Home/BlogAs a follow-up to my controversial post on the effect of infrared “back radiation” downwelling from the colder sky to the warmer surface, the existence of which some dispute (despite the real-time availability of such data), I’ve come up with an experimental setup to see how IR radiation from the sky influences air temperature near the ground. (Yes, I know some of you think there is no such thing, but please indulge my fantasy as if it was true, ok?)
The design is pretty simple and inexpensive, and looks a little like the blackbody radiators that are used as calibration sources. The following cartoon shows the main components:
The idea is to isolate a sample of air and control its environment so that it’s main source of energy gain or loss is through an opening that looks at the sky. You have probably noticed that on a clear evening, dew forms first on the tops of cars and other surfaces. This is because these surface are losing IR energy faster than the air and other surfaces are, so their temperature falls below the dewpoint temperature first.
If we can isolate that effect sufficiently from other sources and sinks of energy, we should be able to get air temperature drops within the cavity in the direction of the colder, effective sky temperature. (We use air since it is very hard to measure the temperature of a cold surface accurately, so we let the cold surface inside the cavity chill the air in contact with it).
The cavity will be lined with aluminum foil, which has very high reflectivity in the infrared, painted on the inside only with high-emissivity paint (Krylon flat white, #1502 if I can find it…apparently, black paint isn’t as good an emitter in the IR.)
The 2 thin polyethylene sheets are in the upward-looking cavity opening to trap a layer of air for thermal conductive insulation, while at the same time passing most IR radiation (something polyethylene is apparently quite good at). The thermal conductivity of the trapped air is a little better (less) than that of Styrofoam, but since convection can occur in an air cavity, I’m sure the actual rate of heat transfer will be more than that for Styrofoam.
SO WHAT KIND OF SIGNALS CAN WE EXPECT?
(…assuming the experiment isn’t a complete failure because of something important I haven’t thought of…)
If you search around on the internet you will find that those who have made such broadband IR measurements of the sky (from what I can tell, usually with instruments that measure between 8 and 14 microns wavelength) report that the effective sky temperature in the infrared is usually 10 to 30 deg. C lower than the near-surface air temperature. Ten deg. C is more typical during humid conditions or cirrus cloud cover, while 30 deg. C would be during clear, low humidity conditions.
Low clouds produce a downwelling sky temperature nearly the same as the upwelling temperature. The sky temperature increases as you scan from the zenith down in elevation, due to the greater path length through the atmosphere.
As an example of the theoretically-expected difference in IR energy flows in and out of the cavity, at an emissivity of 1, a cavity at 300 K temperature should emit a broadband IR flux of 459 Watts/m2, while a downwelling apparent temperature of 290 K (10 deg. lower than the cavity) would produce 401 Watts/m2, the difference being 58 Watts per sq. meter.
In a perfect setup with a cavity emissivity of 1 and no other losses of energy under these conditions, the inside of the cavity would then cool to 10 deg. C less than the surrounding air temperature as the insulated cavity comes into radiative equilibrium with the sky. (I am currently monitoring 2 temperatures in my back yard, with the data sent to my computer by wireless. My first design failed due to large conductive energy loses, which led to the 2nd design, above).
Of course, a “perfect” experimental setup is not possible. I’ve run some numbers based upon the thermal conductivity of Styrofoam and I think I can keep the energy loses to about 20% of the signal being sought, but this is uncharted territory for me.
OK, TIME FOR YOUR PREDICTIONS
So, for all of you who think you know what will happen in this experiment, come on and tell the rest of us. Will the temperature of the air in the cavity stay the same? Will it cool? By how much?
I especially want to hear an answer to 2 questions:
(1) If you think the cavity will be the only source of IR radiation, and there is no downwelling IR radiation from the sky, then what will keep the air temperature inside from falling dramatically lower than the air temperature outside of the box?
(2) If you think the temperature in the cavity will not change, then what is keeping the IR radiation flowing out of the cavity toward the sky from causing a temperature fall? Wouldn’t want to violate the 1st Law of Thermodynamics, ya know.
Let the thinking begin.
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Dr. Spencer,
I am delighted that you are actually going to conduct this experiment – my guess is that your chamber will show lots of cooling on a clear night, and that cloud cover will show up as increased temps in the cavity.
Are you familiar with Robert W. Wood’s experiment to investigate the “greenhouse effect?” He used a pair of cardboard boxes painted black inside, a pair of calibrated thermometers, and all packed in cotton. One box had a plate glass window and the other one made of salt. Salt is amazingly transparent in the IR. The boxes were exposed to sunlight. The salt window box got up to about 65C, and the glass window box to 55. Wood placed a glass window between the sun and both boxes, whereupon both temps reached about 55C. He concluded that radiation played only a minor part in cooling the salt window box. The windows weren’t double. I can’t locate a web reference to this, albeit I think that’s where I read about it. Hope I remembered the numbers correctly.
Your polyethylene might not be as good a window as salt – I think there is absorption around 13 or 14u, not far from your blackbody peak at 300K. Not sure about polystyrene – your illustration shows that material instead of polyethylene. Much of this info is hard to come by on the web – locked up behind pay windows, which I’ve determined to be opaque to visible radiation.
-Jerry
If WalMart sold salt windows, I would use them instead.
I think someone needs to do a new experiment regarding Wood’s setup. I have computed the theoretical energy flows, and if Woods’ two boxes are not essentially identical in their rate of non-radiative energy loss to the surroundings for a given interior temperature rise, it is difficult to see a difference one could ascribe to IR alone. Also, did he take into account the different thermal conductivities of salt versus glass?
Wood’s analysis was rather poor, he admitted so himself. For your experiment I’d suggest using premium Saran wrap, it’s polyethylene and very thin which should minimize absorption. It’s better than polystyrene which has more absorption lines in the region of interest. I’d make sure that both temperature sensors were shielded to avoid direct radiation loss. The styrofoam box will probably work well but I’d be inclined to to use a wide-necked thermos flask.
I was wondering whether Saran Wrap would be better, but didn’t know what it was made of, thanks.
Another thought, a control experiment with the top layer of Saran wrap replaced by Al foil, that would eliminate the radiation loss but keep everything else constant.
yes, when I was making the new cavity last night, I realized it was just as easy to make it twice as long, then separate the two cavities so one could be a control. Haven’t decided what that control should be, though. Yours is a good suggestion. Turns out Glad cling wrap is very thin (0.5 mil = 12.7 microns), and is pure polyethylene, so I’m going with that.
Yes it’s the thinness that makes it attractive given the exponential dependence of absorption with thickness. I’ve rethought my suggestion about Al foil, the metallised Mylar film would be even better, it’s thin like Saran but reflects 99% up in to the IR. It’s also incredibly cheap, it’s sold as emergency thermal blankets in outdoor stores for about $1!
How about this for sky radiation?
“The Diurnal Atmospheric Bulge, giant 1200º bulge of rapidly heated and expanding gases circling the Globe 24/7.”
You really do have to be kidding (Anon at 1:45 pm)? Please do tell us who you are anon…..and then scurry back to WUWT.
Good grief, people seem to be willing to believe any rubbish if it is on the Internet and written in vaguely scientific language.
It is astonishing that somebody is willing to waste so much time trying to disprove something that has been well proven for the last 200 years. Has he not heard of a “spectrometer”?
If you have one you simply have to pass light through a gas and measure how much is absorbed at each frequency. No wine glasses required.
It is like trying to discuss the measurement of temperature while wilfully ignoring the existence of the “thermometer”.
Is this a joke? The fact that the author’s name is “Pratt” may be a clue.
LastMan, yes, I frequently point to the AIRS (on Aqua) satellite instrument that measures at thousands of IR wavelengths.
Now, I can’t speak for “Pratt”, but the reason I am addressing this issue is that there is an increasing number of people who are advancing the notion that there is no such thing as “back radiation” (or that “more CO2 is incapable of causing any warming because…”), and as a result I am deluged with e-mails from them and the public asking me for my opinion. Also, working through the details helps me to better understand radiative transfer, energy budgets, and temperature change.
I sympathise. And I do hope that posting these thought experiments will lead to a reduction in your email inbox, although I suspect otherwise!
