-simplified form, this may be expressed if I can see you, then you can see me.”
You defiantly cannot see me if your eyes are brighter than I.
! ! !
“In ultra-simplified form, this may be expressed if I can see you, then you can see me.”
Such grandiose incompetent posturing!
Please identify when and where any higher luminosity receptor can ever detect, observe, measure anything exhibiting only “lower luminosity”?
Don’t worry about the ultra-simplified version. It is, as advertised, ultra-simplified. Read Kirchhoff’s 1860 version.
Christopher Game says: September 5, 2016 at 6:38 PM
“Dont worry about the ultra-simplified version. It is, as advertised, ultra-simplified. Read Kirchhoffs 1860 version.”
I certainly have, Have you? Have you even tried to convert the 1860s EMR terminology into 21 century photometric or radiometric terminology for a wee bit more understanding. It appears that you only wish for the unwashed to remain unlearned and unable to violently distroy your absurdity!
Such grandiose incompetent posturing!
Please identify when and where any higher luminosity receptor can ever detect, observe, measure anything exhibiting only lower luminosity?
Christopher Game says: September 5, 2016 at 6:38 PM
“Dont worry about the ultra-simplified version. It is, as advertised, ultra-simplified. Read Kirchhoffs 1860 version.”
Such grandiose incompetent posturing!
Please identify when and where any higher luminosity receptor can ever detect, observe, measure anything exhibiting only lower luminosity?
This same idea holds for the radiometric. The higher radiance matter simply has no radiometric flux dispatched in its direction from any lower radiance matter. None at all has ever been detected!
Dear Dr. Spencer,
I have a lot of respect for you and appreciation for what you have done in keeping the skeptical science alive. I hope that you keep it alive for it will be around for years to come, it is important Sir.
The best to answer your question is the maker of the IR thermometer that you are using. They are the experts.
Dr. Spencer,
Nabil is righT on the money!! Read my reply here:
Will Janoschka says:
September 4, 2016 at 5:58 PM
Nabil Swedan says: August 28, 2016 at 8:40 PM
This is a fair and square experiment that met its realistic objectives.
Yes indeed objective Demonstrate that thermal flux from any surface is NOT but proportional to only some power of its own surface temperature, but always must be limited by any opposing field strength (radiance) at each frequency and in every direction. Gustav Kirchhoff. (much more above)
It is the fascist driven US academics that pound such nonsense into the heads of innocent children> Such must stop! Some semblance of science must be restored, by one means or another!
“Consequently, the energy striking the hot plate should increase progressively with time, and the temperature of the hot plate should increase indefinitely”
I don’t follow. If the hot plates temperature increases it will radiate more (T^4!)–until it will reaches a steady state with equal incoming and outgoing.
Why do you persist in objecting to the simplest and most logical and BTW most physical explanation? Occams razor!
“I dont follow. If the hot plates temperature increases it will radiate more (T^4!)until it will reaches a steady state with equal incoming and outgoing.”
Nate, this is just theoretical and I have no answer to it. It appears that no one has ever addressed it before. When the hot plate radiates more, the cold plate will receive more energy and radiates back to the hot plate more and that causes progressive increase in the temperature of both plates. Again this is just pure theoretical for theoretical physicists to address. I do not think that at the engineering level it can be answered for it would be impossible to eliminate the effect of surrounding air, walls, and equipment.
Nabil,
It is straight forward to answer for simplest situation.
Put everything in vacuum close together. Insulate the back side of the hot plate. The hot plate (area A)is heated with a constant power input, P. Maintain the cold plate near 0 K with a cooler. The hot plate will reach a steady temp,Th, with A*sigmaTh^4 = P.
Insert a plate in between. It will reach a temp, Tm such that 2Tm^4 =0+Th^4 (since it is emitting on 2 sides and one side absorbs 0). Hence Tm^4 =1/2 Th^4
Meanwhile Th will increase a bit, since it now receives heat from Tm. Its input heat is P+ AsigmaTm^4 = P +1/2 AsigmaTh^4, and output is still AsigmaTh^4.
Setting these equal, it reaches a temp Th, such that 1/2 AsigmaTh^4 =P.
I.e. its temp is 2^(1/4)x or 20% higher (in K)
Everything remains finite. If the cold plate is not at 0 everything still remains finite. Work it out
Yes, Nabil, radiation is energy, but the temperature of the hot plate does not increase indefinitely. Work through Stef-Bolt on your calculator which should be piece of cake for you since you have classic engineering education, as do I, and I know we covered this topic rather thoroughly.
I would just like to make the point that the greenhouse effect states that the surface of the earth is warmer than it would otherwise be without downward IR so showing that there are absolute differences in temperature between places does not disprove the greenhouse effect. The greenhouse effect depends on assumptions made of how the earths heat balance works and these assumptions may not be correct. Why does a sky that is full of rain look black does this not lead us to the conclusion that less visible radiation is getting to the surface?
Pretty much but not generally, it is relative. Beyond a certain rain cloud optical thickness, the diffuse irradiance decreases, and a sufficiently optically thick rain cloud can be darker than the other clouds. Also, the background illumination affects your eye’s judgement so the amount darkness perceived is relative.
An example of dark non-rain clouds sometimes can be seen well after sunset. Low lying clouds not illuminated by direct sunlight may be dark in the faint twilight sky.
— don penman says:
August 29, 2016 at 12:56 AM
I would just like to make the point that the greenhouse effect states that the surface of the earth is warmer than it would otherwise be without downward IR so showing that there are absolute differences in temperature between places does not disprove the greenhouse effect.–
Well that is not theory, it’s not even beginning of hypothesis.
And earth without an atmosphere would certainly be hotter in daylight. Of course it lack air, what hotter is the ground surface-
so hotter ground surface vs ground surface with atmosphere.
The Greenhouse Effect theory states without atmosphere Earth’s average temperature would be 5 C. Or were earth an ideal blackbody
it would have not an average temperature of 5 C but rather an uniform temperature of 5 C. Everywhere the surface temperature is 5 C: daytime, nighttime, being in the 6 months of night at one of the poles. It also doesn’t matter if such a earth had rotation or what angle the tilt of any rotation was.
This ideal blackbody is a thought experiment, or no such thing is actually possible. It’s imagining the Moon was made from cheese.
And what kind of cheese does not matter, as there are not kinds of ideal blackbodies.
Then the boneheads added stuff to to this ideal blackbody and arrived at silly idea that an ideal blackbody with their add-ons
would have uniform temperature of -17 or -18 C.
Let me explain, an ideal blackbody would [if existed] be the perfect refrigerator. Imagine taking a million square miles lunar surface which was 120 C in daylight, and reducing the temperature to instead being 5 C [about the temperature of your refrigerator]. With side benefit of making the Moon “disappear”- unless you can see is infrared.
So these monkeys thought that through their combined genius level intelligence, that could make an even better refrigerator which had average temperature of -18 C [average or uniform is same thing in their monkey minds].
Once they make up this model, all they could think of which could warm it back up to 15 C was by added greenhouse gases.
Or 33 C [or K] of temperature is added by greenhouse gases- and most of warming by water vapor.
So according to GHE theory greenhouse gases don’t merely add some warming. Instead the greenhouse gases precisely add 33 C.
Now there is no such precise of what each greenhouse gas adds,
but the total effect of all greenhouse gases must equal an addition of 33 C to the contrived super perfect planet which otherwise would have average temperature of about -17 C.
And then later monkey was born, who thought that without CO2, earth would have average temperature of -17 C. “Reason” being that CO2 is a forcing greenhouse gas which strictly controls water vapor.
See, wiki:
https://en.wikipedia.org/wiki/Greenhouse_effect
Oh and wiki controlled by true believer of this theory
and idea it’s description does not precisely align
to their gospel, is excessively unrealistic.
Roy you say correctly…
“The phenomenon can only happen, though, if the cool object replaces something that is even colder, and thereby reduces the rate at which the warm object loses infrared energy to its surroundings.”
This is a very important caveat.
It can be easily overlooked and lead to an incorrect conclusion that colder objects must make a warmer object warmer by their very presence and the radiation they emit.
In an ‘ideal’ setup where two objects, one hot and one colder are isolated from the rest of the universe a colder object will always make a warmer object colder still.
For practical purposes this can be demonstrated (if required) by a polystyrene box (e.g.used to carry bulk chemicals) lined with highly reflective metal film.
In the box two objects one hotter than the other are placed and the box closed.
The hotter object will drop in temperature and the colder object increase in temperature until thermal equilibrium equilibrium is reached when the flow of heat will stop.
The internal energy lost by the hotter object will exactly equal the internal energy gain of the colder object.
This then illustrates why all physics textbooks say heat transfer only occurs spontaneously from a higher to a lower temperature.
Back in the real world surrounded by space at nearly zero Kelvin all your experiments draw valid conclusions.
Anyone who denies that two way radiative flows exist between the hotter and colder objects and that these radiative flows are absorbed must ignore all modern physics textbooks.
However the net of the radiative flows which is called heat transfer is a one way process and can be conveniently dealt with as such.
Roy, congratulations on performing an actual experiment. My objection to the previous two postings is that they were simply mind experiments which prove nothing and illicit a free-for-all of speculative comments.
Before performing an experiment it is always advisable to have some idea of what magnitude of temperature variation to expect, based on theoretical calculation. This gives a guide to whether the measured results are spurious or genuine
The ice is generally below 32 deg.C . Well, I should hope so. Did you mean minus 32 deg.C ? Even so, minus 32 deg.C is very cold for a freezer temperature; are you sure that is right?
..and the heated plate has a temperature of 110 deg.F. Why are we mixing degrees Celsius with some other archaic not-used-in-science temperature scale?
If the ice is generally at plus or minus 32 deg.C, why does the FLIR indicate only 24.1 degree F when looking at the ice? Or was the FLIR picture taken this late in the day when the ice has warmed considerably?
MikeB says:
August 29, 2016 at 3:58 AM
Roy, congratulations on performing an actual experiment. My objection to the previous two postings is that they were simply mind experiments which prove nothing and illicit a free-for-all of speculative comments.
B I N G O !
that was a typo…I meant 32 deg. F. Fixed now.
Hi Roy,
I am impressed by your persistence (and by your daughter’s beauty!). But neither this experiment, nor any other actual experiment will influence those who are hopelessly lost. Had they a bit of rationality, a thought experiment would be enough. But they don’t, so no experiment, thought or otherwise, will change their minds.
Roy W. Spencer, Ph. D. says:
August 28, 2016 at 3:15 PM
“EVERYTHING emits infrared radiation”
Dr. Spencer, I read somewhere that molecules that can receive certain longwave radiation, can/will emit the same longwave radiation also. But I also understood that N2 and O2 are no greenhouse gases. So O2 and N2 don’t emit infrared radiation? An exemption to the above sentence? Or do O2 and N2 emit, but they can’t receive photons?
(I am not well schooled in physics, I am just trying to understand how things work)
I recall looking this up once…apparently O2 and N2 do absorb/emit IR, but at such low levels that it can be ignored in atmospheric computations compared to water vapor, CO2, methane, etc.
I understand there are good emitters and less good / bad emitters. And the same for absorbers.
Another 2 questions:
1. by what material / molecules is radiation (photons) blocked?
2. by what material / molecules is radiation (photons) just reflected? (and not absorbed and re-emitted)
Infrared radiation impinging on a gas can either be transmitted, absorbed, or reflected. Reflection is essentially zero for gases, so that just leaves transmission and absorption. “Blocked” has no meaning…the energy has to go somewhere…so “blocked” probably just means “absorbed”.
Thanks for the answer, Dr. Spencer. In radiation budget figures as in http://scied.ucar.edu/longcontent/energy-budget, reflection is mentioned for the atmosphere, but perhaps that’s reflection by aerosols, though it is a high number: “[Incoming solar radiation] Reflected by Clouds and Atmosphere 79 W m2”.
Gas molecules do not reflect radiation the way a mirror does (called specular reflection), but they do scatter radiation, the way snow and clouds do. It is called Rayleigh scattering and is most effective for short wavelengths. That is why the sky is blue. Scattering is often referred to as reflection, especially when the thing of interest is energy balance. Scattering can usually be ignored over short distances, but can be significant for the atmosphere as a whole. As Wim Rost says, there is also scattering from aerosol particles; that is relatively more important at longer wavelengths.
Correct.
Dr Spencer,
Clouds reflect light of many frequencies, from UV to IR, and beyond. Millimeter radars are specifically designed to reflect from cloud particles, whereas wavelengths out to say 10 cm are reflected by larger particles of ice of liquid water.
Snow likewise is highly reflective, as are gases where there is a refraction boundary. Mirages, and radio wave sub refraction are similar to the phenomenon of total internal reflection.
Any radiation absorbed by a gas will be re emitted if it causes an increase in temperature, obviously. Without external energy input, all gases will continue to emit EMR until they reach absolute zero. CO2 is no different to other gases in this regard.
I apologise for appearing picky, but some people appear to have odd ideas about heat, energy, temperature and a few other things. I would appreciate correction if I have erred in fact. My explanations may have been too economical, and in need of expansion.
Cheers.
“Any radiation absorbed by a gas will be re emitted if it causes an increase in temperature, obviously.”
Gas absorbing radiation doesn’t need to increase it’s temperature-
though could be definition problem. Anything which actually absorb
radiant energy should increase it’s temperature and emit according to this higher temperature. Or re-radiating the energy of radiation doesn’t involve a change in temperature.
Or cloud in space can re-radiate X-ray and it does not mean gas is at temperature needs to emit X-rays- impossible in vacuum of space.
Though around blackhole- possible [or typical].
– Roy W. Spencer, Ph. D. says:
August 29, 2016 at 8:50 AM
Infrared radiation impinging on a gas can either be transmitted, absorbed, or reflected.–
If gas is transparent to radiation it can alter the radiation’s path- diffuse/reflect.
A gas which absorbs a wavelength can re-radiate which can be sort of like being diffused/reflected.
Dr Spencer,
Tyndall found at particular pressure, and particular IR wavelengths, CO2 absorbed IR 750 times as well as air, O2 or N2.
There are approximately 4 parts CO2 for every 9996 parts of air. This means that for every 2000 rays of IR intercepted by CO2, 10000 or so are intercepted by the rest of the air.
Air can obviously be heated by the application of heat directly, by compression, by friction, and so on. No specific frequencies of IR radiation are required. As air cools, it emits photons of ever decreasing energy, not limited to a particular wavelength. CO2 behaves in like manner, as does all matter.
I assumed that physicists know all this. Maybe my assumption was incorrect.
Cheers.
yes, I recently looked at Tyndall’s experimental setup, and it looks like it probably worked as advertised.
Of course, in recent decades thousands of spectroscopic measurements have been made in the laboratory for different gas temperatures and at different IR wavelengths. This information is then used in radiative codes such as HITRAN.
But the absorption by CO2 very definitely is wavelength dependent, especially at low pressures. At pressures closer to 1 bar, pressure broadening causes the absorption to spread out in wavelength around the absorption lines.
Then, the laboratory measurements, when included in radiative transfer calculations, produce brightness temperatures which agree pretty well with both ground based, upward viewing IR spectrometers, and spaceborne ones looking downward.
But, for some reason, none of that makes much difference to those who claim there is no atmospheric “greenhouse effect”.
Mike Flynn says:
August 29, 2016 at 10:57 AM
“There are approximately 4 parts CO2 for every 9996 parts of air. This means that for every 2000 rays of IR intercepted by CO2, 10000 or so are intercepted by the rest of the air.”
Dr. Spencer, do I understand your answer well in the way that the results of Tyndalls experiment are to be neglected?
If not, O2 and N2 indeed should diminish strongly the relative importance of the known greenhouse gases.
“But, for some reason, none of that makes much difference to those who claim there is no atmospheric greenhouse effect.”
A reason is not all of greenhouse effect is caused by trace greenhouse gases.
Or quite simply an actual greenhouse is not warmed by gas greenhouse gases.
Or the greenhouse effect theory is wrong- if partial sort of right- then it’s invalid.
Or greenhouse effect theory does not help understand why Venus is hot or why Mars is cold.
The greenhouse effect theory gives us uncertain of 1.5 to 4.6 [maybe 6] C per doubling of CO2.
The greenhouse effect theory can not even tell what world looks like if it was 6 C warmer.
It’s scientifically invalid, just as Creationism is scientifically
invalid. I think if science was that Earth is 6000 year old- that wrecks science. If you to believe it’s 6000 years old, or doubling Co2 is a threat to our world- that can be your religion.
One have any religion which you like, But not if you insist it’s scientific- or it’s proven to be true rather than a matter of faith.
Wim Rost,
O2 and N2 both emit and absorb IR, but very, very, very little. Both are effectively “transparent” to IR wavelengths. CO2 absorbs and emits strongly at certain wavelengths, but the one (actually a narrow range of wavelengths) that matters most for Earth’s energy balance is between 14 and 15 microns. That wavelength coresponds to a specific energy state transition (a quantum transition) for the CO2 molecule. The reason CO2 absorbs, while N2 and O2 are essentially transparent, is that carbon-oxygen bonds are “polarized” (the oxygens have a partial negative charge, and the carbon has a partial positive charge) but oxygen and nitrogen are not polarized. The polarity of CO2 allows it to absorb a photon or emit a photon, which can be thought of as a rapidly oscilating electro-magnetic field; the “electomagnetic field” of the proton interacts with the molecule’s polarized bonds. I hope this helps.
Thanks Steve Fitzpatrick, clear. But there are still a lot of things I don’t understand. O2 and N2 are effectively transparant for IR wavelengths as you say. But what happens with backradiation to the surface? Our surface isn’t (totally) made up by good LW absorbers -like CO2 and H2O – and if our rocks would be mostly transparant for LW radiation, they should be heated up deeply. Which is not. So, what happens at every square meter at the surface in reality with the enormous quantity of back radiation? (more absorbing of back radiation Wm-2 than by insolation in the earth’ energy balance schemes)
Wim,
I am not sure I understand what you are suggesting. Earth’s surfaces are mostly very good IR absorbers, including rocks, plants, water, clouds, etc. They are also very good IR emitters. But in any case, there is no possibility that an object will ever absorb more IR radiation from a cooler object than it emits toward that cooler object…. there will always be more IR flowing from the warmer object to the cooler object, even though both objects are simultaneously emitting and receiving IR. Net radiative energy flow is always from warmer to cooler. The backradiation from the atmosphere is real, it is just always smaller than the opposite radiation from a warmer surface. May I suggest you do some reading on this subject if you want to understand it?
Steve is correct in his description.
Hello Steve, thanks for the explanations. You are explaining well. I did do some reading, but it is still difficult to imagine what is happening in the atmosphere and to imagine how it works exactly. When I look at the global energy flow’s in an Earth Energy Budget figure and I see that 1m2 surface is only absorbing 161 W/m2 of solar heat but that same square meter is absorbing 333 W/m2 of back radiation as well, it is difficult to have an image about what is happening. The sun’s energy we all can feel. But back radiation?
Even when you know that back radiation will be there 24 hours a day and sunshine just 12 hours, it is difficult to imagine that there is a source of energy (backward radiation) which we cannot feel directly but still should be there and also has an energy flow (333 W/m2) to the surface that is stronger(!) than we get from the sun: 333/2 = 166,5 W/m2. The more while it is an average and cloudy situations should expose us to a higher than average back radiation energy flow. Should be very hot, at least during day time when sun’s energy and back radiation are combined.
When you keep yourself on a sunny day in the shade from the sun, you will experience that the sun is blocked: much cooler.
How can we block the ‘back radiation’ and experience a cooler situation? Or can’t we block a part of the surface or ourselves for back radiation and be ‘in the shade of back radiation’? For me and millions of other people that should be the best proof that backward radiation is there and that it is that strong.
P.S. thinking about the infrared heater we had in the bathroom, half a century ago, we experienced the heating by infrared very well. I admit that the temperature of the source has been much higher which could make a difference. But it was possible to experience that infrared radiation and it was possible to hide yourself for it as well. And I don’t have the feeling that I can hide for atmospheric back radiation. And in the same time I can’t experience atmospheric back radiation very well and certainly not in a way which is stronger than I experience the direct energy of the sun.
All I know is that cloudy nights generally are warmer than clear sky nights. But that could be by ‘insulation’ by clouds as well. Clouds acting as Styrofoam. Keeping the warmth where it is.
Why doesn’t an experiment exists that proofs us – lets us feel – the coolness of a situation without back radiation?
Our skin is most sensitive to relatively high levels of IR radiation. But on a clear evening, try this:
Place your flat hand horizontally so your palm faces the ground, and the back of your hand faces the sky.
Now, flip your hand over…then back again. You should be able to feel the greater IR cooling effects of the sky-facing portion of your hand versus the ground facing portion. It happens immediately.
To answer your last question, there is no real-world source of absolute zero radiation, other than deep space. So, we can’t really experience the lack of sky radiation….although it IS taken into account in the design of spacecraft and spacecraft instruments…instruments not facing the Earth or the sun can get too cold for them to operate when exposed to deep space for too long.
Mr. Spencer, I can imagine that it is as you say, that the upward part of my hands will feel cooler than the downward facing part. But, I will think that I just feel the upward Outgoing Longwave Radiation from the surface. Just 63 W/m-2 from down. Of course it can also be 396 W/m-2 from down and 333 W/m-2 back radiation from above, but I will only feel something like a difference of 63 W/m-2.
But I have seen that the Desert Rock DATA show downward radiation. I suppose the downward radiation as such is really measured by the instruments. Am I indeed right in thinking that the downward radiation directly is measured by instruments? And not is ‘modelled’?
Wim:
You have spent almost all of your life in an ambient field of 300 to 450 W/m2 of longwave infrared radiation. If you are sitting in a room at “room temperature” as you read this, the surfaces of the room are bathing you in over 400 W/m2 of LWIR.
Because you are virtually always in this type of environment, it is easy to think of this as “nothing”. But if you are observant, you can find cases where you can feel the difference.
Go into your kitchen. First put your face close to a wall of the kitchen to see what it feels like. Now, open the door to your freezer. The cold air will immediately fall out to the floor of your kitchen (it’s fun to watch on an infrared camera!). Fan the air at the freezer a little if you want so that the air there is pretty much at room temperature.
Now stick your face where the closed door of the freezer would be. You can “feel the cold”, especially on your cheeks, which are very sensitive. The difference is the reduced “back radiation” from the objects and walls of the freezer, only about 250 W/m2 compared to 440 W/m2 from the room walls.
Wim Rost,
Regarding the measured downwelling radiation, or backradiaiton, it is done by pyrgeometers. They are in essence a set of thermocouples or thermopile. The displayed radiation is not measured, it is calculated using the recommended World Meteorological Organization pyrgeometer calibration equation.
E↓=V/K+5.67 x 10-8 T4d=β
Please see
World Meteorological Organization, 2006: Guide to Meteorological Instruments Methods and Observations, Seventh ed. WMO-No. 8, Secretariat of the World Meteorological Organization-Geneva, Switzerland, Chapter 7, p. 1.7-20.
Also see examples instruction manuals of The Eppley Laboratory, Inc. or Kipp & Zonen type pyrgeometers. They use the above calibration equation.
Pyrgeometers are not radiation sensors in that thermocouples chemical and physical properties do not change with intensity of radiation.
To answer your question, yes backradiation is calculated based on (WMO) calibration equation; it is not measured. therefore, the displayed backradiaiton depends on the correctness of the WMO calibration equation.
Nabil, likewise the temperature readout on a mercury or coil spring thermometer is calculated and depends on the correctness of the calibration. All radiometers, all barometers, all thermometers calculate their reading & depend on correctness of calibration. We manage under these circumstances to get by making useful products and inventions that advance science & our pursuit of happiness.
What if the calibration equation is wrong?
All calibrations are “wrong” in an absolute sense…but how could 300-350 W/m2 of sky radiation be due to a “calibration error”? -Roy
Thank you, Nabil Swedan
The thermometer will then read out a temperature other than 32F for a glass of ice water and other than 212F for boiling water. Sales will be poor.
Dr. Spenser,
A related paper is currently being discussed and I cannot go through the details on this blog.
The correct calibration equation of the pyrgeometer should be the following:
E↓=V/K
With this, back radiation at night should be -200 w/m2 to zero w/m2 based on the manufacturer of the pyrgeometer conclusion. I have it in writing. This is in complete contrast of the the displayed +330 w/m2.
By the way, the whole messy calibration started by engineers and not meteorologists.
strange how studies of the calibration accuracy of the downwelling IR measurements by these devices get agreement of better than 5% between theory and measurements in something that is 350 W/m2,
http://journals.ametsoc.org/doi/pdf/10.1175/1520-0426%281998%29015%3C1229%3AANLACA%3E2.0.CO%3B2
while you claim the entire signal is non-existent. If you can ever prove this to the research community, I’ll personally nominate you for a Nobel prize.
I am not saying it, It is the conclusion of pyrgeometer maker.
Here is Nabil’s WMO p. 1.7-20 eqn. 7.15, 7.16 combined more clearly straight from the document, not sure if just Dr. Spencer’s wordpress site character translation is responsible for the difference:
E↓=L↓= V/K + epsilon*sigma*Ts^4 (WMO eqn.)
L↓= the irradiance measured by the pyrgeometer or calculated from the temperature of the cavity capping the upper receiver (W m2)
V is the output of the instrument (V)
K is sensitivity (V/(W m2))
From Dr. Spencer’s link eqn. 11:
the top of the thermopile is at temperature Ts.
Also the fundamental radiometer calibration equation (11) has another factor constant*(Ts^4-Td^4)
The dome is at temperature Td
——
Nabil’s argument then boils down to both of these second 2 factors in eqn. 11 are 0.0 in reality for:
L↓=E↓=V/K
Nabil is essentially then arguing in reality:
1) the case ts = dome td
Ok, means the calibration factor B in eqn. 11 is really 0.0, it was never needed.
2) Here is the biggie:
Nabil is arguing the WMO (epsilon*sigma*Ts^4) factor is 0.0.
Thus upward longwave irradiances at the surface of the thermopile Rup is 0.0 in Fig. 1 of Roy’s link.
FOLKS: That simply means Nabil is arguing the one stream concept is valid, the two stream is invalid which seems consistent with his other debates (and of course with Kristian’s view).
Gemmun: Let the thrashing about like a dying fish continue.
Hint: My view is two stream, WMO and Roy’s document is the right stuff.
W m2 translated is really = W/m^2
Verify my conclusion by going back to Dr. Spencer’s 8/23 top post:
“Observational Evidence of the Greenhouse Effect at Desert Rock, Nevada”
Inspect “Fig. 2. 3-minute radiation budget data at Desert Rock during July 1, 1998”:
At night atm. calibrated per WMO pyrgeometer LW DWIR shown around +330 W/m^2 and net around -100. If there is nothing to be net of (one stream view) then LW DWIR about -100.
Compare that to Nabil’s 4:16pm: “back radiation at night should be -200 w/m2 to zero….in complete contrast of the displayed +330 w/m2”
QED
—–
FWIW: a more recent 2006 NOAA calibration document:
“The results suggest such a hybrid calibration approach (blackbody and outdoor comparisons with the WISG) can produce +/-2.5 Wm-2 agreement by a field pyrgeometer.
“http://www.arm.gov/publications/proceedings/conf16/extended_abs/stoffel_t.pdf
Better:
http://www.arm.gov/publications/proceedings/conf16/extended_abs/stoffel_t.pdf
Ball4 says:
August 29, 2016 at 6:31 PM
Hint: My view is two stream, WMO and Roys document is the right stuff.
WR: Just an observation. The Potsdam Institute For Climate Impact Research shows for Potsdam measurements of the minimum air temperature near ground at 5 cm above the grass and also the standard minimum air temperature which as we know is measured mostly at 1.80m. Website https://www.pik-potsdam.de/services/climate-weather-potsdam/climate-diagrams/air-temperature-min
The graphs show:
The minimum air temperature annual mean since 2000 is around 6C
The near ground minimum air temperature annual mean since 2000 is around 4C
The Dutch Wikipedia still gives the explanation for the lower near ground minimum temperatures (to explain frost at grass height)in the same way as I ever learned: (translated) Mostly only in the early morning the temperature lowers until below freezing point, because the outward radiation is no longer compensated by upward transport of warmth from the ground.
In the above explanation for the low ground temperatures, down welling radiation is not mentioned at all. Only transport of energy from down in the ground. In case of a huge down flow of radiation (absorption by the surface (!) of 333 W/m2) it is rather strange that at 1.80m the actual average minimum temperature is a 2 degrees higher than the average minimum temperature is 5 cm above grass level. It seems that an eventual down welling radiation stopped (partly) at 1.80 m.
I dont have an explanation for the above. I cannot imagine a huge downward radiation flow and even less a downward flow that stops partly at 1.80 m. The measurement of downward radiation seems to be totally dependent on one calibration formula from an institute (WMO) that takes a position in the AGW debate. Which makes me fearful.
To paraphrase others: More research is needed.
Wim, “down welling radiation is not mentioned at all.”
Experience shows frost, dew form more easily on clear, still nights than cloudy, windy nights. That should give you a hint that indeed the downwelling radiation should be mentioned.
Ball4 says:
August 30, 2016 at 4:53 PM
Experience shows frost, dew form more easily on clear, still nights than cloudy, windy nights. That should give you a hint that indeed the downwelling radiation should be mentioned.
No Ball4, wind means advection from energy from elsewhere. That disturbs the no wind process.
Which reduces the chance for frost, dew as I wrote. On a clear night the DWIR is measured lower by ESRL than a cloudy night enhancing the chances for frost, dew as the surface T becomes less than the air T. As I wrote, cloudy, clear DWIR should be mentioned as a factor along with wind.
Roy W. Spencer, Ph. D. says: August 29, 2016 at 4:56 PM
“strange how studies of the calibration accuracy of the downwelling IR measurements by these devices get agreement of better than 5% between theory and measurements in something that is 350 W/m2,”
Again and again. Just like your experiment here The Pyrgeometers ‘measures’ the reduction in thermal radiative exitance from the instrument to space by the “radiance” of that atmosphere on W/(m^2 sr) never any downwelling flux in W/m^2. Why don’t you go ask the manufacturer not your obviously incompetent research community.
“while you claim the entire signal is non-existent. If you can ever prove this to the research community, Ill personally nominate you for a Nobel prize.”
The field strength signal is clearly evident, it is the flux or power transfer that is clearly absent.. Such truism cannot be demonstrated to someone that believes they know otherwise. You have been presented with alternate points of view that clearly all that you intended in a manner that violates no physical principle.
Steve Fitzpatrick says: ugust 29, 2016 at 6:40 AM
Wim Rost,
“O2 and N2 both emit and absorb IR, but very, very, very little. Both are effectively transparent to IR wavelengths. CO2 absorbs and emits strongly at certain wavelengths, but the one (actually a narrow range of wavelengths) that matters most for Earths energy balance is between 14 and 15 microns. That wavelength coresponds to a specific energy state transition (a quantum transition) for the CO2 molecule.”
What total bull shit! Airborne H2O molecules of 4% by mass of the atmosphere in all 5 phases continuously create EMR exit flux to space from 3 microns to 200 microns wavelength. Airborne CO2 molecules as only a gas of 0.06% by mass contribute nothing whatsoever to atmospheric exit flux. Surface exit flux is even lower than that of CO2.
Earthlings will never know what determines surface temperature until the determination of control of atmospheric column water is understood. All of the rest of CAGW nonsense is a deliberate intentional scam for monetary or political profit!
Wim Rost,
Good reflectors are many metals with smooth surfaces (aluminum, silver, others). Metals are actually extremely strong absorbers of IR and visible photons, but they are such good electrical conductors that an absorbed photon is essentially ‘regenerated’ due to electron movement at the metal’s surface. The conductivity of the metal essentially make them reflect all arriving photons, but only if the surface is smooth. Very finely metal particles (much smaller than the wavelength of an arriving photon) are not reflective at all.
Steve Fitzpatrick, a mirror has a metal coating and this coating reflects short wave radiation – visible light. As I imagine, the visible light is not entering the metal. For LW as you describe above, I understand that a photon is very easily absorbed and directly emitted by a smooth metal – without heating the metal. So in fact entering the metal and leaving it. Am I right?
And going back to the rocks at the surface, they are very amorf, not smooth. Am I right that they mostly don’t reflect but absorb the back radiated photons?
(And if so, couldn’t we construct LW panels to collect all that energy (like with sun panels) as someone proposed somewhere? Perhaps this is a ‘stupid question’ for someone who knows about physics, but I am just interested in the answer because it can help me to understand what is really happening)
Hmmm, perhaps an experiment? Can we make a box which we put on a part of the surface. The only ‘open side’ down. We should construct it in a way that photons can go out (for example by reflection outward via mirrors by small openings / tunnels) and that no photons can go in. If seen from the emitting clouds the total surface of the box would have to be covered with reflecting metal, for example smooth aluminium. Then at night the temperature within the box would have to go down compared to the surroundings, as the surface outside the box can receive backward radiation while the surface inside the box which effectively is protected against backward radiation can not.
Wim Rost,
Yes, most mirrors have a metal coating, usually on glass. The glass has a very smooth surface, so a very thin metal coating on glass also has a very smooth surface, which is required for reflection. You can get the same smoothness via polishing, but it is more difficult (and expensive), so people use glass substrates for most mirrors. There is no difference in reflection between an infrared photon and a visible photon for many metals. Polished aluminim has excellent reflection at all wavelengths from UV through long radio waves. The reflection is in fact just a re-emission of a new photon as the original one is absorbed. The absorbing photon induces electron movement in the metal, and that movement regenerates the ‘reflected photon”.
Most surfaces (water, rocks, Roy’s paint) absorb strongly, but do not reflect much IR. These materials also emit photons with a distribution of wavelengths that depends (mostly) on their temperature… strong IR absorbers are also strong IR emitters, if they were not, then you could just place them in a vacuum and they would accumulate heat indefinitely, until they were much warmer than the surrounding materials… not possible; the net radiative flow is always from warmer to cooler (second law of thermodynamics). Were it possible, then perpetual motion would be a reality.
Wim,
I forgot to add: perfect reflectors do not emit photons, they only refect photons which arrive from elsewhere. If I remember right, this is called “Kirkov’s Law”. If a perfect reflector could emit photons as a ‘blackbody’, then placing a perfect mirror in a vacuum would cause it’s temperature to drop to absolute zero… another impossibility. Net radiative flow must always be from a warmer body to a cooler body, never the other way around.
How thick is the hot plate?
Metal blocks loosing heat are cooler on the surface than internally. My guess is that when you slow the radiant heat loss, the metal surface warms because you decrease the temperature gradient inside the metal.
It’s probably an artifact. There’s no increase in total heat content of the hot plate, just a transient increase in surface temperature. And that increase is coming from inside the hot plate itself.
as I posted, the heated sheet is aluminum flashing, 0.016 inch thick I believe, painted flat black. Yes, the surface facing cooler objects will itself be cooler than the insulated back of the flashing where I’m monitoring temperatures. But it doesn’t matter for the experiment. Any change in the energy budget at the surface affects the temperature thoughout the thickness of the sheet, which happens pretty fast as you can see (especially in Fig. 4) since aluminum has high thermal conductivity and it is quite thin.
OK, yes, of course. Thanks.
As you say, it doesn’t matter, but …
The front of the metal sheet would be the HOTTEST. The front of the plate is where the incoming energy is absorbed. Heat flows out from there — via radiation from the front and via conduction toward the back. For heat to flow back, there must be a temperature gradient with the front hottest.
A thermometer behind the plate will be (marginally) cooler than the front surface, not (marginally) warmer. The difference should be quite minimal (especially with thick insulation around the back of the whole thing and good thermal contact between the sheet and thermometer). [If you switched and used a heating pad behind the sheet, then the back would end up (marginally) warmer than the front.]
Very nice experiment. Its really interesting the effect you get most of the time of adding the room temperature shield and getting a sudden cooling before the warming sets in. Must be the room temperature shield surface warming. The ice surface may already be warm (though you might have trouble measuring it because of a loss of heat as soon as you touch it) or reflecting more than the room temperature shield.
It would really be interesting to start using different materials in a comparison test. Like does a more conductive room temperature shield exhibit more sudden cooling and does it warm as much. Then maybe some transparent shields.
One issue raised by G&T was the insulation value of the CO2. In solving engineering problems for heat loss one must continue the calculation through the barriers, figuring insulation value and eventually finally into a space that will be unaffected by the heat source. . . .like the outdoors or a massive room for equipment. In the simplified greenhouse calculations its assumed to be 50/50. The cooling spikes I mention above gives a hint that it might not be.
As I mentioned in the post, I believe the sudden cooling spikes were due to unavoidable air movements past the hot plate when I put the shield into place over the ice. A plastic wrap cover over the hot shield (Saran Wrap has an IR transmission of about 90%) should reduce if not eliminate the effect.
So, based upon some good comments here, improvements to the experiment would include:
1) cover the hot plate enclosure with plastic wrap to reduce the variable influence of air currents cooling the plate (the initial cold spikes seen in the temperature data).
2) use identical cold and room-temperature plates (again painted with high-emissivity paint) that would be swapped out, so that they are at the same distance from the hot plate. The cold one could be kept in the deep freeze. They probably need to be thicker than aluminum flashing so that they have some thermal inertia.
Dr. Spencer,
I always appreciate your explanation and simple experiments to demonstrate simple physics. Of course, as you ALWAYS point out, there are obvious improvements to the set-up and execution of the experiment to improve results, but then again, it’s no longer simple, is it?
So, in short, without having to read through the comments and responses and tangential discussions, let me, as an engineer of many years with relevant understanding of thermal behavior, reiterate for my own edification what I understand you are saying…
The transfer of thermal energy from one body to another through thermal radiation is from the warmer body to the cooler body, and is a net transfer of thermal energy. This implies that both bodies are radiating thermal energy based on their temperatures at any given time. The only body that is not radiating thermal energy is one at absolute zero…deep space, or as in your simple experiments, a nice clear night sky (or nearly so).
I suggest this thought experiment (no hardware or equations required)… Three identical objects in a vacuum arranged in an equidistant triangular pattern, with temperatures of Object A > Object B > Object C > absolute zero. Obviously, A is radiating thermal energy to both B and C. And, B is radiating to C. But if B is radiating to C, doesn’t it stand to reason B is also radiating to A?
Another words, an object of a lower temperature is not aware of the temperature of the objects around it, it just radiates thermal energy based on its temperature and material properties. It is the direction of net energy transfer that implies flow only in one direction.
Thank you for your time and consideration….
objects A, B, and C are all radiating in all directions, so they are all interacting (as long as they can “see” each other). I basically agree with your assessment (but the sky radiation at the surface, depending on air mass humidity and temperature, is emitting at an effective temperature around 275 K in the global average, not 0 K)
PC…”The transfer of thermal energy from one body to another through thermal radiation is from the warmer body to the cooler body, and is a net transfer of thermal energy”.