I bet you are glad you don’t have to teach. My admiration for those that do is rising by the minute.
The ease of publishing stuff on the internet, and in particular the Blogosphere, seems to have led to a mass outbreak of the Dunning-Kruger effect. When unskilled people lack the ability to realise they have made a mistake they get an inflated opinion of their own abilities.
Publishing “the first thing you dreamt up” on the net seems to “set” some people’s views. They seem to think that because they have written it down, and that it makes sense to them, it must make sense to everybody else.
For those not familiar with Dunning-Kruger, here is the Wikipedia definition:
Sounds like a subset of Climate Scientists!!
The point of the article which you all seem to have ignored, is that according the bogus “greenhouse effect” the atmosphere is supposed to be transparent to incoming electromagnetic radiation.
The Diurnal Atmospheric Bulge proves that this is pure fallacy. The typical tactic of isolating irrelevant points and taking them out of context in order to attack and discredit them is old and tired.
Try addressing the obvious point of the article which is the fact that the atmosphere is clearly heated top-down by incoming solar radiation as opposed to bottom-up as per the silly “greenhouse effect” hypothesis. Try to remember what your bogus theory stipulates and how that flies in the face of reality.
Roy,
Interesting experiment! See this link for polyethylene transmission. If you cover the outside of the styrofoam box with unpainted aluminum foil, it should perform better. One layer of polyethylene may be better than two. Two layers will cause more IR blocking than one. The air near the polyethylene may be warmer, but that is the top. Since the warm air is less dense, the air should remain stratified, without convection.
If the air is still, the thermometer outside the box may also be cooled by radiation and there may not be enough convection to keep it at the air temperature. The difference between the two thermometers would then be artificially small.
It looks like the dewpoint at Huntsville is currently ~72F. If you assemble your box under those conditions, the temp in the box cannot drop below 72F.
As you mentioned, you will have some conductive heat losses.
I predict the temp inside the box will drop from after-dark ambient to 72F and then remain there throughout the rest of the night.
The temperature can indeed drop below 72…it just means that some water will have to condense out in the process.
Firstly if there were no barriers I would expect the cavity to approach the outside temperature but perhaps not reach it at its depth depending on the duration of the experiment.
But, given, it appears two insulating cavities. I would expect the top cavity temp to drop by some amount but remain greater than the outside temperature.
The tricky one, the bottom cavity I would expect its temperature to drop more slowly that the cavity above it whilst remaining greater than the cavity above it, which is greater than the outside temperature.
There are possibly a number of variables the influence of which may effect the experiment such as: duration, whether both cavities are 300k at the start. Although an effort is made to minimise the absorption of the container what is it ? Is there really convective air flow in such a small confined space ?
The temperature in the box should decrease after nightfall. As you mention, Roy, humid air blocks IR more effectively and the final temperature will be higher on a humid night than a dry one with the same air temperature. The dew point is not the limit, however.
I think you need some sort of control. The purpose of the experiment is to demonstrate the existence of IR radiation from the sky (correct?). If there is no IR radiation from the sky, the inside of the box should get very cold. The temperature decrease would only be limited by imperfections in the experiment (conduction through the box, radiation from buildings nearby, etc.) If there is radiation from the sky, the temperature inside the box should still decrease, but the decrease would be limited by the temperature of the layer of the atmosphere that provides most of the radiation. The experiment does not clearly distinguish between the two possibilities.
The control must be a surface colder than the temperature of the atmospheric source. Dry ice ought to work as the control. Make a second box like the first, but with dry ice behind the painted Al foil on the bottom and sides. It would probably be better if the opening on the control is larger. You definitely need a double window on the control. Cover the experiment with the upside-down control, but leave some space for convection from the outside. Since convection may cool the experiment directly, you need to try two arrangements: experiment on bottom with control on top, and control on bottom with experiment on top.
If there is radiation from the sky (at a temperature significantly greater than dry ice), the original experiment should result in a higher temperature than the experiment with control.
Anonymous says:
July 27, 2010 at 1:45 PM
How about this for sky radiation?
Ouch, that made my head hurt. There are so many things wrong with that I don’t know where to begin. But to start with the B.S. about transparency, light vs. dark shadows, the wine glass pictures, etc. Other optical phenomena such as reflection, refraction, (constructive and destructive interference), scattering, etc. come to mind first.
The atmosphere is transparent to visible radiation (i.e., light), which coincidentally comes from the sun. Hint: open your eyes.
That’s quite surprising and interesting turn of events. I’m very interested in results of your experiment. However, I suspect you slightly misread what skeptics of “back radiation” theory say. Or, at least, what I say. I do not contest “back radiation” per se. I contest “back radiation” *theory* that is nicely expressed in the title you previous article “Yes, Virginia, Cooler Objects Can Make Warmer Objects Even Warmer Still”.
The ability of warmer objects to absorb thermal radiation of colder objects and get warmer as the result is the open (for me) question. Showing that temperature of cavity drops to that of the sky above or stays level with outside air neither proves nor disproves the theory.
I look forward to your follow-up article with results or your experiment.
I look even more forward to your next experiment that replicates (as much as possible) your thought experiment from the aforementioned article 🙂
Dear Dr. Spencer,
As you know I have been fairly critical of some of your statements and research on climate sensitivity in the past. So I am especially happy to be able to offer you my support to you and commend you for (trying) to explain the physics behind the well-established (if not inappropriately named) “greenhouse effect” to those in denial about AGW/ACC.
Perhaps you will now appreciate and understand the frustration that climate scientists and those in related discipline shave in trying to communicate the basic, yet complex, underpinnings of the theory of AGW to certain elements in the population. It seems that those afflicted with D-K are now flocking to your site and doing their best to inform you that over 120 years of science are quite simply wrong. I find it telling that none of the people arguing with you have yet received Nobel prize in physics.
Those skeptical of the “greenhouse effect” should note that Drs. Pielke and Lindzen agree with the physics and radiative transfer theory presented by Dr. Spencer (only days ago Dr Pielke wrote on the subject). So the contrarians who are, how shall I say this, out in the cold.
Could it be that there is a huge cavern about to break the already sub-critical mass contrarian/”skeptical” group in two?
Best of luck,
MapleLeaf
For most of those 120 years the AGW idea was considered disproven. Every time it surface a competent peer disproved it for that time period. We are now on the latest outbreak and the result is still in doubt. There is no 120 years of accepted AGW theory.
For those interested in quantitative radiative cooling and water condensation, see the thesis:
Optical Scattering Properties of Pigmented Foils for Radiative Cooling and Water Condensation: Theory and Experiment.
Torjorn M.J. Nilsson
Department of Physics
Chalmers University of Technology
Goteborg Sweden 1994
ISBN 91-7032-941-9
Abstract
“Light scattering properties of inhomogeneous materials were investigated in this thesis. . .
“A 400 um thick ZnS pigmented foil with a pigment volume fraction of 0.15 was prepared and tested in Tanzania. The heating power at noon was 7.2 W/m2, and the temperature of the radiator was 1.5 K above the air temperature. The field experiments agreed well with simulations. . . .
“A 390 um thick foil with TiO2 and BaSO4 pigments was produced and was found to have a fairly high emittance in the thermal infrared; it was therefore used in outdoor experiments. During drought months in Tanzania, the foil condensed 1.19 litre/m2 in a semi-desert. . . .”
“Secondly, the transmittance for 0.33 < lambda < 1.0 um is considerably less than the theoretically predicted value of 92%. Thirdly, there is a "background" absorption in the IR for wavelengths between 2 and 20 um." p 179
He provides detailed heat balance equations etc.
Gerlich and Tscheuschner(2007) has as description of the Woods experiment.
http://arxiv.org/PS_cache/arxiv/pdf/0707/0707.1161v3.pdf
Dear dr. Spencer,
This experiment is too complicated for me. Sometimes, nature does a cheap and simple experiment. Back radiation predicts that when Sun, Jupiter, and Saturn are on one line, outside temperature of the sun will drop with a delay of distance(Sun, Saturn)/c seconds.
…and when Jupiter aligns with Mars, then peace will guide the planets, and love will steer the stars.