Can you explain why heat from a cooler body, even if it is an independent radiator, can be transferred to a hotter body? Do you have concrete proof?
It makes no sense physically. If each body could warm the other their temperatures would rise till they were out of bounds. It’s called perpetual motion.
So…is that true within an iron rod, heated at one end by a flame? Are you claiming there is a net transfer of thermal energy and are you claiming radiation has anything to do with it?
With regard to the rod, Clausius stated that heat can NEVER, of it own, be transferred from a cooler body to a warmer body. He claimed that applied to radiation equally. In your statement, you are not talking about net heat transfer, you are talking about net EM transfer, although I balk at the term transfer.
When Clausius stated that two bodies could never give more heat to the other than they receive, he was talking about a situation in which an external mechanism was in place to supply bidirectional heat transfer. He was very clear that such a process could never happen on its own.
It does not apply to the surface-atmosphere interface where IR from the surface is allegedly warming GHGs in the atmosphere. I say allegedly because I don’t think it has ever been measured directly. In the surface-atmosphere model, net heat transfer does not apply. Rather, the statement of Clausius applies, that heat can NEVER be transferred from the cooler atmosphere to the warmer surface THAT WARMED IT.
Gordon Robertson says:
“Can you explain why heat from a cooler body, even if it is an independent radiator, can be transferred to a hotter body? Do you have concrete proof?
It makes no sense physically. If each body could warm the other their temperatures would rise till they were out of bounds. Its called perpetual motion.”
Lets say the warmer and colder objects are each placed in adiabatic walled boxes adjacent to each other.
To simplify conduction and convection issues lets say there is a vacuum in each box
Each will radiate towards the perfectly reflective walls but the radiation will bounce back and the temperature would stay the same.
Now remove one adiabatic wall so that the hot and colder objects now share the same adiabatic box and can ‘see’ each other.
The more intense radiation leaving the hotter object will be absorbed by the colder object the colder object will radiate less intensively to the warmer object
Now even though the warmer object absorbs the radiation from the colder object it itself is emitting far more radiation and its internal energy drops.
Conversely the colder object gains more radiation than it emits and its internal energy rises.
This process will continue until both are at the same temperature.
The internal energy lost by the warmer object will equal the internal energy gain of the colder object
This is a one way heat exchange involving a two way exchange of energy
Gordon Robertson says:
“Can you explain why heat from a cooler body, even if it is an independent radiator, can be transferred to a hotter body? Do you have concrete proof?
It makes no sense physically. If each body could warm the other their temperatures would rise till they were out of bounds. Its called perpetual motion.”
Gordon, “perpetual motion” would occur if there is kinetic energy which has absolutely no dissipation of energy from friction. Even if such a thing did happen, energy would still be conserved. This is irrelevant to the situation being discussed.
Your statement is like saying putting a lid on a heated pot of water making the water warmer still is somehow going to lead to a runaway temperature rise in the water. Of course not. All you are doing is making the water less able to lose heat, so its temperature rises until energy balance (rate of energy gain = rate of energy loss) is once again restored.
It doesn’t matter whether the lid is acting on radiation, conduction, convection, or all three. The energy transfer concept is the same.
Gordon Robertson,
No, I never implied a “perpetual motion” or run-away system…sorry if I wasn’t clear enough and you interpreted it as such.
Consider a transient thermal system of two spherical objects A and B some distance apart in space, both at the some initial uniform temperature and with equivalent internal heat generation. We agree that both objects are radiating energy to space, correct?
But each object is also radiating energy in the direction of the other, but since both are at the same initial temperature (and have the same internal heat source) the net heat transfer is zero.
It also stands to reason a portion of each object’s surface in view of the other will be at a higher temperature than the “dark-side” of each object?…which by the way I have demonstrated with a simple FEA model, and it is not a run-away model, it does eventually reach equilibrium.
So, is the local temperature rise on each surface due to the radiation of object A to B and B to A even though the “net” energy transfer due to thermal radiation is zero? Or, maybe put another way, each A and B is shadowing the others’ radiation to the surroundings and the radiation heat loss to the surroundings is reduced?
Which do you prefer…or is there a better way to explain the local temperature rise on each surface?
PC says: August 29, 2016 at 10:45 AM
“Another words, an object of a lower temperature is not aware of the temperature of the objects around it, it just radiates thermal energy based on its temperature and material properties. It is the direction of net energy transfer that implies flow only in one direction.”
You state: “it just radiates thermal energy based on its temperature and material properties”. Can you say where you got such an idea, and why you think that would ever be true/correct? Can you say what you mean by ‘radiates thermal energy’? What is being radiated? Does this involve some power transfer? Why? Why would not every thing increase in temperature forever?
‘
It can be shown mathematically a cool body cannot heat a warm body, but the cool body can slow the rate at which the warm body loses heat.
q = ε σ (Th4 – Tc4) Ac
where:
q = heat transfer per unit time (W)
ε = emissivity of the object (one for a black body)
σ = 5.6703 10^-8 is the Stefan-Boltzmann Constant
Th = hot body absolute temperature (K)
Tc = cold surroundings absolute temperature (K)
Ac = area of the object (m2)
Increasing the atmosphere 0.97 C reduces the surface’s radiative cooling rate by 0.38%.
Assuming: εσ(Th4-Tc4)Ac / εσTh4Ac = 0.0038 = 0.38%
JDAM, i’m glad you bring this up, because it begins to get at the heart of the disagreement….
Your example correctly shows that the rate of net IR flow between two objects with different temperatures (Th, Tc) will go down as the temperature difference decreases.
This is true!
But what I think you are IMPLYING is that Th cannot go UP in response to raising of the colder object’s temperature.
This is ONLY TRUE if Th isn’t being actively heated by a separate mechanism (the sun in the climate system. the flood lamp in my experiment), with its loss of energy being controlled by Tc.
When Th is being actively heated, and the whole system is in an initial state of energy equilibrium (Th and Tc constant with time), then if you RAISE the temperature of Tc, Th will ALSO be raised.
THAT is the situation being addressed by the experiment, and which exists in the real climate system.
If people would just try to understand this fundamental issue, we wouldn’t keep debating whether a cooler object can raise the temperature of a warmer object. It only happens because the warmer object has a separate source of energy (the sun), and the cooler object controls the rate at which the warm object loses energy.
If the actively-heated warmer object’s ability to lose energy is reduced, its equilibrium temperature goes up.
Sorry I didnt intend to imply that Th cannot go up its just rudimentary equation for radiative heat transfer.
Dr. Spencer,
“THAT is the situation being addressed by the experiment, and which exists in the real climate system.”
That puzzles me. When is the surface and the atmosphere in a state of energy equilibrium? The closest to that would be just before dawn when the Tc and Th might be be approximately constant with time. In this case a cloud coming over should raise the surface temperature.
What other conditions would make your case?
Chic:
This is another semantic diversion, and please don’t let people trap you into it.
Yes, it is true that for any given location on the Earth at any given time, the local conditions are almost always out of energy equilibrium…usually very far out of equilibrium. As a result, temperatures are continually changing. That’s “weather”.
What I’m talking about is the average state of the climate system as a whole, over years. For example, if we can believe the record of ocean heat content over the last several decades, the global oceans on average have been in energy balance to within about 0.2 W/m2 (out a global average energy input/output of 240 W/m2). That’s about 1 part in 1,000.
That is the energy balance we talk about in the context of climate change: global average conditions, over relatively long periods of time.
Sorry. I was trying to avoid a semantic diversion by asking a question that that I thought would clarify your point. Putting it in the context of global averages over long time periods makes it harder to identify what the cool object (Tc) is that makes the surface (Th) warmer than it would be. Just saying it’s the greenhouse effect doesn’t cut it, IMO. I think your experiment should convince reasonable people that substituting a cool object for a colder one will make the warm object warmer. I don’t think it extrapolates well to the climate system.
as I originally stated, the only intent of the experiment was the 2nd Law objection…not to prove that greenhouse gases warm the Earth’s surface. “Energetic equilibrium” applies to everything that has a constant temperature (barring phase changes or nuclear reactions).
JDAM…”Increasing the atmosphere 0.97 C reduces the surfaces radiative cooling rate by 0.38%”.
No disrespect intended, but could you put away the math and explain this is terms of reality? Or, at least, as close as possible within the framework of what we know about atomic structure?
The EM emitted by an object is obviously related to the environment in which the object is located. I have no argument with that. What I am debating is that the rate of cooling is affected by a cooler object radiating EM to the warmer object.
If you have the atmosphere, which is 99% nitrogen and oxygen, in contact with the surface at 20C near the surface, that temperature will govern the radiation of the surface.
Why?
The surface atoms radiate based on the average temperatures of the individual atoms. The temperatures of the atoms are proportional to the energy levels of the orbiting electrons, especially the valence band electrons. Those energy levels define the kinetic energy of the atoms which is their heat (thermal energy).
We all know that short wavelength EM warms the surface but the transferred heat is absorbed and redistributed to a certain depth. I haven’t seen any figures on what the actual surface temperature is of the surface on which we walk. I have seen days with certain substances where it was difficult to walk on a surface in bare feet due to intense solar radiation.
How can you raise or reduce the average kinetic energy of surface atoms?….by controlling the ambient temperature in which they exist. If you double the ambient temperature to 40C, the atoms will warm and radiate at a different frequency with greater intensity. If you drop the ambient temperature to 0C, the atomic radiation will drop in frequency and intensity.
That’s because there is a conductive and convective interaction between the surface and the atmosphere. Where in your equation does it mention the effects of conductance and convection?
If the radiation intensity is reduced, you could claim that to be a reduction in the cooling rate of the atoms.
None of that has anything to do with the cooler atmosphere (on average) radiating IR to the surface, nor has it anything to do with the atmosphere absorbing IR.
In fact, 99% of the atmosphere cannot apparently absorb IR in the range emitted by the surface. However, that 99% determines the temperature of the surface to a degree hence it’s radiation intensity. Therefore, GHGs like CO2 are having little or no effect on surface cooling.
Incandescent bulb! Oh the horror! LOL
Nice job proving the point with a scientific method Doc! Now days most people in the science community would just model it and take it as fact…..
yes, the experiment kind of relies on the energy inefficiency of incandescent bulbs.
I hear that.
The Sun is not energy efficient in our chaotic atmospheric conditions, or we would not be here discussing this. Just sayin,,,,, I wish we had a cardboard control option in the real world 😉
One of those who disagrees with me, and who I have banned from posting any more comments here, sent me an email in response to the experiment….I paraphrase…
“Hilarious! so, your experiment proves you can cook a turkey with ice cubes!”
I’ll give him the benefit of the doubt and assume he didn’t actually read the experiment results….
Of course, a much closer analogy would be that I’ve shown it’s more likely that you can cook a turkey in an oven whose walls are filled with insulation than with ice cubes.
Ha, thanks for the anecdote Dr. Spencer and all of your experiments.
I understand better now why so many universities use thermodynamics as a “washout” course for many technical majors. A lot of people just cannot grasp the concepts.
Hi Dr.Spencer,
“Hilarious! so, your experiment proves you can cook a turkey with ice cubes!”
Yes, try to cook the turkey into a thin 1mm high thermoconductive box suspended in the outer space.
Say a 2x2x2 meter box with just the turkey and a 1kW heater inside it.
The heater on one side and the turkey on the other side of the box.
The turkey probably never cook (at least on the face not exposed to the heater).
Cover the outside of the box with an another “box” thick 1 meter or more of of water ice, and it will surely cook.
Have a great day.
Massimo
P.S. You wouldn’t use that oven for your thanksgiving lunch of course, it could be hard to recover the turkey from its orbit!
“..would lose IR radiation directly to the cold depths of outer space, which is essentially at absolute zero temperature,..”
Space is an empty place containing nothing, hence, it cannot have a temperature including absolute zero. A parked car has absolute zero velocity, but a non-existent car cannot have absolute zero or any velocity at all.
Pete, I’m referring to the cosmic background microwave radiation, which emits at at effective radiating temperature of about 2.7 K. It represents a radiative energy source which is so small that it is usually assumed to be zero. The radiation comes from deep space.
pete:
The density of matter in space is extremely low, but it is not zero (“nothing”). Its extent is also incredibly vast.
Very careful measurements in all directions from earth show that “space” behaves exactly like an ideal blackbody (to within measurement accuracy) at 2.725K +/-0.001K. This is the “cosmic microwave background” radiation Dr. Spencer refers to.
The magnitude of this radiation is 3 microwatts per meter-squared (0.000003 W/m2), so it is almost always ignored in radiative balance calculations for the earth.
Energy can only be transferred from a colder source to a warmer sink by means of a refrigeration cycle, requiring mechanical work and a net input of energy.
the net input of energy is from the sun, or the flood lamp.
In a refrigerator, a compressor generates heat by compressing a gas into a liquid at high pressure, the heat is removed by a fan through radiator fins, and then the high pressure of the pressurized liquid is partially released in the evaporator resulting in a cold gas which is then used to chill the air.
If you reduce the evaporator (cold side) efficiency, I suspect you will make the compressor side even hotter, by returning a gas to it which is too warm.
Which makes my point.
Pete true, but warming the surroundings reduces the rate of radiative heat transfer (and conduction for that matter).
pete:
No! Clausius spent a great deal of effort 160 years ago anticipating confusions like yours.
He clearly stated that such “ascending heat transmission” was possible if “compensated” by a greater “descending heat transmission” from hot to cold, with nothing else involved.
(“Net” transfer from cold to hot is only possible with a work input as in a refrigeration cycle, but that is not what we are talking about here.)
Subscript test: a1
@Roy…”In a refrigerator, a compressor generates heat by compressing a gas into a liquid at high pressure, the heat is removed by a fan through radiator fins, and then the high pressure of the pressurized liquid is partially released in the evaporator resulting in a cold gas which is then used to chill the air”.
Here’s a more accurate description from Wiki (a good ways down the page):
https://en.wikipedia.org/wiki/Refrigerator
It claims the compressor converts a low pressure gas to a high pressure gas which is liquefied in the condenser.
The point is Roy, you cannot transfer heat from the cold space of the refrigerator without all that ‘external’ rigamarole. The evapourator inside the freezer compartment absorbs the warmer air in the cold space as the high pressure liquid converts back to a low pressure gas.
As Clausius, stated, heat can NEVER, of its own, be transferred from a cooler object to a warmer object. Please note as well that in the process of refrigeration, there is no transfer of heat between bodies via radiation.
yes, Gordon, that’s what I said about the compressor side of the equipment.
And I’m not the one who made the refrigeration analogy…which I think is a poor one.
And I can’t believe that we are still discussing whether “heat, of its own, being transferred from a cooler object to a warmer object” is being violated. The NET heat flow is ALWAYS from warmer than colder. But if that heat flow is temporarily reduced by increasing the temperature of the cold object, and and the same time the hot side is continuously heated by some OTHER process, then the hot side CAN increase in temperature. Surely you must understand this by now?
Instead, you just keep repeating “heat can NEVER, of its own, be transferred from a cooler object to a warmer object”. The Wikipedia entry for the 2nd Law of Thermodynamics, which should be non-controversial, specifically states that the Clausius Statement applies to the NET of whatever flows are occurring in opposite directions.
Now I have definitely understood why Gordon Robertson’s argumentation is not correct.
It was the missing link in a long chain.
Bindidon,
The “missing link” is a term often used to describe fossils that are believed to bridge the evolutionary split between higher primates such as monkeys, apes, and humans.
I think you are correct in using this term to distinguish between deniers, denialists and warmists.
Dr no: I’m not a friend of counterproductive polemics.
Everybody (I hope) understands diffusion, so think of the earth and GHGs in similar terms. Two adjacent objects at different temperatures BOTH radiate IR photons to the other. The hotter radiates more than the colder so there is a NET transfer of photons (and heat energy) from the hotter to the cooler, but the cooler still radiates energy back to the hotter. Take the cooler object away and replace it with nothing (cold space), and the hot object loses more NET photons (energy).
Some comments definitely don’t pass the bodyguard robots…
I applaud the actual science of this endeavor, but I think you might be demonstrating a different atmospheric phenomena — thermal gradient.
This configuration creates a miniature room with the columns of air stabilizing between the plate above and the ice pan below. This gradient has a lower surface temperature when the ice is uncovered and results in a lower temperature at the top plate. When the ice is covered, the gradient changes accordingly and the air near the plate is at a higher temperature.
You stated:
Roy W. Spencer, Ph. D. says:
August 29, 2016 at 8:55 AM
As I mentioned in the post, I believe the sudden cooling spikes were due to unavoidable air movements past the hot plate when I put the shield into place over the ice. A plastic wrap cover over the hot shield (Saran Wrap has an IR transmission of about 90%) should reduce if not eliminate the effect.
I also think the warming ice causes this gradient to lesson which is why your temperature band narrows as the experiment progresses.
It seems like you built a mini-atmosphere up to the tropopause.
While I am not sure it demonstrates what the title claims, I love to see actual science (I.e, experimentation) on display vs. models.
Bravo!!
” FTOP says:
August 29, 2016 at 5:43 PM
I applaud the actual science of this endeavor, but I think you might be demonstrating a different atmospheric phenomena thermal gradient.”
With all this heating one could get thermal gradient near ceiling and/or ceiling itself could be 10 C warmer than “normal”. But in such small spaces you need something to cause a gradient [like a ceiling]. One going to convective currents [or micro thermal uplift and downdraft- https://en.wikipedia.org/wiki/Vertical_draft%5D
but also going convection which doesn’t include air movement-or much air movement.
I think Dr. spencer accurately stated that a vacuum would be required to eliminate convection and ideally isolate IR.
From wiki
“In a stratified building, temperature differentials of up to 10C can be found over a height of 10 metres on average. In extreme cases, temperature differentials of 10C have been found over a height of 1 meter. In a destratified building, temperature differentials can be reduced to 1 – 2C or less from floor to ceiling.”
https://en.m.wikipedia.org/wiki/Thermal_destratification#/search
When Dr. Spencer moves the cardboard, it is like destratifying a room.
By having the warm surface high and cool surface low there’s no convection.
There can be some mixing brought on by moving the white panel, but moving it horizontally will minimize pushing air around.
“By having the warm surface high and cool surface low there’s no convection”. Does this mean that when there is a temperature inversion in the atmosphere (warm air above a cooler surface) there is no downward convection?
Pretty much. In the context of this experiment, there will be more radiation from the warm air aloft and less from the cool air below. So that effect tries to bring the two air masses to a similar temperature. Radiational cooling, primarily from the ground, will win out, cool the surface, and in turn cool the air in the inversion. Inversions can be surprisingly thin.
Here in New Hampshire, I often see good temperature traces where you can see the development of the inversion at night, and then the recovery in the morning. First the sun warms the land, which quickly warms the air in the inversion, then the air warms slower because a thicker layer of air has to be warmed, then convection continues until the windy layer above is able to mix down to the ground as the atmosphere become “neutrally buoyant” and air parcels can move up and down with very little effort.
I’ve considered what it would take to construct an experiment that would show the effect you’re demonstrating.
I had in mind something with copper sheets with water channels on the back side to keep things at constant temperature or to measure the heat gained or lost as the water wended its way – basically way too over engineered compared to your simpler, cheaper, and just as effective design. Probably more effective.
Good job.
I’m a bit disappointed that 142 comments into this that there are people who don’t understand what’s going on. Not surprised, just a bit disappointed. OTOH, I trust that a lot of people here have learned quite a bit from this post. Count it as a winning post.
— Ric Werme says:
August 29, 2016 at 7:24 PM
Ive considered what it would take to construct an experiment that would show the effect youre demonstrating.
I had in mind something with copper sheets with water channels on the back side to keep things at constant temperature or to measure the heat gained or lost as the water wended its way basically way too over engineered compared to your simpler, cheaper, and just as effective design. Probably more effective.–
Pipes rather than channels and something cold like liquid nitrogen.
Or apparently if making nitrogen [difficult and Liquid Nitrogen is
cheap and available] one instead can make “cryogenic-temperature alcohol”:
http://www.wikihow.com/Make-a-Liquid-Nitrogen-Substitute
Anyhow it’s liquid which is cold and can go thru pipes and give
controlled temperature.
I think it is important to clarify that oxygen and nitrogen do not emit or absorb, at all, in the 15 micron region where CO2 acts as a powerful greenhouse gas. That is they absorb and emit nothing, zero, nada, zilch. There is no mechanism for them to do so.
Although it is true that all warm bodies above absolute zero emit electromagnetic radiation in some part of the spectrum, unlike blackbodies which emit some radiation at every wavelength, thin gases absorb and emit radiation only at discrete wavelengths or narrow bands. Oxygen for instance will absorb and emit strongly at 0.7 microns, but this is in the visible region where it acts more to block incoming radiation from the Sun, rather than block outgoing infrared from the Earth
The Earths surface emits 95% of its radiation in the infrared region between 5 and 50 microns. In this part of the spectrum the Earths surface approximates to a blackbody. The so called Greenhouse Gases are the ones which absorb and emit radiation in this region. Water vapour is the biggest contributor to the greenhouse effect (60% in clear sky conditions) followed by CO2 (26%).[KT97, Table 3]
P.S Oxygen, of course, also emits in the microwave region, which is what Roys satellite temperature measurements rely on.
I think it is important to point out to MikeB that if he would plug in 15micron and say 300K, or 250K, or 200K into Planck’s ideal formula for irradiance it will compute a nonzero value for N2 and O2.
I didn’t even have to tell the formula about N2 or O2, so all objects emit an ideal curve at all frequencies at all temperatures all the time. A spectrometer can be used to show the actual N2 or O2 nonzero irradiance & emissivity, reflectivity, transmissivity at each frequency, angle of incidence, temperature if MikeB thinks that might be interesting to know.
This has been done frequently looking up and looking down for Earth atm. across the spectrum.
Firstly, Planck does not have a formula for irradiance, he has a Law for RADIANCE from a blackbody.
Plancks Law does not distinguish the contributions from various gases.
So what have you actually done to come to this wrong conclusion?
The conclusion is correct MikeB. The y ordinate on Planck law is irradiance from an object be it gas, liquid, solid, plasma. The irradiance from the sun at the TOA is well known.
No it is not. If you dont understand anything stop inventing nonsense. Plancks Law applies to blackbodies, nothing else.
P.S. Look up th edifference between radiance and irradiance.
MikeB: Let’s use irradiance in a sentence. Planck’s bb cavity pinhole irradiated his thermopile, it did not radiated his thermopile. Properly he plotted his results at a temperature and frequency as irradiance.
The error in this experiment is due to the fact that, whereas the lamp is capable of raising the temperature of the plate even more if cooling is slowed, in contrast, for the vast majority of the Earth’s surface, the Sun’s radiation is not capable of raising the existing surface temperature, as we can show using the Stefan Boltzmann Law.
Oh really? please use the Stefan-Boltzmann Law and show us then.
the SB equation is only to determine the rate of RADIATIVE energy emission by an object of known temperature and emissivity. It does NOT determine what that temperature is for the Earth’s surface, which is the result of many radiative an non-radiative energy exchanges.
Dr Spencer: NASA energy diagrams do add the solar flux (168) and atmospheric flux (324) and then deduct sensible cooling flux (102) leaving a net of 390 watts per square meter. Is it just a coincidence that this is the precise flux which, when entered into Stefan Boltzmann calculators (with emissivity 1.0) yields almost exactly 288K? I think not. They should have deducted the upward radiation to the atmosphere if they were going to include the downward radiation from the atmosphere. All that radiation back and forth between the atmosphere and the surface does is to cool the warmer surface, thus making it harder for the Sun to warm the surface.
As in the book “Die Kalte Sonne” it is obvious that the Solar radiation on its own cannot possibly explain the mean surface temperature. That, Sir, with respect, is where you are mistaken, just as you are in thinking that, without IR-active gases, the surface would have been the same temperature as that at the radiating altitude. Even that 255K figure (which is indeed calculated using Stefan Boltzmann) is wrong because it assumes uniform flux. In fact the solar flux varies over 24 hours and, because there is a fourth power of T in Stefan Boltzmann, the mean temperature achieved by variable flux would be more like 245K, not 255K.
You need to explain why you think that isENTROPIC conditions in a vertical column of the troposphere would NOT be the state of maximum entropy which the Second Law says will tend to be approached. The whole concept of entropy increasing is based on the concept of unbalanced energy potentials dissipating, and such have fully dissipated when conditions are isentropic – that being different from isothermal when a force field like gravity is involved.
In short, there WILL be a tendency for the troposphere to move towards a stable, isentropic state wherein the sum of ALL forms of internal energy is homogeneous in a vertical column. Hence we deduce that as potential energy due to a force field diminishes, the kinetic energy (temperature) increases in order to retain a constant sum of the internal energy in all its forms. And that is why we don’t need to add in back radiation to explain the surface temperature.
Furthermore, because we don’t do that, we cannot at the margin add a bit more radiation from carbon dioxide and assume that extra bit is warming the surface more, because the surface temperature just isn’t determined by the sum of solar radiation and atmospheric radiation – as NASA energy diagrams very clearly and incorrectly imply it can be.
You have it backwards. The upward IR flux from the surface in those diagrams is not calculated as the residual of the other terms. It is the best known flux, estimated from the global distribution of average surface temperature, and surface emissivity close to one. The other terms are lesser well known estimates from a variety of sources. They are then all compared, and any disagreements lead to subjective adjustments to the lesser-well known fluxes until there is energy balancs. If you read the original works on this you will see there are error estimates on each of the energy fluxes.
I don’t know why you and others bring up the subject of entropy. It has no predictive value for determining the average, quantitative temperatures within the climate system, which is energized from the outside and ends up with temperature gradients based upon many energy transfer processes.
No I don’t “have it backwards” thank you.
The temperature is determined by the INWARD flux if and only if radiation is the only input of energy. But nothing gets hotter than a blackbody would for any given level of flux, regardless as to how much back radiation or insulation there is.
Yes the outward flux lets you measure the temperature, but you have to explain how the temperature got to be what it is. That is why NASA scrapped their initial energy diagrams that did not show backradiation, because they realized there was missing energy INPUT since the Sun’s radiation did not give the right answer in Stefan Boltzmann calculations. As I said, their diagrams had to show 324 watts per square meter of atmospheric radiation or they had no explanation for the surface temperature.
I’ll help you understand why energy is indeed conserved. What happens is that the more the outward radiation is countered by back radiation, then the more is the proportion of inward radiation which undergoes the process first written about this century which physicists now call pseudo scattering. The surplus energy in the incident radiation is not thermalized but instead is immediately re-emitted and used for a portion of the target’s Planck function. The warmer target (surface) thus uses less of its own thermal energy for its radiation, and so it cools more slowly as we both agree it does.
But you need to think outside the square and try to understand the process that I have described which is beyond doubt what is really the case.
Ugh. Those are some serious mental gymnastics you are making with what happens to thermal radiation…incident radiation not thermalized but immediately re-emitted as a portion of the Planck radiation? Really? What magic trick is this?
And please do share with us the original NASA energy budget diagram that doesn’t include downwelling IR radiation from the sky?
I’m still waiting to hear what credientials you have to back up throwing around technical terms.
It’s easy to form sentences with fancy words. It’s not so easy for them to confom to what actually happens in nature.
Hmm.
THIS: “What happens is that the more the outward radiation is countered by back radiation, then the more is the proportion of inward radiation which undergoes the process first written about this century which physicists now call pseudo scattering. The surplus energy in the incident radiation is not thermalized but instead is immediately re-emitted and used for a portion of the targets Planck function.”
Sounds a lot like the gentleman from Australia that is the only other time I have heard this concept “pseudo scattering”. The only difference in no mention of Venus or Uranus.
Roy, respectfully:
If you immersed yourself in a tub of icewater, according to this idea of cold things warming warm things more, why shouldn’t your body temp increase, when we know it will do the opposite?
Or, try it not touching the ice, as in a mostly closed-off ice cave off the Lake Superior shore standing on a wooden plank. How long would you have to stay in there before you broke a sweat from all that ‘heat’ coming off the ice?
If you brought dozens of big tubs of snow and ice into your home, will your house warm up from it?
If you stand in front of your refrigerator with the door open, will you warm up from the ‘heat’ coming off the cold air leaving out of the frig?
I doubt you would claim a person or house would actually become measurably warmer under those scenarios under any duration.
GHGs are not even considered in your or my experiments.
If what you’re saying was truly practical, the real world be a lot different.
Why in practice do these scenarios not actually work according to the idea presented in your experiment?
What makes your results different and superior to everyday common sense experience we all have with ice & cold?
I think what you are missing here Bob, is what would be in place of the tub of ice water if it wasnt there. You automatically assume from common experience that it must be something warmer than the ice. But a true comparison is with NOTHING, that is to say a background at absolute zero.
Do you see how it works now?
In ‘theory’ yes, but we don’t have “nothing” to compare to, as by definition “nothing” cannot be defined or measured, so it’s as ‘unreal’ to this scenario as the concept of infinity.
There is no place on earth with an absolute temp of zero, making your analogy impractical.
If the tub of ice water wasn’t there we’d experience room temperature, not absolute zero.
That’s how I see it working.
Bob, it is not complicated. You seem to be looking for some reason not to understand so that you can continue in the ignorance of your own prejudices.
What you call ice is radiating 315 watts from every square metre of its surface. That is a significant heat source. It provides energy to any other object that that absorbs that radiation. And the fact that you dont understand it is not going to stop it happening.
I asked rational questions and responded to yours with what I think is common sense. Your ad homs are irrelevant.
You lukes get downright angry when confronted with logic.
WTF is “what you call ice” supposed to mean? Ice is ice.
Bob:
You say: “If the tub of ice water wasnt there wed experience room temperature, not absolute zero.”
In this experiment, the ice is the cold “reference” background, and the cardboard is the warmer (but still cooler than the body we’re measuring) background.
In the case of the earth, deep space, with an “effective blackbody temperature of 2.725K — measured with great precision (and with the vastness of space, it is NOT correct to say there is nothing out there, even though the density is incredibly low), and the atmosphere, radiating with an effective blackbody temperature averaging about 255K is the warmer (but still cooler than the body we’re measuring) background.
In Roy’s experiment, the warmer background results in a higher temperature of the body under test. The general level of the earth’s surface temperature CANNOT be explained if it were radiating directly to space.
Bob, you obviously did not read either the post or the discussion, otherwise you would know your example is not analogous to what is being discussed.
And the results DO agree with common sense.
Using a pot of water analogy…an open pot of water on the stove which cannot quite reach the boiling point can be made to boil by putting a lid on it.
The lid reduces the rate of energy loss by the pot, increasing the water temperature. Even though the lid is cooler than the hot pot.
All that is necessary is that the hot object is being actively heated (the sun, for the climate system), and that its rate of energy loss is reduced by some cooler object (the atmosphere) replacing a very cold object (deep space). This will make the hot object (the Earth’s surface) even warmer still.
Roy you are right that I didn’t read that post.
Your explanation to me didn’t address my specific questions here on this post, and if your prior post didn’t discuss my specific examples, then we aren’t getting anywhere, yet.
The common sense analogies I used were not about a boiling pot of water with a cover. They were about ice and snow and our common experience with ice and snow.
Why doesn’t snow and ice warm our bodies?
What you are talking about is extremely counter-intuitive.
I’m starting to see why people get ticked off at you.
You completely dodged my questions.
THIS post is about the effect of ice on a warmer body. So were my questions.
Again, why doesnt snow and ice warm our bodies?
If you cannot effectively answer my direct questions, I will be forced to think you haven’t thought it through.
By the way, no one I know of thinks the atmosphere doesn’t retain some heat overnight, so I feel like you’re constantly beating this dead horse with your experiments.
your analogy does not apply Bob! In the experiment I posted, the ice does not warm the heated plate! It’s the ROOM TEMPERATURE SHIELD that blocks the view of the ice that results in the heated plate getting even hotter.
Why must you divert attention from the subject of the experiment by erecting a straw man example which is nothing like what I am demonstrating? It is you who is dodging the results of the experiment by suggesting I am claiming something I am clearly not.
“Again, why doesnt snow and ice warm our bodies?”
Bob, if you first dipped your toe into liquid N2 then into ice water, your toe would feel warmer and be warmed due to transfer of higher KE.
The experiment of Dr. Spencer top post shows the same concept except using the same transfer of KE through radiant energy transfer. it is pretty simple, you should learn from it. A Cool Object Can Make a Warm Object Warmer Still.
…assuming you still have a toe attached to your foot.
No kidding! Make that a fast dip.
That is such a ‘pat’ answer.
You claim that a cold body can warm a warmer body.
I provided you with real-world examples where according to our senses cold objects do not warm our warmer bodies.
Tell me how that is wrong. Tell me why my analogies are so far off base.
I provided the scenarios where cold situations do not warm our warmer bodies. Are you claiming that ice will warm our bodies? If you cannot provide any evidence of that then you are talking pseudo science.
Bob – If you prefer then, use another pat answer. Put a nice heavily insulated winter coat in the freezer overnight. It will be colder than your body when you take it out and wear it at room temperature. Your body will feel warmer. Or you can read the top post to also find out how a “Cool Object Can Make a Warm Object Warmer Still”.
Let’s go to the claim that I made a strawman argument.
The fact is all you lukes are on the attack, always seeking to destroy anyone who questions the significance of GHE.
What is that significance? That the atmosphere stays warm overnight when out of direct solar radiation. I don’t dispute it, and as I said before, I don’t know anyone who does.
Stop this BS of trying to make us who have serious doubts concerning your use of logic and evidence into enemies to be vanquished or called ‘sky dragons’ or some other label.
YOU claimed cold warms.
Dispute that if you will or can.
I consider that statement to be a belief of yours – a belief that then must be TESTED as a universal principle.
So I applied it to several easy-to-understand scenarios where this supposed universal principle falls apart.
One more time: Why doesnt snow and ice warm our bodies?
If “cold warms” is supposed to be a universal principle, then you should have no trouble providing evidence to an Michigan Tech engineer with heat transfer experience that ice and snow warm our warmer bodies.
If you can’t do that you’re all talking through your hats.
“Why doesnt snow and ice warm our bodies?”
Bob really does want an answer.
A. The snow and ice constituent particle KE in contact is lower than our body constituent particle KE.
Bob says:
“YOU claimed cold warms.
Dispute that if you will or can.
I consider that statement to be a belief of yours a belief that then must be TESTED as a universal principle.”
There was no claim that “cold warms” as a universal principle. There is a claim that “a Cool Object Can Make a Warm Object Warmer Still” (to quote the title).
Or to paraphrase the claim, a “merely cool” object doesn’t cause as much of a temperature drop in nearby objects as a “very cold” object would. I can’t see why this should be even moderately controversial.
You see Dr Spencer, you STILL think that the Sun’s radiation striking the surface is warming that surface to the mean global temperature. I say this because you write “All that is necessary is that the hot object is being actively heated (the sun, for the climate system)”
Yes, indeed, your experiment (which just confirms the well-known fact that cooling by radiation can be slowed by radiation from a cooler source) is not relevant because, whereas Stefan Boltzmann calculations will show that the lamp has not yet achieved the maximum temperature that it could achieve, that is not the case for the Sun. The Sun’s radiation for most of the surface is not what is raising the temperature of the surface. For example, that temperature can rise in the morning even under thick cloud cover. So the rate of cooling is not relevant, because you don’t know how the surface is indeed receiving the required input needed to explain the observed temperature. I can tell you how it is, but we’ll get to that at a later date.
So, when it is clear that in nearly all locations on the Earth the solar radiation is not sufficient to explain the mean temperature over 24 hours, especially where there is cloud cover outside of the tropics, we must conclude that the rate of cooling is not relevant.
Even where the Sun’s radiation has warmed the surface some hot morning, why then is the rate of cooling much faster in the afternoon when the Sun is still shining, than it is around 2:00am when there is no solar radiation and yet the cooling almost ceases altogether?
First of all, can you please stop with the anonymity? Who am I speaking with? You are obviously connected with the book by the same name, so why not tell us who you are and what your qualifications are to be throwing around technical terms?
Regarding your comments, it’s hard to know where to start….
As I have discussed over and over, a surface heated by the sun (or any other energy source) will experience increasing temperature as long as it cannot lose energy to its surroundings. This is simple energy conservation. The solar flux absorbed by the surface of the Earth does NOT limit how hot its temperature can be (except in the absurd extreme of the surface temperature becomes as hot as the sun). Of course, in the real world any heated surface always loses energy to its cooler surroundings, but that loss can be reduced by a variety of methods, including radiative “shields” that are transparent to sunlight but mostly opaque to infrared light (the atmosphere, a pane of glass).
And the more you can reduce the rate of energy loss (initially), the higher the temperature you can achieve. This is simply energy conservation.
For you to say that the sun cannot warm the Earth’s surface to the observed temperature then suggests you don’t understand the 1st Law of Thermodynamics (conservation of energy).
You then confuse the issue by noting how temperatures can remain high even when it’s cloudy. Well, of course they do…there’s (1) the heat capacity of the system which slows the rate of temperature fall, and (2) the greenhouse effect of clouds, which also reduces the rate of cooling.
You clearly are not familiar with the many calculations using known physical processes that result in terrestrial temperatures (geographically and vertically through the atmosphere) through computer simulation. We can start the computer simulation with everything at 100 K or at 400 K, and the temperatures will gradually converge to what is observed in nature. We have actually done this in a 1D time-dependent model of the climate system.
When you accomplish the same quantitative results with whatever alternative physics you seem to believe in, let us know.
(snip) No diversions from the central point being discussed, please. That is all you have been doing. And claiming credentials you refuse to prove. -Roy
Because of professional considerations I will not disclose my name. Please respect that. Goodbye.
actually, I have a lack of respect for that. John. -Roy
I think one real life example that people don’t appreciate is best seen coming home after Christmas vacation up in these northern latitudes. I set the thermostat to 55-60 (yeah, warm, but I don’t like coming home to frozen pipes), and turn it back up when we get home.
The furnace gets the air temp back up to 68 or 70 quickly enough, but dang house still feels cold. I’m pretty much convinced that the reason is the walls are still cold and are radiating less IR than they will be when they finally catch up.
Of course, the thermostat is somewhat affected too, so studying this means bringing in an aspirated temperature sensor from a weather station, FLIR observations of the patterns of warming from the walls, studying interior walls, which warm from two sides, vs. exterior walls which warm from one, etc.
Then there’s dealing with my engineer’s knowledge that there’s no difference in the air’s warming rate whether I set the thermostat to 70 or 80 so it’s very hard for me to set it high. But I could do it in the name of science!