Why not measure the “Back Radiation” (from the cold atmosphere) heating of the Earth directly?
This has been done thousands, if not millions of times.
The device to use is called a Parabolic Mirror.
As I’m sure you are aware, a parabola has a focal point where all Electromagnetic Fields (and even sound waves) can be concentrated at a focal point.
This is the basis for Satellite Dish antennas, parabolic microphones and even Parabolic Mirror Solar Ovens used over the entire Earth.
Here is a link to some measurements done by the Physics Dept. at Brigham Young University where they have used Solar Ovens to COOL water and even FREEZE water when the Solar Oven is pointed at the cold atmosphere during the Day and Night.
——
Solar Cookers and Other Cooking Alternatives
“The second area of solar cookers I looked at was their potential use for cooling. I tested to see how effective they are at cooling both at night and during the day. During both times, the solar cooker needs to be aimed away from buildings, and trees.
These objects have thermal radiation and will reduce the cooling effects. At night the solar cooker needs to also be aimed straight up towards the cold sky. During the day the solar cooker needs to be turned so that it does not face the Sun and also points towards the sky.
For both time periods cooling should be possible because all bodies emit thermal radiation by virtue of their temperature. So the heat should be radiated outward.
Cooling should occur because of the second law of thermodynamics which states that heat will flow naturally from a hot object to a cold object.
The sky and upper atmosphere will be at a lower temperature then the cooking vessel. The average high-atmosphere temperature is approximately -20 °C.
So the heat should be radiated from the cooking vessel to the atmosphere.”
http://solarcooking.org/research/McGuire-Jones.mht
——–
This link shows that heating of the Earth’s surface cannot occur from the colder atmosphere.
In fact, the article shows how to COOL items placed in the Solar Oven at NIGHT AND DAY!
All you have to do is point the Oven away from the Sun during the Day and the Oven will transfer heat from the WARM object in the Oven to the COOLER atmosphere!
It can even be used to produce ICE when the ambient air temp is +6 deg C!
“If at night the temperature was within 6 °C or 10°F of freezing, nighttime cooling could be used to create ice. Previous tests at BYU (in the autumn and with less water) achieved ice formation by 8 a.m. when the minimum ambient night-time temperature was about 48 °F.”
And, this also confirms the validity of 2nd Law of Thermodynamics….heat energy CANNOT flow from Cold to Warm objects.
Further, if the Back Radiation (which the AGW’ers say is available Day and Night) actually could reach and heat the Earth, Solar Ovens would produce heating DAY and Night!
Our energy problems would be over!
…by conduction.
As Dr S. and others here have stated, over and over again, we are talking here about RADIATION.
Any object with a temperature higher than absolute zero will emit electromagnetic radiation in a vacuum. The wavelength of that radiation depends on the temperature of the object. The hotter the object the shorter the wavelength. The Sun emits white light because it is very hot. The radiation from inter-stellar dust is in the far infrared because it is very cold.
When an object is radiating energy, it can make no decision about what that radiation will land on. It cannot say, “oh, I had better not emit any radiation in that direction because there is a hot object over there, I had better emit it in the other direction where it is colder than I am”.
Unless, of course, you think:
a. every object is a sentient being
b. they have the means to direct where they emit radiation
c. they will always make the same decision not to radiate in the direction of a hotter object.
Get a life…
…or better still an education. It is never too late.
The Solar Oven produces heat by concentrating RADIATING Electromagnetic Field energy at a focal point.
The Aperture Area of the Solar Oven determines the Watts available (Watts/m^2 X Aperature Area = Watts)
The Watts received are concentrated in the smaller focal point Area increasing the watts/m^2 which increases the temperature.
(See the Stefan-Boltzmann Law)
The “BACK RADIATION” (which, according to Trenberth’s Earth Energy Budget is 324 w/m^2 at the Earth’s surface) can’t even heat a tiny bit of water at the focal point of a Solar Oven where the “BACK RADIATION” is CONCENTRATED!
Like All Parabolic Antennas, Solar Ovens can be used for Receiving or Transmitting Electromagnetic Fields.
In fact, the RADIATION from the warmer water is being RADIATED to the Colder atmosphere causing the water to FREEZE even when the ambient air temperature is above freezing.
All RADIATED HEAT ENERGY CAN ONLY FLOW FROM HOT TO COLD OBJECTS, just like the 2nd Law of Thermodynamics says:
“Second Law of Thermodynamics: It is NOT POSSIBLE for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.”
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/seclaw.html#c3
And just like the Physics Dept. at Brigham Young University says:
“Cooling should occur because of the second law of thermodynamics which states that heat will flow naturally from a hot object to a cold object.”
And, their measurements PROVE that Back Radiation from a colder atmosphere CANNOT heat up a warmer Earth surface.
———-
Electromagnetic Fields are Vector quantities and one must use Vector Mathematics when summing EM fields.
Electromagnetic radiation
“Electromagnetic radiation (sometimes abbreviated EMR) takes the form of self-propagating waves in a vacuum or in matter. EM radiation has an electric and magnetic field component which oscillate in phase perpendicular to each other and to the direction of energy propagation. Electromagnetic radiation is classified into types according to the frequency of the wave, these types include (in order of increasing frequency): radio waves, microwaves, terahertz radiation, infrared radiation, visible light, ultraviolet radiation, X-rays and gamma rays. Of these, radio waves have the longest wavelengths and Gamma rays have the shortest.”
“EM radiation carries energy and momentum that may be imparted to matter with which it interacts.”
“Electric and magnetic fields do obey the properties of SUPERPOSITION, so fields due to particular particles or time-varying electric or magnetic fields contribute to the fields due to other causes. (As these fields are vector fields, all magnetic and electric field vectors add together according to vector addition”
http://en.wikipedia.org/wiki/Electromagnetic_radiation
Heat flux
“Heat flux or thermal flux, sometimes also referred to as heat flux density or heat flow rate intensity is a flow of energy per unit of area per unit of time. In SI units, it is measured in [W·m-2]. It has both a direction and a magnitude so it is a vectorial quantity.”
http://en.wikipedia.org/wiki/Heat_flux
Vector addition of fields…
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/mulpoi.html#c3
This shows how resultant field vectors are calculated.
Using superposition, many, many sources can analysed to produce a single resultant vector.
The “BACK RADIATION” has an EM field Magnitude of 324 w/m^2 and a Direction of propagation towards the Earth.
The Earth’s Radiation has a Magnitude of 390 w/m^2 and a Direction of propagation towards the Atmosphere.
The RESULTANT EM FIELD has a Magnitude of 390-324 = 66 w/m^2 and a DIRECTION of propagation towards the Colder Atmosphere!
There is ZERO w/m^2 propagating from the Colder Atmosphere to the Warmer Earth.
This totally complies with the 2nd Law (“It is NOT POSSIBLE for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow.”) and with the MEASUREMENT PROOF!
So, how does a photon know not to go faster than the speed of light?? How do triangle with right angles know to have all sides comply with Pythagoreans theorem? How does gravity know to attenuate with distance?
That is a really sad argument. The physical world does what it does whether we understand all the technical reasons or not.
About that temperature and energy bit. Without an increase in energy the body will not rise in temperature. Ever hear of kinetic heat??
Mindert, while you are right in principle, you might want to plug in some numbers for how large that temperature drop will be. I think detecting it will be about as hard as detecting the sea level rise from someone spitting in the ocean.
Thomas, I am not a physicist. Hopefully, some physicists may provide for the details. Note, that this is not my idea but something which follows from the back radiation theory. I guess that some physicists will consider the prediction ludicrous anyway. Perhaps, the experiment can be done in a vacuum chamber.
Gord, now repeat that experiment with a parabolic mirror on the moon where there is no atmosphere and you will find that the item cools a lot more quickly.
Since a clear sky is colder than the surface a solar oven isn’t expected to produce any heating. You may be fooled into believing that a large reflector will gather a lot of extra energy, but it won’t, it will just collect energy from a smaller part of the sky, which doesn’t matter if the sky is isothermal.
Mindert, the “details” is that the effect is far too small to observe. If you instead took an infrared detector on Earth, pointing it at Jupiter when Saturn is passing behind it, you would be able to detect a dip in radiation, which is no more surprising than that you will see a dip in total visible light received.