The mean flux from the Sun that impinges on the surface is about one-eighth of the Solar constant, as is shown in NASA energy diagrams. This low flux (even if it were uniform day and night) would only raise a blackbody in Space to about -40C. The mean temperature achieved by variable flux would be even lower, and we have not even as yet deducted the simultaneous sensible heat losses that make the Sun’s job even harder. So the Sun’s radiation doesn’t have a hope of explaining the Earth’s mean surface temperature on its own. Hansen thought it must be a combination of solar radiation and about twice as much atmospheric radiation, but he was sadly mistaken.
You cannot explain the surface temperature just by using the sum of the fluxes from the Sun and the atmosphere. And, because these fluxes are totally different with Planck functions barely overlapping, the NASA energy diagrams are most definitely wrong where they imply that the fluxes can be added and then used to calculate the surface temperature using the Stefan-Boltzmann Law. A prerequisite for that law to be applicable is that the source itself must be a single blackbody, and the law is inapplicable anyway when the source is effectively cooler than the target after attenuation due to distance.
The surface does not act like a blackbody either, for a blackbody by definition only exchanges energy with other bodies via radiation. So, in practice, it has to be in outer Space.
It is well known that radiation from a nearby body will slow the rate of cooling by radiation of a hotter body. This is all that Roy’s experiment does indeed confirm. However, Roy thinks his IR thermometer measures the rate at which it is warmed by the colder atmosphere, but in fact it measures the rate at which it cools, that rate being slowed by less-cold bodies such as a cloud.
Finally, Roy does not appear to know that the Second Law of Thermodynamics is stated with words to the effect that in a natural process the total entropy of the bodies participating in that process will increase towards a maximum that is determined by the constraints of the system. The Second Law is not about heat from hot to cold, because there are exceptions in force fields due to the fact that entropy is affected by changes in other forms of internal energy, not just by changes in kinetic energy. This is critically important, because the Second Law can thus be used to disprove Roy’s contention that the troposphere would be isothermal without IR-active gases. Such a state would not have maximum entropy and thus would be unstable.
Hmmm, interesting confusion.
Where to start.
First, a blackbody is by definition simply an object which absorbs all radiation falling on it, irrespective of wavelength. It will also happen lose energy by radiation (in proportion to the 4th power of its temperature in kelvins) , but that does not mean it will not also lose heat by other means, conduction and convection. In fact, unless it is in a vacuum, it will. And this is why you cannot use the Stephan-Boltzmann Law to determine how hot an object will become for a given irradiance (i.e. incoming radiation). It cant be used that way and NASA do not do that; the Law can only tell you how hot the blackbody already is by measuring its radiation output!
The steady state temperature attained by any body depends not just on the heat input, but on the heat output as well. The two have to be taken together. Thus added insulation will raise the temperature for any given heat input.
Stefan-Boltzmann can and is be used to show that without an atmosphere the Earth temperature would be about minus 18 degree Celsius. The mean surface temperature is in reality is about plus 15 degrees. This 33 degree difference needs to explained. The Greenhouse Effect is the explanation (although there are plenty of crackpot theories around).
I would not suggest that you think you can “teach” me about this field of physics which I have been teaching for decades and carrying out extensive post-graduate research.
A blackbody is “any object that is a perfect emitter and a perfect absorber of radiation” and Stefan Boltzmann calculations are only strictly accurate where there are no sensible heat transfers involved and the incident source is itself a blackbody. As we know from Wien’s Displacement Law, the temperature is proportional to the modal frequency in the Planck function, and so this necessitates a single blackbody source, which is not the case when climatologists think they can explain Earth’s surface temperature with the sum of solar and atmospheric radiation. They DO assume this, and that is why they assume that extra radiation from carbon dioxide will add to the assumed warming effect of the rest of the atmospheric radiation. But they are mistaken.
NASA energy diagrams imply that we can adjust for the fact that the surface loses sensible heat if we just deduct that outward flux (102) and use the net (168+324-102 = 390) to get the temperature of 288K using Stefan Boltzmann. The underlying assumption that radiation from the colder atmosphere is helping the Sun to raise the surface temperature each morning is totally incorrect, but you cannot explain the surface temperature with radiation in any other way. There is a totally different way to explain it, however, but that will have to wait for another day.
Please read all my comments to Dr Spencer, because I don’t wish to have to repeat the rest of the explanation here.
as I’ve replied to you elsewhere, you are totally misrepresenting how the UWIR of the Earth has been estimated from observations. It is based just on observed surface temperatures and an emissivity close to 1. It is not the result of trying to balance other energy fluxes. Please stop misrepresenting what the global energy budget researchers do.
Also, if you really do have the credentials and experience you claim, quit hiding. What is your name?
With respect, Dr Spencer, it seems you have not studied NASA energy diagrams and noted that the flux figures are indeed adding to a net of 390 watts per square meter INTO the surface and that is indeed the exact flux that yields 288K (15C) in Stefan Boltzmann computations. This is no coincidence. In fact they “adjusted” the atmospheric radiation figure by overstating it (324) to ensure that the calculations did work. There is no way that an atmosphere with a mean temperature below zero could emit 324 watts per square meter. Even a blackbody at 0C would not emit that much, and gases do not emit like blackbodies.
Only the Solar radiation should have been used because the radiation back and forth between the surface and the atmosphere can only cool the surface.
But, as I have correctly explained, the solar radiation is nowhere near sufficient by itself to raise the Earth’s surface temperature to the estimated global mean. So the real world is nothing at all like your experiment.
For most of the world’s surface, the temperature has been achieved (and is maintained by) the Second Law process that is maximizing entropy and creating isentropic conditions with an associated non-zero temperature gradient due to the force field of gravity.
Experiments this century have demonstrated this same temperature gradient being created by centrifugal force, and of course we see it on other planets, even those without solid surfaces. That is why Jupiter is so hot in its core, and thus I rest my case.
Dr. Spencer, fairly obvious he who must not be mentioned has acquired a new screen name and no amount of testing and sound 1st principle science will change these political views.
By the way, MikeB, you don’t even have a correct understanding of Pierrehumbert’s textbook explanation of the Greenhouse effect. The 255K (-18C) temperature is NOT based on an Earth “without an atmosphere” because that would receive a mean flux of one-fourth of the Solar constant. Instead, Pierrehumbert considers the existing atmosphere without greenhouse gases. He deducts 30% of the Solar constant, that being the amount supposedly reflected by the atmosphere, and from that result he determines the 255K temperature, even though only a uniform flux would achieve such.
This of course is his first error, because the solar flux varies over each 24 hour period at any given location.
The second error is that without greenhouse gases there would be no clouds reflecting two thirds of that 30% albedo.
His third error is that which I have discussed in an earlier comment to Dr Spencer regarding the state of maximum entropy.
So I will leave you to study all the comments I have written – and I may not return here for some time. So Goodbye.
Thank God for that. Goodbye
so, we still don’t know exactly who this “expert” is, now that he’s left.
Except he never leaves, Katy bar the door. Political science is better received elsewhere.
Whoa, didn’t I just wade through several dozen ColdSun posts above with timestamps from this afternoon?
The more of his posts I read the more convinced I am it is the one and only physics expert from Australia, studied physics at Sydney University.
Here are a few: “The underlying assumption that radiation from the colder atmosphere is helping the Sun to raise the surface temperature each morning is totally incorrect, but you cannot explain the surface temperature with radiation in any other way. There is a totally different way to explain it, however, but that will have to wait for another day.”
This would be (words combination that must be banned)
He pulled up Jupiter but was careful not to give it away with Uranus or Venus.
Here is another: “I would not suggest that you think you can teach me about this field of physics which I have been teaching for decades and carrying out extensive post-graduate research.”
Used several times in the past by the gentleman from Australia.
Finally this one: “This is critically important, because the Second Law can thus be used to disprove Roys contention that the troposphere would be isothermal without IR-active gases. Such a state would not have maximum entropy and thus would be unstable.”
If that is not him he has an identical twin that never learned about Venus and Uranus.
h*e*a*t*c*r*e*e*p*
I think you are right. They cannot be two people so deluded
Coldsun, Are you saying that the earth is spinning faster than the atmosphere and friction between the atmosphere and earth is responsible for the temperature increase? Isn’t this effect that causes Jupiter’s red cloud?
— August 30, 2016 at 6:31 AM
Hmmm, interesting confusion.
Where to start.
First, a blackbody is by definition simply an object which absorbs all radiation falling on it, irrespective of wavelength–
A blackbody is mostly or definitionally about what is emitted from
a warmed body. Or it’s about the thing that emits rather than the thing that absorbs.
Or in analogy it’s about suns rather than blackholes
No, it is both. Look it up.
TheColdSun says: “because these fluxes are totally different with Planck functions barely overlapping, the NASA energy diagrams are most definitely wrong where they imply that the fluxes can be added and then used to calculate the surface temperature using the Stefan-Boltzmann Law.”
Oh, where to start???
The first thing is that he is trying to use the Stefan-Boltzmann Law “backwards”. The SB Law allows you to calculate the radiative output of a body given its temperature. In other words, Qout is a function of T.
It does not allow you to calculate the temperature, even the steady-state temperature, of a body given its radiative input. T is not a function of Qin. Temperature must be computed using all of the energy inputs and outputs, with the outputs often being a function of the body’s temperature and ambient properties.
Steady-state temperature is achieved when the inputs and outputs (including the temperature-dependent outputs) match.
Second, I am astounded at the number of people who do not realize that with a “conserved” quantity (energy here), your analysis must allow the flows of this quantity to add. So it is entirely proper to add the inflows of solar and longwave infrared radiation to the surface. If you could not, energy would not be conserved!
By the same token, conservation of mass means that in analyzing fluid flow problems, you can add different inputs to a control volume together, even when coming from very different pressures.
Yup.
Mr. Cold Sun fancies himself as an expert, but won’t prove his qualifications and his real identity. Not very professional, so I’m assuming an anonymous amateur wannabe.
I think what you are missing here Bob, is what would be in place of the tub of ice water if it wasnt there.
You automatically assume from common experience that it must be something warmer than the ice. But a true comparison is with NOTHING, that is to say a background at absolute zero.
Do you see how it works now?
Sorry, that was a reply to Bob Weber
Dr Roy,
I agree with your findings.
However, I feel some people may be confused with your conclusion. You say
“There is no violation of the 2nd Law of Thermodynamics in the experiment; a cool object can make a warm object even warmer still through infrared radiative effects”.
The words “a cool object can make a warm object even warmer still” imply, to me, that the cool object is actively heating the warm object, and as we know this is not possible.
Perhaps words like – the temperature difference between the warm object and the cool object influence the temperature (cooling rate) of the warm object. The warm object is dumping its heat via radiation, how quickly depends on the temperature of adjacent objects.
That said, many here have commented on how a solid object in the path equates to gaseous CO2 or any other GHGs.
The earth system with solar in, surface IR out to space, anything in between will have an effect upon surface temperature. With GHGs acting as a conduit or insulator for IR, we desperately need some good experiments demonstrating how changing GHG levels influences surface temperature.
I was purposely provocative with the phrasing, Steve. Yes, how you phrased is more accurate. I’m trying to get people to think through the details so they understand what is happening.
I quickly translated Roy’s statement into “a cool object can keep a warm object warm longer.” Perhaps better would be “a cool object keeps a warm object warm longer than a cold object will.” Or “a cold object cools a warm object faster than a cool object.”
I agree about collecting information about GHG vs surface temps. I suspect one reason why GHGs may be less effective than models have been told is that surface warming due to GHGs will lead to more convection that gets heat off the surface. While things like an accurate temperature profile are “easy” to collect, measuring convection may be massively difficult and probably requires things like releasing tracer chemicals in the atmosphere. Perhaps SO2 from coal plants could be tracked and correlated with the GHG H2O, but having to track all the other weather happenings would make it be a tough study.
I think we agree that the effect of more CO2 will be smaller than advertised, and the ways in which it can be made smaller are numerous.
Hi Ric,
you wrote
-I quickly translated Roys statement into a cool object can keep a warm object warm longer. Perhaps better would be a cool object keeps a warm object warm longer than a cold object will. Or a cold object cools a warm object faster than a cool object.-
If you exclude that a cool object can effectively keep warmer a warmer object forever than a colder one, then I don’t agree.
The warmer object should reach an higher temperature forever indeed.
It’s not a transitory event.
In case I misunderstood you I apologize.
I think what Ric meant is that the cooler object doesn’t generate its own energy, it is redirecting energy created by the lamp. -Roy
Have a great day.
Massimo
Well, from the phrase Steve Richard quoted, a cool object can make a warm object even warmer still I was approaching the thread as not including the lamp, but as the thermal mass of the cool object radiating photons that are absorbed by the warm object. Of course, the warm object is radiating photons with a higher average energy and in greater numbers, so it’s still cooling, just not quite as fast as it would if the cool object was replaced by a cold object.
Of course I exclude “exclude that a cool object can effectively keep warmer a warmer object forever than a colder one” in my system without a lamp, In Roy’s system with a lamp, it’s still not a closed system, but the heat of the lamp and surroundings will lead to melted ice and generally warmer temperatures until an equilibrium is reached with as much heat going through walls, etc as the lamp is adding to the system.
If Roy’s room was a truly closed system, then yeah, everything keeps heating up forever and Roy should be selling whatever the walls are made of as R-infinity insulation.
Yes, Dr. Spencer, invoking those same words to prove a point is useful.
May I propose this?
It is a study using instruments able to ‘detect the unique spectral signature of infrared energy from CO2’.
http://newscenter.lbl.gov/2015/02/25/co2-greenhouse-effect-increase/
You can read there that CO2 backradiation actually reaches, with 0.2 Watt/m2/decade, about 10% ‘of the trend from all sources of infrared energy such as clouds and water vapor’.
It is indiscutably few but not negligible.
I imagine that this “unique spectral signature” is obtained by comparing, line by line, the different downwelling IR backradiations with e.g. the HITRAN database, and by eliminating all lines CO2 shares with H2O, N2O, CH4 etc etc.
But there are, as it seems, not only factors increasing CO2 in the tropospheric layers…
http://onlinelibrary.wiley.com/wol1/doi/10.1002/2015GL064696/full
In this paper it is reported about rather huge transfers of CO2 from the troposphere toward the lower thermosphere, where it manifestly has the inverse effect of what happens above surface: the authors write that “CO2 generally
produces infrared radiative cooling above the tropopause”.
Interesting!
Bindidon
The effect of CO2 in the atmosphere is to warm the surface and cool the upper atmosphere. This is not a secret! However, we live on the surface and this is what counts. When we talk about global warming we mean at the planetary surface where life exists, develops and evolves. This is what matters and this is what global warming refers to; not what happens in the stratosphere!
The magnitude of back-radiation is measured by a network of stations around the globe. I mean MEASURED, not estimated by some model. They tend to return a figure of about 340 watts per square metre. To break this down, If you want to see the magnitude of CO2 back radiation in terms of its spectral composition, then it is necessary to use a Fourier Transform Infrared detector, which measures the back-radiation at each infrared wavelength. The following result is an example.
https://scienceofdoom.files.wordpress.com/2010/04/longwave-downward-radiation-surface-evans.png
To understand the significance of this result it is necessary to know that CO2 emits radiation around a wavelength of 15 microns. We can see from the measured result that this where most of the back-radiation cones from. This is the unmistakable fingerprint of CO2 as a major component of back-radiation. It is really game, set and match.
This is why, as far as CO2 is concerned, the science is settled. The debate is over amongst all scientists and those able to understand what they are looking at!
Or do you have some other explanation?
This is why, as far as CO2 is concerned, the science is settled.
Sorry, MikeB…
Although lots of people guess I’m a warmist, I’m not at all.
Thus, for me, the science in that corner isn’t settled at all yet.
It seems to me that you just discovered SoD… I began to read there about five years ago. Excellent site!
But manifestly, you did understand neither the reason for my comment nor its scientific background.
Please
– notice that the SoD plot you show is dated 1996 (!!!);
– read carefully, before commenting again, the two articles mentioned in my double comment:
http://newscenter.lbl.gov/2015/02/25/co2-greenhouse-effect-increase/
http://onlinelibrary.wiley.com/wol1/doi/10.1002/2015GL064696/full
So “The effect of CO2 in the atmosphere is to warm the surface and cool the upper atmosphere.”
I see … and does the greenhouse gas water vapor do likewise? Does it thus make the lapse rate steeper?
You will find it does the opposite, and I have explained why in a paper yet to be presented at a well known institution.
If you choose to believe Dr Spencer’s assertion that solar radiation just keeps on raising the temperature (even up to that of the Sun) if the “insulation” is good enough, then go your way.
This is my final goodbye I assure you, because you people are not interested in learning. Some of you do not even realize that the input of radiation does indeed determine the equilibrium temperature of a blackbody in Space. The other point about which you show no interest in learning, is the fact that there are inputs other than radiation that are determining the surface temperatures, and they must be considered in the overall energy balance.
I suggest that you note in particular my summary of the errors in Pierrehumbert’s writings.
I repeat: “Does the greenhouse gas water vapor do likewise? Does it thus make the lapse rate steeper?
That’s all – take it or leave it.
TheColdSun
How is life in Australia? I haven’t seem you post on this blog since the 3000+ thread with at least 50% yours.
Hope all is well with you.
Norman:
I don’t think it’s *him*.
Not the same posting style.
Look up the author of “A Cold Sun”.
Toneb
I did look up the author of the Cold Sun. But a lot of things he says were said by “him” in other posts.
Also I think he is being very careful in word choice to try and disguise himself so Roy does not ban him again.
Who else uses “pseudo scattering” besides him and Postma?
His other mechanism is h*e*a*t*c*r*e*e*p for keeping the surface warm but he will not say it on this blog.
I could be wrong but keep reading and his mask will fall off and he will refer to Venus or Uranus if he posts long enough.
Norman:
You may be right.
Look out for mention of KE as well.
There’s very little learning to be had from statements like “I have explained why in a paper yet to be presented at a well known institution.” Perhaps I could learn from your book, it should certainly be more clear than the jumble of comments you’ve left here. Blogs are like that, it’s one reason I have my own web site.
Oops – that should have been a reply to the ColdSun’s comment at August 30, 2016 at 5:47 PM that, for the moment, is right above.
Roy W. Spencer, Ph. D. says: August 28, 2016 at 3:15 PM
“because the ice (and the room-temperature sheet) emit IR back to the hot plate. Its the NET (2-way) transfer of IR radiation that matters. EVERYTHING emits infrared radiation. To claim otherwise is just plain silliness.”
Your claim than is the thermal EM flux is physically emitted radiance in a direction of higher radiance. This is your false claim of ‘EVERYTHING emits’. Have you ever witnessed such physical flux? I do claim that any object completely within an environment of higher radiance (higher temperature) can only absorb thermal EM flux, never emit such flux. In fact to claim otherwise is just plain silliness.
“Your claim than is the thermal EM flux is physically emitted radiance in a direction of higher radiance. This is your false claim of EVERYTHING emits. Have you ever witnessed such physical flux? I do claim that any object completely within an environment of higher radiance (higher temperature) can only absorb thermal EM flux, never emit such flux. In fact to claim otherwise is just plain silliness.”
And how can anything do that pray/
Is it sentient and able to say to those photons “look ere mate you’ve come from something hotter than me so ***er off!”
Anyone whose DK dragon-slayer mind says “to claim otherwise is just plain silliness” about empirical science is firmly down he rabbit-hole along with Alice….. where his arrogance/ignorance sent him.