To doubt that downward radiation exists is folly. My question is since CO2 absorbs/emits at about 15 micro that equates to a temperature of roughly 200 K per Wien’s Law. The CO2 back radiation of 15 mirco to a 300 K surface (10 micro) will not cause the surface to get hotter. i.e. go to 301 K.
You say metal cools faster than air due to losing IR energy faster than air. It is just as true that steel has a specific heat about .5 so has a low capabiltiy to hold heat just as CO2 has a specific heat of .87 and has a low capibility to hold heat. Conduction and convection could account for the rapid loss down to air temperature.
It will if the 300k object is being continually heated to keep it at that temperature.
Another thought experiment for you, based on the extreme case. Imagine your 300 K object is the only object in the Universe. It is in a vacuum surrounded by pitch black everything. It has an internal heat source meaning that despite this hostile environment it is at equilibrium at that temperature. The rate at which it emits radiation is balanced by the heat it is internally generating.
Now assume that spontaneously, out of nowhere, it is surrounded by 1,000 similar objects all at an equilibrium temperature of 299K, i.e. colder than it is.
Leave the objects for a day to reach a new equilibrium. Will the temperature of the 300k object have risen?
If you think it cannot go above 300k you would have to assume that being surrounded by 1,000 objects similar to yourself all emitting energy at a slightly longer wavelength is no different from being surrounded by total cold blackness.
Like many here you are confusing heat with temperature. The temperature of an object will rise if either:
a. Its ability to lose energy is impaired
b. The amount of energy it receives increases.
In this case it is receiving radiated energy from 1,000 objects only 1k colder than it is.
The temperature of the object will rise, in other words it will get “warmer”.
In your original idea the back radiation will cause the 300k object to rise in temperature. Maybe not even by as much as 1K, but rise it will.
So if I take a blow torch at 5000F and place another blow torch also at 5000F beside it so the flames converge at a point will the temperature be 10000F? I say you still have 5000F.
Also, isn’t this a feed back loop? If the cold ones cause the hot one to raise temperature the hot must cause the cold one to raise to temperature. Then the cold one having raised to a new temperature now will cause the hot one to raise and then we go round and round.
Possibly, possibly not. But that has nothing whatsoever to do with heat radiation and nothing to do with the comment I made that you are replying to. A blow torch transfers heat through conduction and convection, not radiation.
Bingo! By George he’s got it!
The final bit missing in your comment after “round and round” could be “in ever decreasing circles”. The temperature increments get smaller and smaller each time until the temperature of both settles again, and both at a higher equilibrium temperature than they started.
It is irrelavant that the increments get smaller or not. What you seem to be saying is that we have an endless supply of radiant energy because at no time into future do they ever achieve the same temperature. A billion years from now there is no telling how hot the objects would be.
No it is not irrelevant, when the addition is decreasing you have a convergent series and rapidly settle at a new steady state, check out Zeno’s paradox.
Each increment is smaller but still positive. The system would never be stable but continually increasing in temperature forever. Not reasonable.
The original increment would have to be so small as to be almost nonexistent to fit reality which means no AGW!!
That is not how convergent series work in the real world. Eventually, and usually quite quickly, a new equilibrium is reached. You would find that in the temperature feedback loop you describe the increments are actually small, so quickly resettle because the final increments are swamped by random variations (read up on quantum theory for more on this point).
In reality, the smaller the increments in a positive feedback loop the quicker the system will reach a new equilibrium.
“In your original idea the back radiation will cause the 300k object to rise in temperature. Maybe not even by as much as 1K, but rise it will.”
Original GHE theory claimed 33K warming. Now we down to (maybe not even as much as) 1K. Interesting.
I’m a little confused about the Wood experiment data. It seems to imply a reverse greenhouse effect if the salt window box got hotter. A window that allows IR radiation to escape should cool the box and the glass that captures IR radiation and re-radiates part of that energy back to the box should reduce that cooling effect. Maybe I misread the data.
On another issue, some are obviously having a very difficult time understanding the concept of a cooler object heating a warmer object. It might help to see it not as heating, but as reducing the cooling by radiation. In the simplest of terms, all bodies at greater than 0 Kelvin radiate energy (heat). If Earth had no atmosphere then all of the energy radiating from the surface would be lost into space. Add an atmosphere and then some of that energy is absorbed by the atmosphere, which then radiates part of that energy back to the Earth and part of it into space, reducing Earth’s heat loss. The atmosphere also directly absorbs some of the incoming sunlight, complicating the process. The greenhouse effect occurs because certain materials are transparent to visible light, the frequencies at which most sunlight energy is transmitted. Some of these materials, including glass and CO2, are opaque to IR radiation, the frequencies at which Earth radiates energy. It then becomes a simple energy balance equation. If most of the incoming energy passes through the atmosphere, but most of the outgoing energy is captured and re-radiated with roughly half going back to Earth the Earth heats. It is still the Sun that heats the Earth, but less of that heat can escape back to space because of the greenhouse effect.
yes, this has already been explained…they don’t believe it, though.
If you could isolate the loss of heat from radiation I think that it could work. Not sure if that would be feasible, however. The top of the atmosphere does that for us though! The Earth can only cool radiatively to space, since there is no matter to conduct to and again, not matter to “convect” through. Measuring from the bottom of the atmosphere does not allow one to as easily deal with those effects.
What one would really want is to somehow create an isolated column of vaccuum extending from the ground to the top of the atmosphere, and compare the atmospere-less area to the surrounding area. Although then the problem is the difference in pressure…What would that mean?
Andrew, the moon is at the same distance from the Sun as the Earth. Make your vacuum measurement there!
Thomas-Not that simple, seeing as the moon is not really a good approximation of the conditions the Earth would experience if you just got rid of the atmosphere. You could not just do a single measurement, since, due to the vastly different lengths of day, at any given time the Moon will have been cooling down or heating up for a lot longer than the Earth has (imagine half a month of sunshine!) The moon I also tend to think has a different emissivity and albedo than would an atmosphere-less Earth. However, if that is not the case, then one would still need to measure their temperatures compared to one another over full diurnal cycles of both.
However, I do find that the mean of the Moon’s day and night mean temperatures is 250.15 Kelvin, which should be compared to the estimated mean for the Earth, ~288.15 K, which is 38 K difference. This is roughly close the “33 K” quoted value for the Earth’s GHE.
Note that the Moon’s albedo is 0.12, while with an atmosphere the Earth’s is 0.3. I’m not sure how this factors in, but note that an atmosphere-less Earth would not have clouds, so it’s albedo would be different. Note also that my earlier comment was incorrect on an important point there-namely that the vaccuum column out to space would be show a contrast of greenhouse effect present and not-there is in fact a subtle difference between no atmosphere and no greenhouse effect, and the absence of clouds would mean that the vaccuum column would tend to reflect the difference made by albedo from clouds AND the greenhouse effect.
I personnally don’t doubt downwelling back-radiation exists and is measured by monitoring equipments.
What I doubt is what Andrew S suggested : that the 324 W/m2 back radiation depicted in the notorious IPCC-Trenberth diagram would heat the surface.
I find it hard to accept this 324 W/m2 back radiation would heat the surface as much (if at all) as 324 W/m2 direct sun radiation, otherwise thermal concentrating solar plants would work during nighttimes. I’m ready to accept it would warm somewhat the surface but how much ??? Maybe Dr Spencer can explain how to calculate such warming.
(BTW, if 324 W/m2 back radiation cannot heat the surface as much as 324 solar radiation, what’s wrong in Trenberth’s radiative energy diagram).
324 W/m2 is 324 W/m2, no matter where it comes from. In fact, the downwelling IR it is reducing the energy losses from solar collectors. To the extent the solar collectors get hot, they lose energy through IR radiation, and that loss is partly off set by sky radiation.
Interesting comment. So, during the day the earth is being heated by 648w/m2 and rising?? Why don’t we cook??
> thermal concentrating solar plants would
> work during nighttimes
They would work at night about like they’d work work during the daytime if set up with the mirrors angled so they were were _not_ reflecting sunlight onto the target.
Try this:
http://mynasadata.larc.nasa.gov/P18.html
Get an infrared thermometer and go point it at
— the North Star at night, on a clear cold night
— the North Star at night, on a clear warm humid night
— the same point in the sky, on a cloudy cold night
— the same point in the sky, on a cloudy humid night
— the same point in the sky, on a clear day
— the same point in the sky, on a cloudy humid day
Dr. Spencer, I think you’ve attracted a passel of sophomore boys. Most of these folks are just having fun with you. If you can sort out the ones who come here from WUWT and CA by name and speak just to them, you might make progress. But the other kids are just trying to keep the confusion level high for the fun of watching everyone else struggle to understand, I think.
Interesting theory. Sounds like a trick I would pull, actually.
Hank, aren’t there mirror shapes for concentrating diffuse radiation? Why wouldn’t something like that work?? Not nearly as efficiently as a shape for a direct source of course.
Basically we are told that there is an average of 390w/m2 coming back at the ground during the day. Shouldn’t you be able to at least warm something up with that radiation??
There is a company using solar ovens as ice boxes by pointing them at open sky in Africa!! Doesn’t sound like much energy in the backradiation to me!! They actually produce ice depending on conditions!
Infrared Thermometers use thermistors, thermocouples or cold-junction semiconductors.
These devices, like the thermistor, change in resistance with temperature changes.
They transfer heat energy to the atmosphere the same as the Solar Oven cools water when pointed at the cold atmosphere.
Direct measurement of “Back Radiation” requires COOLED IR detectors.
All instruments, like the AIRS instrument used by NASA, to directly measure Back Radiation use cryogenically COOLED IR detectors (far below the -20 deg C average temp atmosphere) in all their instruments, to make these measurements possible.
Demesure, you can’t concentrate the heat radiation from the atmosphere since it isn’t a point source. Make some diagrams to see how a concentrating mirror works and you’ll see why.
UPDATE ON THE SKY RADIATION EXPERIMENT: The Box (ver. 2.0) is now operating in my backyard, next to my regular weather station.
It is built basically like I described in my original post…including the cavity painted with Krylon 1502 Flat White high-emissivity paint, and genuine Glad Cling Wrap (12 microns thick, & 100% pure polyethylene…da good stuff!) for the 2 IR windows.
When I left home the cavity temperature in The Box was 140 deg F and still climbing…there is some sunlight shining in, which is why.
Assuming all goes as planned, and tonight is relatively cloud free (which it usually is in N. Alabama in the summer), tomorrow I will post temperature traces. I’m archiving temperatures at both stations, which are about 2 to 3 feet apart, every 5 minutes.
Can’t your ol’ NASA friends get you some 3M black velvet coating 9560 series optical black paint?
Now that the experiment has begun, I checked the downwelling IR at Goodwin Creek, MS (about 165 mi west of you). It is currently running ~400 W/m^2 when clear. This would be an effective temp of 62F. The low temp there last night was 72F. In Huntsville it was 76F. Both sites have a dew point of 72F.
Assuming the box cools after the daytime super-heating, this should be interesting.
Dr. Spencer, beware of the moon!
Svante Arrhenius used the heat radiation from the moon as measured by Langley for the first quantitative study of the greenhouse effect:
http://www.globalwarmingart.com/images/1/18/Arrhenius.pdf
Obviously Arrhenius considered back ratiation in his simple mathematical model so in some way the radiation from the moon was corrected for that back radiation. Perhaps the radiation from the moon was simply much stronger,
If you want to measure the unbiased back radiation during the night you should probably avoid to get the light from the full moon into your instument.
Wasn’t the cavity supposed to start at 300K ?
Or was that just hypothetical and you were just inferring that it would initially be higher than the external temperature ?
Can someone tell me what WUWT and CA are ?
300 Kelvin is about 80F, ambient in some areas.
WattsUpWithThat and ClimateAudit. Web sites that donot shall we say hew the AGW line
300 K was just hypothetical…just as the 290 K for the sky temperature was. The point is that the IR temperature of the sky is usually less than that from the surface.
How about real data collection and analysis instead of meaningless thought experiments. Try this experiment -go somewhere the air temperature is less than 0C and there is lots of packed ice, measure air temperature, air humidity, wind speed and your body temperature (around 37C). Build an igloo and get inside with no food or heating. Measure inside the igloo, the air temperature, the humidity and your body temperature regularly (say every ten minutes). You can stop the experiment when your body temperature drops near 35C and the onset of hypothermia or signal to a friend to get you out. Now answer the question has the wall temperature of the igloo increased your body temperature from back radiation?
If still a problem?- repeat the experiment with no clothes on so that your body will radiate more heat. Is there back radiation? Will you last longer than if you had clothes on?
Did you seal the entrance to the igloo? 8>)
Roy: I applaud this effort to FINALLY get some empirical evidence to support the GHG hypothesis. This has been my primary focus for years, and it is extremely troubling to me that so many scientists vehemently support the GHG effect with no supporting empirical evidence, whatever. In most fields of science, this kind of hypothesis would be a joke. Some folks (like me) might even say that the lack of this evidence suggests that All the empirical data that I have found show that the GHG hypothesis is junk-science. It still has not been explained to my why the temperatures in Atlanta in Summer are much lower (even at night) than the temperatures in Phoenix, which is at the same latitude and elevation. Atlanta has at least 3 times as much greenhouse gas.
I bid you good luck!!!
BTW: These statements don’t seem to mesh:
“Ten deg. C is more typical during humid conditions or cirrus cloud cover, while 30 deg. C would be during clear, low humidity conditions.
Low clouds produce a downwelling sky temperature nearly the same as the upwelling temperature. The sky temperature increases as you scan from the zenith down in elevation, due to the greater path length through the atmosphere.”
BTW, here is the 30-year average temperature, humidity, solar radiation, etc. data for Phoenix and Atlanta for any doubters out there: http://rredc.nrel.gov/solar/old_data/nsrdb/redbook/sum2/state.html
Dr. Spencer;
With respect, this is an interesting experiment that I believe will eventually reinforce the critical concept of the “Speed of Heat” (aka Thermal Diffusivity). The real reason a Solar Oven is more effective at cooking things is that it increases the speed of heat by replacing absorption (by the dirt) with reflection (by the mirrors). You can also bake a cake by putting it on a rock in New Mexico in July it just takes longer. A Solar Oven does not create extra energy; it simply deprives the Earth under the Oven of energy and delivers this energy more quickly to the item being cooked.
If I understand your experiment you will be increasing the speed of heat inside your cavity. Ok that’s fine. If you increase the speed of heat things may (if all other effects are considered) reach a higher temperature.
To make your experiment more accurate you must instead measure the temperature underneath (instead of next to) your cavity to demonstrate that more ENERGY (note: NOT POWER, NOT TEMPERATURE but ENERGY) has been retained in the TOTAL SYSTEM. Please report the temperature of the Earth under your cavity to demonstrate that you have indeed retained more ENERGY inside your cavity than the Earth’s surface sans experimental cavity would retain. Thanks for that data.
Just a few comments on your text (intended to be respectful, please ignore or e-mail me if the “flavor” is too strong, I can de-spice my comments if desired):
Regarding dew formation you wrote: “This is because these surface are losing IR energy faster than the air and other surfaces are, so their temperature falls below the dewpoint temperature first.” BINGO!!!!!!!!!! DING! DING! DING! (FLASHING LIGHTS GO OFF NOW) – THIS IS A DAILY DEMONSTRATION OF THE “SPEED OF HEAT”…… Most metals have thermal diffusivities much higher (be certain to convert values to “per unit volume”, most references state them in “per unit density” values) than plastics and naturally occurring materials (dirt). This actually comes down to the atomic structure of the materials, in a metal the atoms are all dancing together in a little conga line that is at a faster beat than in materials with long atomic structures like plastics and glasses (like fiberglass, a common practical insulator). Materials like diamond actually have all the atoms precisely placed in the conga line (a crystalline structure) and have damn wicked speeds of heat.