Toneb says: August 31, 2016 at 5:57 AM
(Your claim than is the thermal EM flux is physically emitted radiance in a direction of higher radiance. This is your false claim of EVERYTHING emits. Have you ever witnessed such physical flux? I do claim that any object completely within an environment of higher radiance (higher temperature) can only absorb thermal EM flux, never emit such flux. In fact to claim otherwise is just plain silliness.)
“And how can anything do that pray/”
Maxwell’s equations demand such! If I keep an enclosed object 20 C below its enclosure by removing sensible heat (perhaps a measurable flow of yet colder water), the exact amount of heat power removed must be added by whatever source is maintain the temperature of the surround. A one way heat flux. Such is demonstrated every day in labs across the world.
They demand that a molecule/atom can reject an IR photon because it came from a cooler object?
Oh, OK, if you say so.
PS: Maxwell’s demon was a hypothetical mind experiment my friend.
That reflect when cold argument made me realize that the Dragonslayers will never learn.
The idea that a photon can measure the temperature of an object, i.e. measure the kinetic energy of molecules vibrating near the point of impact, is pretty impressive. If that happened, it would be really easy to make a great thermometer that could measure the temperature of a point.
Note – that is not how current non-contact thermometers work!
Toneb says: August 31, 2016 at 3:09 PM
“They demand that a molecule/atom can reject an IR photon because it came from a cooler object? Oh, OK, if you say so.”
There is absolutely no need to reject what was never produced!
“PS: Maxwells demon was a hypothetical mind experiment my friend.”
Maxwell’s demon remains thousands of times more philosophically viable than your incompetent guess of some fantasy ‘photon’!
Ric Werme says: August 31, 2016 at 3:45 PM
“That reflect when cold argument made me realize that the Dragonslayers will never learn. The idea that a photon can measure the temperature of an object, i.e. measure the kinetic energy of molecules vibrating near the point of impact, is pretty impressive. If that happened, it would be really easy to make a great thermometer that could measure the temperature of a point.”
What total Bull Shit, even if your so called ‘photon’ did physically exist such must have fourspace dimensionality, some power\energy\action finite density. Thus can never be a point!
Hmmm. Maybe we should start calling the resurrection of Maxwell’s Demon theory, Sky Dragon’s Demon.
Cripes, the “pseudo” is strong with this one.
“There is absolutely no need to reject what was never produced!”
No photons eh?
Err even more magica – the cold object knows beforehand that there is a warm one within range (potentially the whole Universe) and decides at that point not send any photons over.
Or maybe there is no such thing as a photon?
In that case what is that physicists are using when (eg) they fire single photons in the demonstration of the doube-slit experiment?
PS: please forgive me if I misinterpret your “pseudo” as, to be honest it’s bollocks, and I gave up the will to live reading.
Roy W. Spencer, Ph. D. says: August 31, 2016 at 4:43 PM
“Hmmm. Maybe we should start calling the resurrection of Maxwells Demon theory, Sky Dragons Demon.”
Challenge on:: stupid Idiotic insane Dr. Roy Spencer! Just try to defend your idiocy against those who have painfully learned repeated (soldering iron up your ass) truth of the physical! -will-
Is air made of molecules, Will?
How do you know?
Have you ever seen molecules, Will?
I’m told there are quantum well infrared photodetectors which can measure IR radiation from an object colder than the detector itself. But I doubt that will convince you, either. https://en.wikipedia.org/wiki/Quantum_well_infrared_photodetector
Do you agree that the net IR flux between two object of differing temperature depends on BOTH objects’ temperatures? If so, then a cooler object can indeed increase the temperature of an actively heated warmer object.
How do you explain this otherwise than with 2-way IR flows, with the cooler object emitting IR flux in the direction of the warmer object?…
Does the cooler object take the warmer object’s temperature before deciding to emit any IR at all?…
Does the warmer object take the cooler object’s temperature before deciding just how much to emit?…
If the two objects happen to have exactly the same temperature, do they get together and mutually decide not to emit?…
How do you rationalize in your mind what you apparently believe, Will?
Roy W. Spencer, Ph. D. says: August 31, 2016 at 6:45 AM
“Is air made of molecules, Will? How do you know?”
1. Yes they can have a temperature! (sensible heat)
2. They can be observed with the proper instrumentation.
“Have you ever seen molecules, Will?
With a SEM certainly!
“Im told there are quantum well infrared photodetectors which can measure IR radiation from an object colder than the detector itself. But I doubt that will convince you, either.”
I have never witnessed such, in spite of the claims! If the temperatures are close enough there is finite probability of power transfer in a reverse direction, but this can never be called a thermal EM flux.
“Do you agree that the net IR flux between two object of differing temperature depends on BOTH objects temperatures? If so, then a cooler object can indeed increase the temperature of an actively heated warmer object.”
Yes indeed, You claim ‘net’ I claim ‘only’! Anything that interferes with the dispatch of the entropy of a heat powered object, must result in an increase in potential to dispatch. Such is definitely not what this Earth’s atmosphere does!
“How do you explain this otherwise than with 2-way IR flows, with the cooler object emitting IR flux in the direction of the warmer object?”
The radiance (field strength) of the lesser temperature object reduces the thermal flux emitted from the higher radiance object, exactly as the (complete) S-B equation demands! That is not reverse flux nor ‘back radiation’ as you claim! Such has never been discovered.
“Does the cooler object take the warmer objects temperature before deciding to emit any IR at all?”
Depends upon field strength or spectral radiance, not temperature!
“Does the warmer object take the cooler objects temperature before deciding just how much to emit?”
Depends upon differential field strength or differential spectral radiance, not temperature!
“If the two objects happen to have exactly the same temperature, do they get together and mutually decide not to emit?”
No! They merely conform to all 22 of Maxwell’s equations!
“How do you rationalize in your mind what you apparently believe, Will”
I just did by answering all of your questions!
Dr. Spencer, how do you rationalize in your mind what you propose as a perpetuum mobile of some kind?
Interesting! no reply from Dr. Spencer who likely thought about EMR, his life work, almost as much as I. I truly respect and value his opinion. Not so much the hangers here that cannot tie their own shoelaces.
“no reply from Dr. Spencer”
That’s because he agrees with the “hangers on” my friend.
When I was about 30, I discovered I had been mis-tying my shoelaces for my whole life. I was making a granny knot, so they weren’t staying tied well.
After that and a couple other events, I started checking my long-time beliefs for other errors.
Does “used to not be able to tie his own shoelaces” count?
‘Does used to not be able to tie his own shoelaces count?’
U betcha!
Will asks, “Have you ever witnessed such physical flux?”
See the experiment in top post. The data shows the answer to Will’s question is Yes.
EM flux is physically emitted radiance in a direction of higher radiance as witnessed by the top post test.
Trick obviously has never learned the difference between ‘radiance’ and ‘flux’.
Fill me in Will, according to what you witness in the top post since what you write here is wrong according to that experiment.
Trick, and Dr. Tim Folkerts,
As per Dr.Roys written description, with absolutely no witnessing, the opposing ‘radiance’ (field strength), not flux, properly creates interference with the actual thermal radiant emittance (exitance) from Dr. Roy’s heated above plate. The cardboard has higher spectral ‘radiance’ than his ‘ice’ does. With the cardboard in place its opposing higher ‘radiance’ forces the above heated plate to increase its ‘radiance’ to achieve the same differential radiance to achieve the required one way downward ‘flux’ that prevents then above heated plate from increasing temperature even further until achieving the temperature of the illuminating fixtures, where all flux must cease. This Dr. Roy illustrates with his toy FLIR Systems gadgets.
Electromagnetic radiative flux, although the most relativistic phenomena ever discovered never conflicts with the painfully learned physical Laws of Thermodynamics. BTW ‘radiance’ can never radiate, as it is a normalized ‘field strength’ never flux.
Only Dr. Spencer’s interpretation, with his fixation on the Schuster Schwarzschild two stream approximation, for star ‘brightness’, is clearly in error.
The experiment is clearly not in error Will, it is you making the errors, you are simply wrong.
“properly creates interference with the actual thermal radiant emittance (exitance) from Dr. Roys heated above plate.”
This is wrong Will, incoherent photons (or if you prefer EMR) do not interfere. You are simply wrong about that. Totally.
Ball4 says: August 31, 2016 at 8:38 PM
“The experiment is clearly not in error Will, it is you making the errors, you are simply wrong.”
PLEASE explain where in any of Maxwell “equations” is some coherency ‘interval’ ever expressed?
(properly creates interference with the actual thermal radiant emittance (exitance) from Dr. Roys heated above plate.)
“This is wrong Will, incoherent photons (or if you prefer EMR) do not interfere. You are simply wrong about that. Totally”
Trick, Please describe any possible situation that may demonstrate the nonsense of which you claim!
Will Janoschka says:
August 31, 2016 at 9:09 PM
Trick: August 31, 2016 at 8:38 PM
The experiment is clearly not in error Will, it is you making the errors, you are simply wrong.
PLEASE explain where in any of Maxwell equations is some coherency interval ever expressed?
(properly creates interference with the actual thermal radiant emittance (exitance) from Dr. Roys heated above plate.)
This is wrong Will, incoherent photons (or if you prefer EMR) do not interfere. You are simply wrong about that. Totally
Trick, Please describe any possible situation that may demonstrate the nonsense of which you claim!
Trick, Thermal EMR has near zero coherence interval. Such has even less coherence length. This only means that the interference with emission (exitance) is non stationary and hard to detect. The effect on exitance from a lower radiance has been measurable for over 100 years.
Will asks,”where in any of Maxwell equations is some coherency interval ever expressed?”
“Please describe any possible situation that may demonstrate the nonsense of which you claim!”
“The effect on exitance from a lower radiance has been measurable for over 100 years.”
Witness, describe, see where the Maxwell effect comes from in the measurable experimental data in the top post, proves Will is wrong. Maxwell, Planck and S-B are right as is yet again experimentally demonstrated in the top post confirming data.
Quit whining Will, go replicate the top post experiment, find science is repeatable. You will learn the correct difference between radiance and flux. And irradiance!
Will, Maxwell’s equations predict “superposition” of EM fields — ie they *add* together, they don’t somehow “cancel”. So the wave from the “hot” side simply pass through the waves from “cold” side.
Check out http://www.acs.psu.edu/drussell/Demos/superposition/superposition.html
especially “Two sine waves travelling in opposite directions”.
This would be like two waves of equal intensity from two objects of equal temperature. They don’t “cancel” or destroy each other. They don’t stop the other. They form a standing wave with an even larger amplitude than either wave separately.
If one wave was slight large (slightly “warmer” source), then there is an almost standing wave that heads toward the cooler surface. There is NOT simply a small wave heading from warm to cool.
Tim Folkerts says: September 1, 2016 at 12:46 PM
“Will, Maxwells equations predict superposition of EM fields ie they *add* together, they dont somehow cancel. So the wave from the hot side simply pass through the waves from cold side.”
The vector summation of ‘ALL’ Poynting vectors at any location and frequency is called the singular Poynting Vector at that location and frequency.
“This would be like two waves of equal intensity from two objects of equal temperature. They dont cancel or destroy each other. They dont stop the other. They form a standing wave with an even larger amplitude than either wave separately.”
The existence of such a standing wave (say within a waveguide) is precisely your effect of no flux in either direction. The effect on coupling impedance to the source is also very measurable.
“If one wave was slight large (slightly warmer source), then there is an almost standing wave that heads toward the cooler surface.
There is NOT simply a small wave heading from warm to cool.”
If you have ever observed what you have just described, you seem to be unable to describe what you observe. Which is the field and which is the flux?
Will, in the top post there is a field of radiation between the ice and the hot plate, the net photon flux toward the ice (being absorbed, reflected, transmitted by ice) transferring the KE in the hot plate to KE in the ice.
Really simple. Many would think you could get it right. No Poynting vectors needed. No mention of heat needed.
“Trick obviously has never learned the difference between radiance and flux.”
That’s the entire point right there Will.
They both happen.
However in the GHE the “flux” is ALWAYS from hot to cold.
Cold still radiates to warm however ….. and slows down cooling.
The GHE.
End of.
Give me even one example of thermal radiative flux in the direction of higher temperature.
“Give me even one example of thermal radiative flux in the direction of higher temperature”
Roy’s.
The GHE.
All things.
It is a fact that all things radiate in the em spectrum when above 0k. But, as I said, the Flux is always warm to cold”.
That is the 2nd LoT.
BUT there is a constant exchange of energy.
A cold object will slow a warm object from cooling when screening a colder object.
Happens all around us my friend.
Try looking.
Fell into the trap.
Not Flux ….
Radiation exchange.
Will, “Give me even one example of thermal radiative flux in the direction of higher temperature.”
The top post experiment.
Toneb says: September 1, 2016 at 2:14 AM
(Trick obviously has never learned the difference between radiance and flux.)
“Thats the entire point right there Will.
They both happen. However in the GHE the flux is ALWAYS from hot to cold.”
Thank you! and the magnitude of that flux Is always proportional to the difference in the two opposing radiances!
“Cold still radiates to warm however .. and slows down cooling.
The GHE. End of.”
‘radiates’ just what? certainly no power is exchanged with or without ability to influence temperature.
Will Janoschka says: September 1, 2016 at 3:43 AM
(‘Give me even one example of thermal radiative flux in the direction of higher temperature.’)
Toneb says: September 1, 2016 at 5:28 AM
“Roys. The GHE. All things. It is a fact that all things
radiate in the em spectrum when above 0k.”
There is absolutely no evidence for your claim. All of this physical is completely explained by the one way flux proportional to the difference in opposing radiances at each frequency. Any increase in surface temperature is completely explained by the gravitationally induced lapse rate never by some imaginary ‘back radiation’ that transfers no power.
“But, as I said, the Flux is always warm to cold.
That is the 2nd LoT. BUT there is a constant exchange of energy.”
But radiatively in one direction only.
“A cold object will slow a warm object from cooling when screening a colder object.” Happens all around us.”
But that is not what happens in this atmosphere. This Earth’s atmosphere is heated by forced mechanical convection not surface radiative flux.
Toneb says: September 1, 2016 at 5:29 AM
‘Fell into the trap. Not Flux . Radiation exchange.’
Persistent nonsense! What can you possibly mean by ‘radiation exchange’ when nothing is exchanged?
Ball4 says: September 1, 2016 at 10:28 AM
(‘Will, Give me even one example of thermal radiative flux in the direction of higher temperature.’)
“The top post experiment.”
All of this physical is completely explained by the one way flux proportional to the difference in opposing radiances at each frequency. There is no need nor any evidence for Roy’s claimed perpetual motion.
Will, your conclusions are wrong, thru misuse of terms you have never learned to use correctly.
It is foolish for you to write there is perpetual motion in the top post experiment. It is powered!
The ice emission irradiates the hot plate, the ice radiates photon flux calculated from SB across the spectrum, absorbed, reflected when incident on the hot plate. The hot plate irradiates the ice, hot plate radiated photon flux from SB, the ice absorbs, reflects and transmits the hot plate radiation.
The net flux is the simple arithmetic difference between the two fluxes. The KE from the hot plate electrical power is thus transferred to the ice by radiative energy transfer (not very much by conductive energy transfer and not much by convective energy transfer). This is really simple, no mystery and, as simple as it is, you can’t ever get it right.
Try to think that thru and write down what really happens when the cardboard is in place, to show us improvement, we’ll help you get it right if you really want to learn. But I think you really want to just stir the pot with incorrect writing. I didn’t even have to misuse the word heat!
Ball4 says: “September 2, 2016 at 7:12 AM
“Will, your conclusions are wrong, thru misuse of terms you have never learned to use correctly.”
Trick, Please try to find even one error not associated with your brainwashing about “everything with temperature radiates”. in the presence of higher radiance in every direction an frequency. No `flux’ whatsoever can be emitted toward that higher radiance.
Whoever inculcated you with you nonsense truly learned less than 3% of what is known of Maxwell’s equations. Of course all western academics all have that same 3% allowing 97% ton agree on the same nonsense. Only the outside 3% ever learned how to think!
“It is foolish for you to write there is perpetual motion in the top post experiment. It is powered!”
With both the cardboard and the ice they only absorb power “actually emitted from the above plate, always proportional to the difference on opposing radiances. That is the only (singular flux)!!~
“The ice emission irradiates the hot plate, the ice radiates photon flux calculated from SB across the spectrum, absorbed, reflected when incident on the hot plate.”
The whole S-B equation, not the fictitious splitting, clearly indicates that one way flux. The hot plate never reflects anything not emitted. Your claimed ice emission is exactly that fake perpetual motion that does nothing.
“The hot plate irradiates the ice, hot plate radiated photon flux from SB, the ice absorbs,”
Correct and melts the ice! Again only that flux allowed by the whole S-B equation.
“reflects and transmits the hot plate radiation.”
That makes no sense at all!! Why is the ice melting?
“Please try to find even one error..”
Just read Will’s error filled comments and compare to the data in the top post experiment, plenty of obvious Will errors and misuse of terms. The data conclusively show the ice is irradiating the hot plate proving Will wrong. By test. No inculcation needed, fully in compliance with Maxwell & Planck EMR testing. It is Will not in compliance with Maxwell, Planck results not the test.
“The whole S-B equation, not the fictitious splitting, clearly indicates that one way flux.”
The experiment shows this is completely wrong Will, but give us more laughter and try to demonstrate your proof.
“Why is the ice melting?”
It is room temperature. Will should have known that but has to ask.
reflects and transmits the hot plate radiation. That makes no sense at all!! .
Of course not, to a commenter like Will that can’t understand even the most basic experimental dats results in this field in the top post. Will has not accomplished enough to understand. Try to learn from the experiment Will, plenty of help around here. It is ok to make your mistakes obvious to all and learn from them.
Ball4 says: September 2, 2016 at 4:14 PM
(wj: Please try to find even one error..)
“Just read Wills error filled comments and compare to the data in the top post experiment, plenty of obvious Will errors and misuse of terms. The data conclusively show the ice is irradiating the hot plate proving Will wrong. By test. No inculcation needed, fully in compliance with Maxwell & Planck EMR testing.
It is Will not in compliance with Maxwell, Planck results not the test.”
Please again attempt to find any physical error. All you have is fantasy nothing ever physical!
(The whole S-B equation, not the fictitious splitting, clearly indicates that one way flux.)
“The experiment shows this is completely wrong Will, but give us more laughter and try to demonstrate your proof.”
Roy’s experiment attempts to justify some opposing flux with only fantasy! I need no proof whatsoever, I only challenge your inane concept of opposing EM flux at the same frequency. Where oh where is even one demonstration of what you fictitiously claim.
Why is the ice melting?
Why is the ice melting? Even faster with Roy’s extra heat lamps?
Dr. Spencer’s test is not fantasy Will. Only Will’s untested comments are fantasy.
“I only challenge your inane concept of opposing EM flux at the same frequency. Where oh where is even one demonstration of what you fictitiously claim.”
Dr. Spencer’s top post demonstration.
Roy W. Spencer, Ph. D. says: August 31, 2016 at 4:43 PM
“Hmmm. Maybe we should start calling the resurrection of Maxwells Demon theory, Sky Dragons Demon.”
Challenge on:: stupid Idiotic insane Dr. Roy Spencer! Just try to defend your idiocy against those who have painfully learned repeated (soldering iron up your ass) truth of the physical! -will-
Challenge on:: stupid Idiotic insane Dr. Roy Spencer! Just try to defend your idiocy against those who have painfully learned repeated (soldering iron up your ass) truth of the physical! -will-
Yasir massa I be good, don beet me no more!
Dr. Roy these are the folk that must rape your daughters! then dispose of all that plus you into the top of the volcano.