You wrote: “(We use air since it is very hard to measure the temperature of a cold surface accurately, so we let the cold surface inside the cavity chill the air in contact with it).” I’m sorry and I will give you the benefit of the doubt on this. You have probably not been tasked with selecting temperature sensors much. But this statement is not only incorrect, but the exact opposite of the truth. Many types of temp sensors exist and the accuracy of those that measure lower temps are not materially different. For example, Platinum RTD’s can measure the temperature of a surface from almost absolute zero up to temperatures just below where they melt (a couple of thousand degrees Kelvin if my memory is correct). This can be done with accuracies of about 0.1 degree F easily. With a bit of very careful calibration accuracies of 0.01 degree F can be obtained over a smaller range. In contrast, measuring the temperature of air is fraught with errors with typical defendable accuracies of 0.5 degrees F hard to obtain.
You wrote:
“Wouldn’t want to violate the 1st Law of Thermodynamics, ya know.” As an engineer I have been fighting against the laws of physics my whole career, so far they are batting 100%. There have been times when, to my embarrassment, I predicted an outcome that did not occur. I long ago humbly submitted to the laws of physics and am now contented by getting as close as possible to breaking them as I can within given financial and time allocations. I find your comment just a bit hubristic (perhaps a new word, I hereby claim the copyright if indeed it is new). Many of us with much engineering experience are trying to communicate that this “greenhouse” effect that makes the temperature or energy (apparently totally interchangeable terms in the climate science world) of the Earth slowly ratchet upwards at the end of each day does not “jive” (not a new term) with what we have empirically observed in the real world. Ignore us if you want too, but those laws that you claim we are misinterpreting will indeed determine the outcome. We have an old saying: “if your hardware does not perform as predicted by your model, you must improve your MODEL”. I have never seen somebody improve their hardware until it does what the model says. PLEASE TAKE THAT STATEMENT TO HEART, I REPEAT: “I HAVE NEVER SEEN SOMEBODY IMPROVE THEIR HARDWARE TO MAKE IT PERFORM AS THE MODEL PREDICTS!!!!!!!!!!!! And no I am not some crackpot; some hardware that I have helped design is currently orbiting around the Earth and is seeing all of the heat transfer mechanisms (conduction, convection and radiation) every day as it flies over our heads. I have seen lots of models, measurements and understandings improved to make the correlation between models and reality higher. But the hardware (i.e. the Earth) is the reality. Have you ever heard of the German scientists that determined that a bumblebee could not fly because their model said so?
So here are my answers to the questions;
(1) – I do not deny the presence of ”downwelling” or ”back” radiation so I cannot answer this question.
(2) – I predict that the temperature will change, without the time to analyze all the speeds of heat of all the objects I will not at this time predict the direction or magnitude of the change. But I eagerly await your measurements. Perhaps we can deploy thousands of these from the North Pole to the South Pole.
P.S. Investigate the potential of synthetic diamond as an optical and thermal material. It has very high speed of heat and is “nearly” transparent over a very broad wavelength range. Disclaimer: I do not work for or own stocks in any synthetic diamond companies.
Cheers, Kevin.
I should have said it is easier and cheaper (for a backyard experiment) to measure air temperature accurately than to temperature of a surface accurately.
Yes, the thermal conductivity, as well as the total mass, of the various “things” changing temperature do impact the rate at which the temperature changes…I have alluded to this at least twice before….but not the direction it changes, nor the final equilibrium temperature.
Dr. Spencer;
you wrote:
“Yes, the thermal conductivity, as well as the total mass, of the various “things” changing temperature do impact the rate at which the temperature changes…I have alluded to this at least twice before….but not the direction it changes, nor the final equilibrium temperature.”
THERE IS NO FINAL EQUILIBRIUM TEMPERATURE…….
THE INPUT (IE THE SUN LIGHT ARRIVING AT THE SURFACE) CYCLES FROM LOTS TO NONE…….
THE SURFACE AND THE ATMOSPHERE IS ALWAYS LAGGING BEHIND WHAT THE RADIATIVE MODELS PREDICT…………
TO CAUSE THE TEMPERATURE OF THE SURFACE TO RISE THIS LAG TIME MUST BE DECREASED………..
THIS LAG TIME IS DETERMINED BY THE THERMAL CAPACITIES OF THE MATERIALS (WATER AND DIRT MOSTLY, A TINY BIT BY GASES)………..
CHANGING THE RATIO OF HEAT TRANSFER MECHANISMS CANNOT CHANGE THE THERMAL CAPACITIES OF THE MATERIALS……
THE “GREENHOUSE” EFFECT DOES NOT DECREASE THIS LAG TIME……
PERIOD………….
Cheers, Kevin.
You mention that you can get 12.7micron film, well polythene has a refractive index of about 1.5 so it will have an effective thickness of about 19 microns and hence its reflective properties will dynamic in the band of interest.
Polythene has I believe about 10% reflectance per surface but with full reinforcement from the second surface that would be ~20% and I think that would occur at every wavelength equalling 2*19/n microns (n=1,2,3,…).
Which could be useful as it could supress the downwelling peak at around 500/cm (20 microns). That is if I have got my reinforcements and cancellations the right way around!!! Two layers of cling film could give around 30% rejection. Of course it cuts both ways (up and down), but if it is tuned to suppress a band where the sky is brigthest and pass were the sky is cooler, it could assist (or hinder if I have it the wrong way around).
Using thicker films would give a more even response.
I think that condensation on the top layer could be a problem if it occurs, as even sub milimetric layers of water have an IR emissivity close to 1. But by and large conduction through (air – surface – core – surface – air) insulating layers is mostly a surface layer effect so with two layers of cling film the top surface should only cool by about 1/4 of the difference between the ambient and the internal cavity.
I believe that polythene has important absorption/emission bands at around 750,1500,3000 (1/cm) plus a small IR continuum. But all in all its IR emissivity is low.
I am not sure what shape the cavity should be but as the insulated surfaces are warming (conduction) not cooling (radiation) I would have thought that a shallow cavity maiximising the radiative area would be of benefit.
The uninsulated thermal mass of the cavity needs to be as low as possible if it is to equilibrate on some reasonable time scale.
I suspect that the total emissivity will be governed more by the reflectance of the polythene rather than surface coatings, and that will effect both the equilibrium temperature and the response time of the cavity.
I think that the wide field of view may be a problem as the IR flux will be brighter away from the zenith. Perhaps only a high emissivity bottom to the cavity with reflective sloping sides would be better.
One thing that does bother me is that the experiment would depend a lot on the nature of the surrounding ground surface. As I see it what you are making is a better pyrometer than the surrounding ground surface. Given a long enough night the ground would equilibrate as well, not to the same temperature, but nevertherless, quite low. I am not sure that the surrounding ground surface is the correct control. The ideal control would be to point the cavity at something with a really low radiative temperature, thereby showing that the sky is warm by comparison, but I am not sure how that could be achieved.
My guess is that the main effect will be an initial cooling rate much higher than the control will be achieved but then the control will largely catch up given enough time. But that would be mostly due to the lower thermal mass of the cavity and its insulation from the ground. I suspect that in order to substantially reduce the cavity temperature below the control and keep it there you would have to reduce the field of view of the cavity so it only sees the zenith.
Alex
Now I think about it a bit more, I can see that because polythene has quite a high refractive index, (higher than water). It should have high reflectance at angles significantly away from the zenith, so the cavity should have a restricted field of view and hence a better performance, but yet again it cuts both ways (up and down) but I think in general it should be a benefit.
Unfortunately reality tends to get rather complicated if you look at it too closely. I know that to work all this out properly one has to work with complex refractive indices as polythene is slightly conductive (hence the small continuum absorption/extinction) but that would have to wait for another day, or another life.
Alex
Re: TheLastMan post July 28,2010 at 6:30AM
The 2nd Law does not say that Radiation is excluded! What Physics textbook has EVER said that?
– EM fields are Vector quantities and must use Vector Mathematics when adding or subtracting fields.
– Hotter objects produce larger Magnitude EM fields than Cooler objects
– There can only be one resultant EM vector field.
If two opposing EM fields are added, the Magnitudes are subtracted (Hotter-Cooler) and the Direction of propagation is in the direction of the Hotter field.
Zero mass Photons don’t propagate by themselves, the Photon Energy is CARRIED by EM fields.
– If you don’t understand the Vector nature of EM fields you should be able to understand the vector nature of FORCES.
– EM fields are Force fields too, in fact, the Electromagnetic Force is one of the four fundamental forces.
When two opposing EM FORCES are added, the Magnitudes are subtracted and the Direction of the zero mass Photon movement is in the direction of the Larger Force (Hotter).
Just like a block of wood with two opposing forces on it it does not “make a decision” which way to move, it ALWAYS moves in the direction of the Larger Force.
It is IMPOSSIBLE for the block of wood to move in the direction of the weaker Force.
And, just like the block of wood, the zero mass Photons always move in the direction of the Larger Force. Hot to Cold.
It is IMPOSSIBLE for the Photons to move in the direction of the weaker Force. Cold to Hot.
Like you said…Get a life…or better still an education. It is never too late.
“There can only be one resultant EM vector field.
If two opposing EM fields are added, the Magnitudes are subtracted (Hotter-Cooler) and the Direction of propagation is in the direction of the Hotter field.”
The vector you need is the Poynting vector S although it is commonly substituted with its time averaged vector . The Poynting vector itself is not constant and when considered at a surface between a hotter and a cooler body both its amplitude and “direction” vary in time and with the position of the surface. Superimposing opposing beams does not produce a Poynting vector that always points away from the hotter body. This is true of the time averaged vector which expresses the fact that on average the energy flux will be from the hotter to the cooler body.
In the case were the bodies have the same temperature and the Poynting vector is taken across a surface midway between the two is zero as one would expect but the vector S itself is not, it varies in both amplitude and direction, expressing the fact that energy is flowing in both directions.
That the average flux is always from the hotter to a cooler body does not conflict with it being composed of two real fluxes moving in opposite directions.
Removing the cooler body from a two body system will remove the flux from the cooler to the hotter body thus increasing the rate of energy loss of the hotter body and if it has a heat source with a constant rate of heating, lowering its temperature.
Reintroducing the cooler body will reestablish the flux to the hotter body, thereby reducing the vector, and the rate of energy loss of the hotter body and as the hotter body has a constant heat source it will increase in temperature.
Unfortunately is not uncommonly stated as the Poynting vector without reference to its being a time average which does give rise to the appearance of the sum of the real flows being unidirectional and of constant amplitude when this is not so.
Alex
Ooops that won’t make a lot of sense as greater and less than signs bracketing the time averaged Poyning vector have been treated as tags and eaten. I will try {S} for the time averaged vector instead.
“There can only be one resultant EM vector field.
If two opposing EM fields are added, the Magnitudes are subtracted (Hotter-Cooler) and the Direction of propagation is in the direction of the Hotter field.”
The vector you need is the Poynting vector S although it is commonly substituted with its time averaged vector {S}. The Poynting vector itself is not constant and when considered at a surface between a hotter and a cooler body both its amplitude and “direction” vary in time and with the position of the surface. Superimposing opposing beams does not produce a Poynting vector that always points away from the hotter body. This is true of {S} the time averaged vector which expresses the fact that on average the energy flux will be from the hotter to the cooler body.
In the case were the bodies have the same temperature and the Poynting vector is taken across a surface midway between the two {S} is zero as one would expect but the vector S itself is not, it varies in both amplitude and direction, expressing the fact that energy is flowing in both directions.
That the average flux is always from the hotter to a cooler body does not conflict with it being composed of two real fluxes moving in opposite directions.
Removing the cooler body from a two body system will remove the flux from the cooler to the hotter body thus increasing the rate of energy loss of the hotter body and if it has a heat source with a constant rate of heating, lowering its temperature.
Reintroducing the cooler body will reestablish the flux to the hotter body, thereby reducing the {S} vector, and the rate of energy loss of the hotter body and as the hotter body has a constant heat source it will increase in temperature.
Unfortunately {S} is not uncommonly stated as the Poynting vector without reference to its being a time average which does give rise to the appearance of the sum of the real flows being unidirectional and of constant amplitude when this is not so.
Alex
-A propagating EM field is the Poynting Vector and has the units of w/m^2
-The Poynting Vector always “points” in the direction of propagation and is normal to the radiating surface.
-The Poynting Vector is A VECTOR and must be treated as such.
The effect of opposing propagating EM Fields (Poynting Vectors) are seen in polar radiation patterns of Antennas.
There are Nulls and Peaks in the radiation paterns that are easily predicted and designed using Vector Mathematics.
The fact that the Resultant Vector produces Nulls and Peaks exist proves that there cannot be fluxes moving in opposite directions.
There is only ONE resultant Vector.
In fact, any single point measurement made in space will ONLY detect the resultant Vectors Magnitude and Direction!
Just like opposing forces on a block of wood.
The block of wood cannot move in two directions at the same time and any measurement made will only detect the resultant force!
And, if two opposing EM fields are equal there is no resultant EM field …and any measurement done will show ZERO w/m^2.
This effect is easily seen in a Soap Bubble when the walls become very thin producing reflection.
As the walls thin, more and more light within the bubble will cancel until all light is cancelled.
“Given a certain thickness of the bubble wall, a certain wavelength will be cancelled and its complementary color will be seen. Long wavelengths (red) need a thicker bubble wall to get out of step than short wavelengths (violet). When red is cancelled, it leaves a blue-green reflection. As the bubble thins, yellow is cancelled out, leaving blue; then green is cancelled, leaving magenta; and finally blue is cancelled, leaving yellow. Eventually the bubble becomes so thin that cancellation occurs for all wavelengths and the bubble appears black against a black background.”
This is at www
exploratorium edu /
ronh / bubbles / bubble_colors .html
Gord
Nearly there Gord, one more step…
Are we not saying the same thing?
Situation 1: the hot object is emitting 300 units and the cooler object is emitting 1 unit. The hot object is then emitting a net 300 – 1 = 299 units in the direction of the colder object.
Situation 2: the hot object is emitting 300 units and the cooler object is emitting 200 units. The net emission from the hot object is then 300 – 200 = 100 units in the direction of the colder unit.
Remember my explanation of what causes the temperature of an object to rise. Either:
a. the rate at which an object loses heat is reduced
b. the rate at which an object absorbs heat is increased
From the perspective of the hotter object, if you move from situation 1 to situation 2 the hot object is prevented from emitting as much radiation by the presence of the colder object. Therefore the rate at which it can lose heat (radiation) in the direction of the colder object is reduced from 299 to 100 and therefore it rises in temperature.
The difference between our two perspectives is down to quantum theory and the fact that the behavior of a photon can be correctly predicted assuming it is a particle AND assuming it is a wave. You are describing the “wave” explanation and I am describing the “particle” explanation. The net effect is the same.
Sorry about the “education” jibe. Clearly you have some, whereas there are plenty here that seem to be distinctly lacking!
No, we are not saying the same thing.
Like I said..the Direction of propagation is in the direction of the Hotter field.
There is ZERO EM field propagation from cold to hot.
You have the Magnitude right but the DIRECTION wrong.
Here is how it works: (Ignoring the surface area and emissivity of each body for illustration purposes only)
Assume a colder enviroment beyond the two bodies.
The 300 body would radiate 300 units to the colder enviroment by itself.
(That’s important for The Law of Conservation of Energy considerations)
Now insert the cooler 200 body.
Between the 300 and 200 body there is 100 units in the direction of the 200 body.
There is zero units back to the 300 body.
The 200 body absorbs the extra 100 units and is heated to 200+100 = 300.
Now both bodies are at equilibrium and the same temperature.
Both bodies can be considered to be a single body.
This single body is Still radiating 300 units to the colder enviroment, complying with The Law of Conservation of Energy.
No energy was created or destroyed.
All that happened was that the warmer 300 body radiation heated the colder 200 body to 300 on it’s way to the colder enviroment.
———–
You can do the same calculation with a 300 body and a 1 body.
The 1 body will be heated to 300 and 300 units are still radiated to the colder enviroment.
———–
If the colder 200 body or the colder 1 body heated the 300 body, creation of energy would result.
The colder body heats the warmer body, the now even warmer body heats the colder body that heats the warmer body even more….etc.