Methinks that your “pseudo” has infected your brain Will and sent you insane.
Now if this were your blog and someone posted on it posts of that quality, would you be surprised if you were banned?
“Just try to defend your idiocy against those who have painfully learned repeated (soldering iron up your ass) truth of the physical.
That would be science then (well apart from the hot iron thing), since as Newton said “If I have seen so far it is because I stand on the shoulders of giants”.
Precisely what EM theory and consequently the GHE has evolved from. Empirical science, that works and hasn’t been proved to be false despite repeated observation, testing and prediction.
You seem to be some daintified creature that has not been (caned) fired
for some innocent “aw shit”. Happens every day!
EM theory is precise and works. The absolute corruption of that with CO2 CAGW is what is truly politically obscene!
Daintified?
No.
Just someone who is fascinated by the hubris displayed by “pseudos”.
And how, no matter what they do not shift.
Like “the whole world is wrong, and I insist it conform to the way I want it”.
Often (always?) from some overwhelming anger sourced from an ideological objection to AGW.
There are several good psychological studies of your kind if you’d care to look.
I definitely do not want to look into your septic tank. Myone is bad enough. Poor kitten shudders if I even consider!
You really ought to find a way of displaying the air movement within the system
There is convection from the hot plate, so air MUST be moving in from the sides and around the ice.
When you cover the ice, the moving air is warmer because the upward moving air is no longer cooled by the ice..
It is very possible that it is a CONVECTION/AIR MOVEMENT EFFECT that you are observing.
Try to find some way of re-doing the experiment with the ice and hot plate at the same level so convection effects are totally removed.
Andy,
Dr. Roy attempts to do ‘radiation’ as that is his thing. To him any conduction/convection simply cannot be as that is not what is the intent of this fool display of ignorance.
Radiative effects are so minor that they can be demonstrated only in a vacuum environment. Why oh why is radiative power transfer ever even considered within this Earths atmosphere?
“Why oh why is radiative power transfer ever even considered within this Earths atmosphere?”
Err ….. because we receive energy radiatively from the Sun….
And lose it radiatively to space??
In between radiative *stuff* happens, you know, like when you sit in front of a log fire in winter and feel the radiated warmth, even though the air temp is still cold.
And objects receiving that LW energy get warm.
Do you think that is the reason Will?
We can measure the total insolation in Watts outside Earth’s atmosphere
we measure the exact same Watts exitance from the Earth’s atmosphere into every other direction. Again: Why oh why is radiative power transfer ever even considered within this Earths atmosphere? Is this only to scam folk into putting up with meteorological nonsense?
Andy, convection results when a fluid in a gravity field is warmed from below. The hot plate in top post isn’t creating much convection since it is warming a fluid in a gravity field from above. Thus convection is not the source of the data.
Like approx. 10km of the stratosphere just above the tropopause where convection is minimal and you get a smoother ride on commercial jets.
That is but insane meteorological nonsense. In this planets atmosphere there is continual fluid mass motion in any direction for any reason, but mostly from the mechanical rotation of this planet.
The scammers of meteorology claim that ‘convection’ is the transport of heat vertically via temperature/density nonsense. While river fluid advection convects bottom silt downstream horizontally. What Bull Shit! Check with any HVAC expert upon thermal convection, fluid heat transport in any direction that the fluid is moving independent of any reason for such mass motion. All atmospheric convection must be circular in space with angular mass motion more important than radial mass motion. The outward mechanical radial convection transports heat, both sensible and latent to higher altitude where such entropy can be exited to space much more efficatively and effectively than can be done from any rocky surface.
“but mostly from the mechanical rotation of this planet.”
Nope – you talk of the Coriolis effect.
A definition:
https://en.wikipedia.org/wiki/Coriolis_force
“…. the Coriolis force is an inertial force (also called a fictitious force)[1] that ACTS ON OBJECTS THAT ARE IN MOTION relative to a rotating reference frame…..”
The air has to be set in motion FIRST.
Then it is acted upon by the Earth’s rotation.
You put the reaction before the action!
And convection is THE major action in moving air through the troposphere.
Oh, and I don’t give a to** about your “HVAC expert upon thermal convection”.
He/she has no expertise over planetary physics.
As your spouted bollocks is testament, to those who are.
Toneb says: September 3, 2016 at 2:24 PM
wj(but mostly from the mechanical rotation of this planet.)
“Nope you talk of the Coriolis effect.
A definition:
https://en.wikipedia.org/wiki/Coriolis_force” . the Coriolis force is an inertial force (also called a fictitious force)[1] that ACTS ON OBJECTS THAT ARE IN MOTION relative to a rotating reference frame..
Coriolis effect is not anything mechanical at all, it is an illusion.
At the equator, because of surface roughness the lowest airmass must be traveling tangentially eastward at 1000 MPH. The coherent mass inertia can be orthogonally separated into angular and outward radial components. That upward outward mass momentum is mechanically not thermally powered. This and similar outward motion at 60 degrees latitude provide over 85% of all atmospheric mass movement and convects 90% of lower power atmospheric heat upward to levels where such sensible and latent can be dispatched via EMR with greater efficacy than can possibly be done from the surface.
“He/she has no expertise over planetary physics.”
And you obviously have no expertise in any kind of science!!
of lower atmospheric heat upward to levels where such sensible and latent power can be dispatched via EMR with greater efficacy than can possibly be done from the surface.
“Coriolis effect is not anything mechanical at all, it is an illusion.”
Well done, yes, just what I said.
“That upward outward mass momentum is mechanically not thermally powered.”
No. As explained above nothing is “powering” the CF. It is an APPARANT force that we see as a reult of the Earth moving beneath as the air moves….. and it can only move due to deltaT >deltaP >wind.
Also there is no upward componant to Coriolis as the Earth moves eastwrad under an air parcel and does not move downwards. THINK! It is only a movement relative to the Earth rotating.
“And you obviously have no expertise in any kind of science!!”
Wrong again. I worked for the UKMO for 32 years the last 21 of them as an on-the-bench Forecaster.
Toneb says: September 5,2016 at 9:16 PM
wj(That upward outward mass momentum is mechanically not thermally powered.)
“No. As explained above nothing is powering the CF.”
Just what may be your “CF?”? Is that fictitious Coriolis nonsense?
With the air-mass 400 meters vertical and 200 km latitudinal and 1000 MPH eastward, from any inertial reference. Just what is your calculation of total outward atmospheric mass momentum? Has anyone ever taught you how to do that calculation? Can local gravitational attraction reduce the resultant outward mass velocity, with such continual applied momentum force applied from below all the way around the equator? Air-mass goes outward at over 100 MPH at a nominal altitude of 3.5 km ASL.
Toneb says: September 5,2016 at 9:16 PM
“It is an APPARANT force that w”e see as a reult of the Earth moving beneath as the air moves.. and it can only move due to deltaT >deltaP >wind.”
TOTAL RELIGIOUS BULLPUCKY! The Air-mass motion and its radial momentum are easily measured by those doing science rather than scam.
“Also there is no upward componant to Coriolis as the Earth moves eastwrad under an air parcel and does not move downwards.”
There can be nothing such as an air parcel with a well mixed troposphere! Such concept is but another religious scam!
“THINK! It is only a movement relative to the Earth rotating.”
Earth’s Atmosphere moves relative to both the inertial and to the rotating (accelerating) reference frames. Adherence to one can only result in scam for the other.
(And you obviously have no expertise in any kind of science!!)
“Wrong again. I worked for the UKMO for 32 years the last 21 of them as an on-the-bench Forecaster.”
Forecasting is a fine and a valuable service! Such has absolutely no place in science whatsoever! Forecasting is but the splitting of the deterministic from the statistical and chaotic. Forecasting gives what is and what may be likely! Such fall completely on its ass; when trying to put together WHY! Meteorology has more unsubstantiated myths than does astrology.
“Just what may be your CF?? Is that fictitious Coriolis nonsense?”
Not fictitious my friend because without it there would be no storms, as air would simply pass from regions of high to low pressure (at all altitudes)- no jet streams to carry them either
More teaching required I see. OK:
Imagine an air parcel over the equator. And another one over 45 deg N.
The parcel over the equator is travelling at (say) 25,000 miles/day – lets call it 1000mph
At 45 deg N :
Radius of Earth (ignoring oblate speheriod flattening) x Cos45 x Pi
4000 x 0.7071 x 3.142 ~= 9000miles circum = 370 mph
So:
When air over the equ is FORCED TO MOVE north to 45 deg N it finds (conservation of momentum) that the ground bolow it is moving west increasingly quickly, such that it displays motion to the right/east.
Same thing happens to the air at 45N when FORCED TO MOVE south. It’s slower eastward speed is met with the ground’s increasing eastward speed and so it moves to the right/west.
The SH is a mirror image.
https://en.wikipedia.org/wiki/Coriolis_force#/media/File:Corioliskraftanimation.gif
BTW:
https://www.theflatearthsociety.org/forum/index.php?topic=64533.0
Toneb says: September 6, 2016 at 10:50 AM
wj:(Just what may be your CF?? Is that fictitious Coriolis nonsense?)
“Not fictitious my friend”
First: of all a I am not your friend. Second: No Coriolis, no such force. I write @ Will Janoschka September 3, 2016 at 5:14 PM,
I write only of Centrifugal force, Earth’s centrifuge, centrifugal convection. This is the power source for both Hadley and Polar convection cells, nothing thermal. The Ferrel Cell is gravitationally powered, in the opposite direction. Please point out ever one scientific error in my 09-03-16 post!
“Imagine an air parcel over the equator. And another one over 45 deg N.The parcel over the equator is travelling at (say) 25,000 miles/day lets call it 1000mph”
Pritty close, and 100MPH upward at 3.5 Km.
“At 45 deg N :Radius of Earth (ignoring oblate speheriod flattening) x Cos45 x Pi 4000 x 0.7071 x 3.142 ~= 9000miles circum = 370 mph”
No math capability what so ever. Eastward surface velocoty 45 N 770 MPH lowering to 500 MPH at 60 degree latitude.
The airmass at the equator expands as it rises, splits to lower static tropopause pressure @ 30 lat and begins to descend resulting in surface trade winds near 15 lat and returning to surface low pressure doldrums near the equator.
Do they teach nothing at the UKMO except Wikipedia trivial nonsense meteorological lies? Coriolis effect is your illusion as you are standing on this accelerating (rotating) reference frame, that is nowhere flat!
“Seems like you do school to get brainwashed to be able to feed at the public trough!”
No I trained in physics to become a UKMO meteorologist and sat on the bench at both RAF fighter training stations and in public/commercial offices, using that knowledge …. which incorporates Coriolis.
Without CF the atmosphere would not work the way we see it and NWP models could not predict it..
It is BASIC empirical physics my friend.
And that you are ignorant of it, and what is worse deny it without any attempt to learn, says everything I need to know re my (past) *discourse* with you.
Ta ta
PS: I definitely don’t have the patience of Norman.
Life’s to short to get dragged down the rabbit hole too frequently.
Toneb says: September 8, 2016 at 3:10 AM
wj:(Seems like you do school to get brainwashed to be able to feed at the public trough!)
“No I trained in physics to become a UKMO meteorologist and sat on the bench at both RAF fighter training stations and in public/commercial offices,”
Trained in the pseudoscience of religious meteorology, to becomem a dreg on society! Point out even one technical error in my posts! You seen unable even to calculate planetary surface velocity.
“using that knowledge . which incorporates Coriolis. Without CF the atmosphere would not work the way we see it and NWP models could not predict it..”
So your catechism can predict the deterministic! Why do you call this any more scientific than the farmers almanac?
“It is BASIC empirical physics my friend.”
You cannot even describe what your ’empirical physics’ may be!
Please tell Dr. Spencer what you may mean.
You obviously have no comprehension of the scientific. When have you ever designed a empirical test of any of your religious fantasy?
And that you are ignorant of it, and what is worse deny it without any attempt to learn, says everything I need to know re my (past) *discourse* with you.
“I write only of Centrifugal force”
I write only of Coriolis.
Different thing my friend.
Do try to keep up.
“Do they teach nothing at the UKMO except Wikipedia trivial nonsense meteorological lies? ”
No they teach empirical science – you know? stuff that has been worked out for centuries (in this case) via observation, calculation etc ….. but more importantly in this case via COMMON SENSE.
“Coriolis effect is your illusion as you are standing on this accelerating (rotating) reference frame, that is nowhere flat!”
Err, have you ever looked down on a sphere?
Did it not look flat to you?
It is the component relative to a slice through the equator that is the driving force. It travelling around the axis at greatest speed and that at the poles Least.
Heck man, it’s not difficult!
http://www-pord.ucsd.edu/~ltalley/sio210/dynamics_rotation/lecture_dynamics_rotation.pdf
Look here my friend……
http://www.shodor.org/os411/courses/_master/tools/calculators/coriolis/
Notice the Greek character in brackets (Theta).
It’s the LATITUDE.
The monumental D-K inspired hubris on display – one day certain people will realise that “argument from authority” has merit.
That’s why we go to school/university, do jobs dealing with the matter. FFS.
Toneb says: September 7, 2016 at 2:42 AM
wj(I write only of Centrifugal force)
“I write only of Coriolis.”
Then you write only of fantasy!
“Different thing my friend. Do try to keep up.”
Please stop deliberately insulting me! If anything I remain your most feared enimy!
wj(Do they teach nothing at the UKMO except Wikipedia trivial nonsense meteorological lies? )
“No they teach empirical science you know? stuff that has been worked out for centuries (in this case) via observation, calculation etc .. but more importantly in this case via COMMON SENSE.”
This is the exact same verbiage as used by all the violent religious groups. We know!! We are skyintests!, you Rat Boy know nothing!
wj(Coriolis effect is your illusion as you are standing on this accelerating (rotating) reference frame, that is nowhere flat!)
“Err, have you ever looked down on a sphere?
Did it not look flat to you?”
The exterior of a spheroid, to me, appears as a Yang-Mills convex manifold, never with zero curvature.
“It is the component relative to a slice through the equator that is the driving force. It travelling around the axis at greatest speed and that at the poles Least. Heck man, its not difficult!”
Fantasy, but religious dogma in the Church of Meteorologie (French)!!
“Thats why we go to school/university, do jobs dealing with the matter. FFS.”
Seems like you do school to get brainwashed to be able to feed at the public trough! Is that not what weather forecasters do? Such has absolutely no scientific method nor scientific merit.
— AndyG55 says:
September 1, 2016 at 4:38 AM
You really ought to find a way of displaying the air movement within the system
There is convection from the hot plate, so air MUST be moving in from the sides and around the ice.–
Yes, it MUST.
One could choose to view this experiment as proving convectional heat is the only way to heat a surface above what it should be heated via radiant heat.
My premise is Greenhouse gases do not raise a surface temperature
above what the primary source of radiant energy [the sun] can achieve.
The highest temperature that sun can raise a surface temperature
to on Earth is about 80 C, though one can alter the sunlight by magnifying it, and thereby that magnified sunlight can melt bricks-
or exceed 1000 C.
So if talking about sunlight which is not magnified, the highest temperature the sunlight can heat a surface is about 80 C, and way
to get this temperature is to prevent convectional heat loss.
There are number of way to reduce convectional lose. Having a box
which is insulated and having the sunlight shine thru window which prevents the heated air from escaping. A greenhouse.
Another way is simply to have warmer air- the warmer the air, the less convectional heat loss is possible from a surface.
Deserts can have a high air temperature, deserts can have a higher
temperature of it’s surface.
By using saltwater which can prevent convectional heat loss, solar
ponds bottom surface temperature can be heated by sunlight to 80 C. The trick of solar pond reaching such high temperature is by
having enough water to create a salt gradient and not making the water so deep the energy of the sunlight is not absorbed by the water before reach the floor of the pond of which the sunlight is
heating.
Earth land surface can get to 70 C, lunar surface 120 C, and Mars surface 25 C. It gets to these temperature from radiant energy of the Sun. By preventing convectional loss sunlight at earth sunlight can cause a higher temperature by preventing convectional heat loss. The Moon has no convectional heat loss, and a calm day on Mars has very little convectional heat loss- preventing convection heat loss could increase the temperature on
Mars a little bit.
Oh I didn’t get to another variable regarding experiment, which is evaporation of the ice.
Evaporation of ice will make water droplets and water vapor.
Also evaporation is the creation of gas, which is another element adding a mechanical force [like a small fan] to convectional effect of the air.
Of course another variable unrelated to convection is the heated plate reflecting light from filament which +3000 K. Or you could use a heat lamp which heated to plate to same temperature but it’s element would instead be about 2000 K, to see if this alters it- changed how much is reflected. And with less intense visible light you could better see this reflected light.
It would seem to me that folks asserting radiative heating is a myth would be hard pressed to explain how a relatively cool magnetron cooks food in a microwave.
The magnetron converts electrical energy into EMR. The EMR is absorbed by the food. The food constituent KE increases until cooked. No radiant heating term required.
So the kinetic energy of the water molecules is increased by what if not the microwave energy from the magnetron? The 2nd and 3rd sentence above seem to be in direct contradiction. The energy transfer is not precisely the same as IR heating of Co2 [H2O has its vibrational KE energy raised versus Co2 electron energy level], but the [disproved] arguments that a cooler object cannot add energy to a warmer object seem to be of the same class in both cases.
The third sentence just shows I didn’t need to use radiant heating term in describing a mw, whether radiant heating is a myth or not is irrelevant.
CO2 electron level jumps are about 100x the energy level of collisions at Earth temperature so that mode is rare. CO2 rotational and vibrational quantum energy level jumps have about the energy difference found at Earth T so they are prevalent, translational KE is not quantized.
Ball4 says: September 2, 2016 at 8:30 AM
‘CO2 electron level jumps are about 100x the energy level of collisions at Earth temperature so that mode is rare. CO2 rotational and vibrational quantum energy level jumps have about the energy difference found at Earth T so they are prevalent, translational KE is not quantized.”
Very good Trick,
The work function for change in sensible heat of most mass is so low as to be immeasurable. The singular gauge boson ‘photon’ at the location of reception of EMR flux gets to decide which Planck unit of action is reflected, transmitted, or just ‘what the hell, close enough for government work’, absorbed as sensible heat!
“gets to decide which Planck unit of action is reflected, transmitted, or just what the hell, close enough for government work, absorbed as sensible heat!”
Complete pseudo bollocks!
How on Earth does it “get to decide”??
Is it Maxwell’s demon again ??
Again:
You do know that it a hypothetical mind experiment?
That it is a “piss take”.
And you’ve bought it?
Toneb says: September 3, 2016 at 2:29 PM
wj(gets to decide which Planck unit of action is reflected, transmitted, or just what the hell, close enough for government work, absorbed as sensible heat!)
“Complete pseudo bollocks!How on Earth does it get to decide??”
Why don’t you go somewhere buy a clue and learn just what a sub atomic gauge boson photon is considered to do. If and only if the power\energy\action fourspace density of the wave packet is high enough (2ev for nickel) can an electron be liberated from the mass (photoelectric effect).
“Why dont you go somewhere buy a clue and learn just what a sub atomic gauge boson photon is considered to do. If and only if the power\energy\action fourspace density of the wave packet is high enough (2ev for nickel) can an electron be liberated from the mass (photoelectric effect).”
We are not talking of the PE effect.
That involves the production of an electron, and most materials have their threshold frequency in the UV range of the EM spectrum …. The opposite end to that of IR !
We are talking of thermal energy, that produced when a tri-atomic molecule is sent into one of several vibrationary modes.
We are talking of thermal radiation. If not thermalised then the photon must be reflected or be transparent to it. It cannot be “rejected”.
I say you are wrong, but more importantly so does empirical science, so I return to you your phrase….. (With knobs on)
“Why dont you go somewhere buy a clue”
Toneb says:
September 4, 2016 at 7:58 AM
Why dont you go somewhere buy a clue and learn just what a sub atomic gauge boson photon is considered to do. If and only if the power\energy\action fourspace density of the wave packet is high enough (2ev for nickel) can an electron be liberated from the mass (photoelectric effect).
We are not talking of the PE effect.
That involves the production of an electron, and most materials have their threshold frequency in the UV range of the EM spectrum . The opposite end to that of IR !
“We are talking of thermal energy, that produced when a tri-atomic molecule is sent into one of several vibrationary modes.
We are talking of thermal radiation. If not thermalised then the photon must be reflected or be transparent to it. It cannot be rejected.”
Again: learn just what a sub atomic gauge boson photon is considered to do.
After the near surface CO2 downward field strength limits the upward 15 micron flux Some still exits. Because the elevated CO2 molecule bending moment amplitude is thermally high from mechanical convection effects the 15 micron lower flux passes through just like Kirchhoff demanded. however that elevated temperature CO2 molecule, though at lower temperature than the surface can still add to the outward exitance in the direction of an even lower radiance above. And so it goes! With exit flux at every wavelength accumulating with altitude all the way to 200km. This is just as F Miscolczi correctly calculated from the HiTran data base.
“I say you are wrong, but more importantly so does empirical science,”
Where do you have any, that anyone qualified could possibly call ’empirical science’? All you ever produce is computer generated fantasy.
” so I return to you your phrase.. (With knobs on)
Why don’t you go somewhere buy a clue”
Lead me to your vendor O wise one!! I always have much more to actually learn. But you have nothing but BS!
“Because the elevated CO2 molecule bending moment amplitude is thermally high from mechanical convection effects the 15 micron lower flux passes through just like Kirchhoff demanded.”
Unscrambling (perhaps)
A convectively lifted CO2 molecule has an elevated temp compared to other around it?
LWIR from the surface is not absorbed??
(If) My translation being correct – that is more pseudo-science.
Now that you have taken up the correct idea of vibratory modes of a GHG molecule being the agents of the GHE – what happened to your “PE effect” then?
More reading required from you I fear.
Oh, and just how did Kirchoff “demand”?
Toneb says: September 5, 2016 at 9:37AM
wj(Because the elevated CO2 molecule bending moment amplitude is thermally high from mechanical convection effects the 15 micron lower flux passes through just like Kirchhoff demanded.)
“Unscrambling (perhaps) A convectively lifted CO2 molecule has an elevated temp compared to other around it?”
No I clearly stated: Just that above. All of the elevated molecules have the same local temperature via collision. Without convection the non-condensing thermal lapse rate is calculated at -(14-17) Celsius/km. greater negative slope that the -10 Celsius (dry) or -5 Celsius (condensing) lapse. This for every tropospheric altitude the temperature is above that of radiative equilibrium.
test
“Oh, and just how did Kirchoff demand?”
See my above explanation! And please stop with your “More Holy than thou”. If you can identify any technical error, and I make many, please point such out from your own work not some absurd writings from someone who likely knows way less than you.
Toneb says: September 3, 2016 at 2:29 PM
wj(gets to decide which Planck unit of action is reflected, transmitted, or just what the hell, close enough for government work, absorbed as sensible heat!)
“How on Earth does it get to decide?? Is it Maxwells demon again ??
You do know that it a hypothetical mind experiment?”
There are clear physical evidence of your claimed ‘hypothetical’ at every subway terminal at theme park both on entrance and exit called ‘turnstiles’.
That is your piss take.
The wHOLE ANSWER IS HERE: http://joannenova.com.au/2016/09/weekend-unthreaded-132/#comment-1835015
Respond where-ever if you wish! I hab been notified of bad happening soon! I is eltz where/when, now!
“Without convection the non-condensing thermal lapse rate is calculated at -(14-17) Celsius/km. greater negative slope that the -10 Celsius (dry) or -5 Celsius (condensing) lapse. This for every tropospheric altitude the temperature is above that of radiative equilibrium.”
I’m sorry Will, but there is no LR “without” convection. Because without convection the atmosphere would not stir to allow it to become a “heat pump” where the LR is formed via the g/Cp relationship and the physical expression of the gas laws.
Are you advocating a thermal gradient maintained by gravity?
This isn’t one, but before I go through it (again) here (loath that I am to link to the site) is a refutation of it.
https://wattsupwiththat.com/2012/01/24/refutation-of-stable-thermal-equilibrium-lapse-rates/
actually, Toneb, this is correct. The greenhouse effect acting upon solar heating destabilizes the atmosphere (e.g. Manabe and Strickler, 1964)…the “pure radiative equilibrium” temperature profile is extremely superadiabatic. It is never realized in the real world, of course, because of convective overturning. It’s not just from radiative heating of the lower atmosphere, but radiative cooling of the upper atmosphere. I’ve blogged on this many times.
“it is never realized in the real world, of course, because of convective overturning. ”
Yes indeed Roy.
Roy W. Spencer, Ph. D. says: September 6, 2016 at 10:29 AM
“actually, Toneb, this is correct. The greenhouse effect ating upon solar heating destabilizes the atmosphere (e.g. Manabe and Strickler, 1964)the pure radiative equilibrium temperature profile is extremely superadiabatic. It is never realized in the real world, of course, because of convective overturning. Its not just from radiative heating of the lower atmosphere, but radiative cooling of the upper atmosphere. Ive blogged on this many times.”
Thank you Dr. Spencer!
Perhaps at some later time we can discuss some of the more scientific
interpretations to the meteorological concepts: (super)adiabatic, air parcel, and greenhouse effect; that have no science and no physical basis whatsoever.
Toneb says:
September 7, 2016 at 2:28 AM
rs(it is never realized in the real world, of course, because of convective overturning. )
“Yes indeed Roy.”
Does this mean you claim some understanding of how/where EMR exitance is generated, or how such propagates through this atmosphere?
Would you please try to evaluate the ‘brightness temperature’ of that narrow band magnetron might mean? The effective ‘brightness temperature of a 600 Watt MV oven is greater than 800 Kelvin.
In Dr Roy’s experiment if you took the average temperature of the room with the ice shield in place, and then compared it with the average temperature of the room without the ice shield, would not the former result in a lower temperature?
Not the whole room temp, as there is too much volume.
Just the air between the Heat Source & the target.
The IR thermometer is not very good at that kind of thing, as it would have difficulty focusing on just the air in that area.
There also seems to be some inaccuracies with the time base and the temp values as the Heating plate is “anticipating” the adding and removal of the ice shield.
The removal is anticipated by around 1 minute and when adding it at 8:10 it is also by about 1 minute
I have a question that I don’t think Postma can answer. He is fond of a thought experiment featuring two parallel blackbody plates. To eliminate radiation laterally through the gap, imagine the plates extend to infinity. The first plate is heated (let’s say by a power source delivering 100 W/m^2) but is insulated on the side farther from the second plate. So all of the 100 W/m^2 is delivered by radiation to the second plate, which then reradiates 100 W/m^2 to space (i.e., to the side away from the first plate.
Postma maintains that, at equilibrium, the second plate has a temperature of T2 = (100 W/m^2/sigma)^0.25 (no argument here). He also claims that the first plate has a temperature of T1 = (100 W/m^2/sigma)^0.25, since the first plate is emitting 100 W/m^2. And therefore there is no heat transferred between the plates at equilibrium, since they are the same temperature.
He believes this to be correct because the first law of thermodynamics is satisfied.
What he needs to answer is: What happens if the second plate has a lower emissivity than the first (e2 T2)
By Postma’s definition of equilibrium, T1 must also go up, until once again there is zero heat exchange between the plates. So T1new > T1.
But plate 1 is still a blackbody. So plate 1 is now emitting a flux sigma*T1new^4. This is more than plate 1 emitted at temperature T1 – in other words, more than 100 W/m^2.
My question to Postma: Since back-radiation does not exist, where is plate 1 getting this extra power from?
Somehow I typoed “(e2 T2)” above. It was supposed to be “(e2 < e1; e1 =1)".
“My question to Postma: Since back-radiation does not exist, where is plate 1 getting this extra power from?”
It doesn’t.
Your problem is you are not thinking about the power source. If your power source can only emit 100 W/m^2, then it won’t emit more than 100 W/m^2.
That is a strange power source,
A human body makes about 100 watts. Direct all heat of human body so
all it’s energy is lost or emitted thru a 1 meter square.
That pretty difficult and complicated- nevermind.
Burn a similar amount of fuel a human uses, and have it heat a pot of water which boils at some pressure and maintain this pressure.
Tricky part involves “maintain the pressure”. Isolate this, and use the water which always the same temperature [it’s always at it’s boiling point]. The simple point is, if play with amount it’s able to lose than the system will simply not radiate as much heat- but the heat source doesn’t increase in temperature- it’s controlled by pressure- and so one gets less than 100 W/m^2 being emitted.
Oh, same amount of fuel is equal to about 1 candle burning which is heating the water.
Actually a whole paragraph is missing:
“By my understanding, the temperature of the second plate must now settle at (100 W/m^2/(e2*sigma) )^0.25, which is higher than the old value we calculated for T2 assuming a blackbody. (T2new > T2)”
That paragraph should precede “By Postma’s definition…”
Scott says: September 2, 2016 at 1:36 PM
“I have a question that I dont think Postma can answer. He is fond of a thought experiment featuring two parallel blackbody plates.”
Postma believes ,although he has been corrected by everyone, that the temperature required for some power to be radiated to space by some surface area is a maximum temperature. It is not it is the minimum temperature. The maximum temperature approaches infinity as reflectivity plus transmissivity of the mass approach unity.
Because of the rethermalization of the un-powered plate the actual reactive impedance to space has essentially doubled to 754 ohms. Such
rethermalization effectively decreases original emissivity to 1/root2 of its value.
Please note such radiative rethermalization never occurs in this Earth’s atmosphere as mechanical convection always swamps any thermal radiative effects.
“Postma believes ,although he has been corrected by everyone, that the temperature required for some power to be radiated to space by some surface area is a maximum temperature. It is not it is the minimum temperature.”
You seem to think that’s the worst of Postma’s errors, or am I reading too much into that quote?
Scott says: September 2, 2016 at 6:25 PM
Postma articulates, (in words, not dimensional flexibility), though in some error.
You Scott seem to have no capability for articulating anything, whatsoever in any sense of such word!
Will,
I’m sorry to hear of your comprehension issues, which you project onto others as “no capability for articulating anything”. I know physics doesn’t come easily to everyone, but reviewing my comment, I realize that I articulated it quite well.
Reviewing your posts, however, I find a lot of unintelligible nonsense.
Postma does not articulate clearly either, and is wrong on top of that.
Scott says: September 2, 2016 at 9:30 PM
” Will,Im sorry to hear of your comprehension issues, which you project onto others ”
Giggle!
BTW have you seen some of those 8D Korean girly articulated robots?
Can blow your mind while you think you are being entertained!
Again,
I must send much respect to Dr. Roy Spencer! He gets much better each passing day! He has left my comments appear As I remain a contrarian never a Slayer. Pick a subject, pick a side, Ill argue, best I can!
To some here I have been over emphatic in my choice of wording! The PC wording leads only to yes sir, yes sir, three bags full type compliance, never learning. Stir it up, make people start thinking and review what TV brainwashing has lulled you into.
Thank you again, Dr. Spencer.
Congratulations !
Dr Spencer has invented insulation !
You are fooled by convection. Warm air rises and adding ice will cool the air below the plate and reduce the convection rate passing the hot plate. This must be done in absolute vacuum before you can conclude the effect is a result of IR.
And the total area of ice is larger than that of the plate.
The atmosphere is not totaly closed so the surface will be able to transfer energy to space. Therefor you should use a perforatet plate.
A cool object will not reduce the output from a hoter one. The maximum potensial will not change in the hotter object.
Adding more mass with lower energy level will increase reservoar and add time before the object cools down. The maximum potensial will stay the same. By removing convection you will come closer to maximum potensial. Thats what insulation does.
“This must be done in absolute vacuum before you can conclude the effect is a result of IR.”
Incorrect, Dr. Spencer minimized convection just as Dr. Planck did when he developed his law near 1bar and room temperature. Planck didn’t need a vacuum either.
Ball4 says: September 5, 2016 at 2:33 PM
(This must be done in absolute vacuum before you can conclude the effect is a result of IR.)
“Incorrect, Dr. Spencer minimized convection just as Dr. Planck did when he developed his law near 1bar and room temperature. Planck didnt need a vacuum either.”
Dr. Planck never once promoted anything of radiative flux W/m^2 All writings are on Specific (spectral) intensity. W/(m^2 x cm x sr). This can easily be seen from the S-B equation where not only are all the 1/cm integrated, but also the PI steradian solid angle is included as for a Lambertian surface.
Mr. Pettersen says:
September 5, 2016 at 2:03 PM
You are fooled by convection. Warm air rises and adding ice will cool the air below the plate and reduce the convection rate passing the hot plate.
I agree
Dr. Spencer’s own observation confirms that convection is the main factor in the experiment. He says:
The hot plate was so hot that just a small breeze of air from moving the room temperature sheet around the apparatus was found to cool the hot plate by a couple of degrees.
I suggest that the ice causes a region of cool air near it. This reduces the volume of circulation of air near the hot plate. This causes the hot plate to cool more slowly.
The ice affects the cooling of the hot plate by reducing the heat loss by convection and not by radiation from the ice.
Mr. Pettersen also says:
This must be done in absolute vacuum before you can conclude the effect is a result of IR.
A vaccuum is not necessary. The experiment should be repeated with the hot plate shielded from convection by a plastic film.
For those with eyes to see…
http://climaterealists.com/index.php?tid=36
http://climaterealists.com/index.php?id=9004
Maxwell, Woods, The Gas Laws and all ‘dense’ solar system atmospheres, all formed or formng stars, are right! Radiation is ineffectual within such atmospheres.
Brett,
Both you and Hans Schreuder are correct!
Dr. Spencer cannot deny some GHE in his position!
At the same time he could at least admit that there is no possible way to detect/measure some claimed reduction in EM radiative exitance from this planet and atmosphere due to CO2 composition in that atmosphere. ‘Tis but a SCAM, for monetary or political gain!
wrong, Will. The NASA AIRS instrument on the Aqua satellite has measured the decrease in outgoing IR radiation as CO2 has increased: https://www.youtube.com/watch?v=6-bhzGvB8Lo
Roy W. Spencer, Ph. D. says: September 8, 2016 at 5:19 AM
“wrong, Will. The NASA AIRS instrument on the Aqua satellite has measured the decrease in outgoing IR radiation as CO2 has increased: https://www.youtube.com/watch?v=6-bhzGvB8Lo”
You simply cannot measure Earth’s EM radiative exitance from that close. Your AQUA scanner demonstrated that! A Lunar instrument could give a fair estimate! Both wideband and spectral sensors at both L1 and L2 would result in far better accuracy. But that would blow the whole scam! Why no published data from polar regions?
“A Lunar instrument could give a fair estimate!”
What a total scam! Will should know the lunar instrument would have to cost more due to signal attenuation and to even get there at that increased distance. Will is simply in the pay of a lunar instrumentation company to write that.
Ball4 says: September 8, 2016 at 9:55 AM
wj(A Lunar instrument could give a fair estimate!)
“Will is simply in the pay of a lunar instrumentation company to write that.”
Trick,
Please get a clue, I am not “in the pay of”, instead, I am the lunar instrumentation company!
From the moon, Earth’s diameter subtends 32 milli-radians (mr), so a 1 mr instrument could do an 800 point survey of Earth’s spatial/temporal radiance from its location as the Earth spins, while providing a repeatable 28 day history of the Sun’s insolation upon the earth. Still can’t do the poles well!
Pray tell what you think is “a fair estimate”? Lets all see if Dr. Spencer responds, as he is king of remote sensing.
Why all the extra effort when it has been known that the net power output from the surface of the warm body is found by the Stefan-Boltzmann power output of the warm body minus the Stefan-boltzmann power output from the surface of the cooler body to which the warmer radiates. To more easily see that the net power from the warmer’s surface is less than when the cooler surface is absent, look at it as a warm body radiating to a surrounding cooler shell. If it is an isolated system, the warmer will not get any warmer but will cool more slowly, perhaps what you mean by warmer than it would be otherwise, similar to nighttime Earth. Its temperature will not increase. In an open or closed system the temperature may increase due to external energy being added to the warm body and not being able to be radiated fast enough, similar to daytime Earth.
As for the Earth, the whole problem is difficult due to there not being any temperature data for the solid surface, just for the air temperature at a standard level above the surface. With phase changes for water and ice and heat capacity for water and ice, there is much more difficulty in determining the net power output from the surface.
There should be not question about the radiation.
My problem is how to go from surface emission to temperature to heat content of surface material since conduction transfers heat to depth and internal energy to surface. There would be a temperature gradient through the surface.
Point the ir thermometer at the ice and measure the temperature then do the same with the shield. The shield is hotter, meaning it radiates more heat back to the emitter, which lessens the heat lost from the emitter.
You can’t heat a hotter source with a colder but you can prolong the cooling with insulation, which is exactly what’s taking place here.
I’m not really sure what this experiment intends to prove. Can anyone fill in the blanks for me?
You have to pay attention to the view factor.
http://web.mit.edu/course/16/16.unified/www/FALL/thermodynamics/notes/node137.html
What I like most about this ‘activity’ or ‘demonstration’ is the fact that High school physics or physical science teachers can reproduce it in the classroom encouraging a terrific discussion to follow.
Dr. Spencer,
A simple experiment to prove/disprove CO2 effect: Take a box of air with heat lamp on top. Place thermometer on floor of box.
1) Turn on lamp for specific time interval, and measure temperature.
2) Add another 400 ppm of CO2, while removing 400 ppm O2. Repeat test.
3) Add another 400 or 800 ppm of CO2, while removing same amount of O2. Repeat test.
Do this as many times as you wish until you reach 2000 ppm CO2.
Let everyone know the results.
I’m actually surprised no proponent of ‘CO2 Theory’ has done this already. The burden is upon them to prove theory.
Thank you!
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Nice experiment, but very far from proving what it advertises. This is well explained by gbaike, Mr Pettersen, Roger Clague. If we combine this, with the absurdity of the thought experiment of the two plates, one of which heated (when pressed in contact they reach one temperature, while at a distance the heated one heats up further, due to the “back magic” from the second plate ) we can safely conclude that the GHE is a total scam. Simple persons get it already. Phd’s will just need more time.
I have made a simple calculation model of Dr Roy’s set-up that takes into account both radiation as well as convection. Obviously with the many uncertainties due to the dimensions being guesstimated from the pictures. What turns out is that radiative changes due to the shield (if any) are minimal with respect to convective effects (the latter being in the order of 3-5 degrees F, more or less as measured, while the former more like a few tenth of degree, barely measurable). Therefore this confirms earlier comments that I had highlighted with my previous post. If anyone is interested I can share the relevant excel spreadsheet. So again, no GHE whatsoever, convection rules !
Turn the experiment on its side to avoid convection. Put the heat source behind the black metal. Put the cardboard on the top of the ice. Remove the ice and try the experiment. Replace the ice with a mild heat source. Replace the ice with frozen CO2. Place another temp sensor in he air 2 feet below the black metal.