It would produce a perpetual motion machine in a postive feedback loop, resulting in an infinite energy and temperature for both bodies.
Gord
“Therefore the rate at which it can lose heat (radiation) in the direction of the colder object is reduced from 299 to 100 and therefore it rises in temperature.”
Uness the radiation is increased in other directions giving the same AVERAGE temperature.
Cement a friend:
Your igloo thought experiment could be revealing. Consider also an igloo with wall temperature 70K or one of some material with thermal properties similar to snow but at a temperature of 300 K. Where would you least like to spend time and why?
There are so many confusions and errors in this post it is difficult to know where to start.
Sorry, this is just plain wrong or, more accurately, self contradictory. The propagation cannot at the same time be “in the direction of” the hotter field AND “from” hot to cold.
The correct phrase should be “The field propagates from the Hotter field. There is zero EM field propagation from cold to hot.”
The main point you seem to be missing is that both bodies are heat sources at equilibrium temperatures. My units are units of heat energy not temperature. That means that the 300 body is emitting 300 energy units continuously at the same time as it is receiving (either internally or externally) as much energy as it is emitting. The same goes for the 200 body.
More accurately, we are talking about the heat emitted from the hotter object in the direction of the colder object being 300 and the heat emitted from the colder object in the direction of the hotter object being 200. That is not to say it does not count the energy emitted from either objects other surfaces.
You are confusing ENERGY and TEMPERATURE. Two bodies producing the same amount of energy can have different temperatures. No energy needs to be created or destroyed for a body to rise in temperature, all that is required is a change in the energy balance. For instance if the energy is prevented from escaping as quickly as it is being received, then the temperature will need to rise. As you know, a hotter body emits more energy by radiation than a cooler one and a new equilibrium will be established once the temperature has risen far enough.
Another example to help you distinguish between Energy (heat) and temperature is that a larger body with an internal heat source of 300 units will have a lower temperature than a smaller body with an internal heat source of 300 units. That is because the larger body has more surface area from which to emit that energy.
Conversely a larger body at the same temperature as a smaller body will need to have a more powerful internal heat source in order to maintain that temperature.
That does not add up. If there are only 100 units being received by the 200 body, but the 300 body continues to emit 300 units in the direction of the 200 body, what happens to the other 200 not received by the 200 body?
In order for them to be received in another direction, other than towards the 200 body, the 300 body would need to rise in temperature so that it can increase the rate at which it emits energy from its other surfaces. Remember that a given body cannot increase its rate of EM radiation without increasing in temperature.
Again, confusing heat with temperature. All that matters with the 300 and the 200 bodies is that when combined in this system that they both continue to emit the same amount of energy by radiation.
A body emitting 300 units of energy to the blackness of space will have a lower temperature than the same body with part of the path of that radiation blocked by the presence of the 200 unit. That is because it needs to radiate more from the rest of its surface to in order to make up for the energy it cannot radiate to space in the direction of the 200 body because that radiation is blocked by the 200 body. And in order to do that it has to be at a higher temperature.
Not so. It is a converging series positive feedback loop. No extra energy is created, but the temperature rises in order that the energy can be emitted at the same rate as it is being generated.
In reality the temperatures will not swing back and forth in this way, both bodies will just gently increase in temperature until the heat output is again in balance with heat input.
That is the last from me on this subject. If you do not get it by now you never will.
– The Direction of the hotter field is away from the hotter body and it is always going from hot to cold
– I also assumed the “units” were units of heat energy and the more units absorbed corresponds to a higher temperature (see Stefan-Boltzmann Law)
– There is no “confusion” about heat energy units and temperature. Simply use the Stefan-Boltzmann Law and you will see the higher the “units” are the greater the temperature will be.
– The 300-200 = 100 is the Vector addition of 300 unit field and the opposing 200 unit field giving a Resultant field with Magnitude 100 units propagating in the direction from the 300 body toward the 200 body.
There are zero units propagating from the 200 body to the 300 body.
——
If you are not familiar with Vector addition of Fields I have provided a link for you in my reply to your post dated July 28, 2010 at 6:30AM.
My post is right below yours “Anonymous says:” and is dated July 28, 2010 at 5:11 PM.
The vector addition of EM fields follows the same rules as vector addition of mechanical forces, which is not surprising since EM fields are Force Fields that move Photon Energy.
If you want an analogy, assume 300 units and the opposing 200 units are Force Vectors that move zero mass Photons.
The Resultant EM force vector has a Magnitude of 300-200 = 100 units which move zero mass Photons from the 300 body to the 200 body.
There is no movement of photons from from the 200 body to 300 body because the force is in the opposite direction.
Thus, there is no heat energy flowing from the 200 body to the 300 body.
To make the analogy complete, replace the zero mass photons with a “block of wood”, the opposing forces will only move the wood block in the direction of the stronger force.
The “block of wood” will never move in the direction of the weaker force.
——
I completed the calculation of the two bodies, showed that they would both be radiating 300 units at equilibrium and that it complied with The Law of Conservation of Energy.
I suggest that do the same, with units of energy flowing from the 200 body to the 300 body.
Compute the equilibrium units of energy for both bodies and show that it complies with The Law of Conservation of Energy.
Good Luck.
Gord
You don’t seem to understand the implication of what you have calculated here.
Try another tack. Two identical cube shaped black bodies. Each has 6 sides of 1m2 (and obviously a volume of 1 cubic metre) and both with a total surface area of 6 sq m.
Both are at the same temperature of 269.7K (-3.4C) with the same radiative flux at 300w/m2 (a total emission of 6 x 300w = 1800w). The net radiative flux is therefore zero between the two opposing faces. Neither body is able to lose any energy in the direction of the other.
Therefore they both need to rise in temperature in order to emit from all their remaining surfaces the total of 1800w that they are both producing. Because they are both identical black bodies as they rise in temperature at the same rate the net radiative flux between the opposite faces will always be zero.
Therefore all the energy they produce will have to be emitted by the remaining 5 faces in each case. We know that they are both producing 1800w so they need to emit a revised radiant flux of 1800/5 = 360w/m2
Using the Stefan-Boltzmann constant which is 5.6704 x 10^-8 (or 0.000000056704) the S-B formula gives the following temperature
(360/0.000000056704)^(1/4)= 6,348,758,465 ^ 0.25= 282.27K
In other words both bodies will rise in temperature -3.4C to +9.12C. No energy is created in the process. Both bodies are still producing and emitting 1800joules each second just as they were at the start.
To me it is blindingly obvious. No Poynting vectors needed. The net radiative balance formula is all that is needed to understand the point.
Think about the way a Thermos flask works. In New Jersey, with an ambient temperature at 22C, put a heating element in coffee in a Thermos flask and turn up the power sufficient to keep the coffee at 90C constant. Once it has settled at that temperature bury it in ice at -10C.
Take an identical Thermos flask with an identical heating element at the same heating rate and bury it in sand at +50C. Bearing in mind the energy input in each case remains the same as it did at 25C, do you think the other two will rise or fall in temperature?
I suspect the one in the sand will rise to a higher temperature than the one at ambient and the one buried in the ice will fall to a lower temperature than the one at ambient.
Phew! That’s enough of that.
A body that has a temperature will emitt energy energy according to the Stefan-Boltzmann Law.
The emitted energy will be in all directions.
A cube at 269.7K emitts 300 w/m^2 in all directions and does not emitt a total of 1800 Watts any more than a an Eight sided body will emitt 2400 Watts or a Sphere, that will have an infinite number of faces, emitts an infinite amount of watts!
Need more Watts, no problem… just add more sides!
Where did you get that incredibly hilarious idea from?
Gord
You have not understood my comment at all. At no point am I adding any “watts”, or adding any “sides”, in fact I am taking away sides. I am saying that by blocking the heat radiating from one of the six sides of a cube you force the internally generated energy to be emitted from the remaining five.
A cube with a volume of 1 cubic metre has 6 sides of 1 square metre.
If it has in internal heat source generating 1800 watts then the radiant flux, which is the total energy divided by the surface area, is 1800 watts divided by 6 square metres which gives you 300 watts per square metre. According to the S-B law a body emitting 300 watts per square metre has a temperature of -3.4C. If you stop heat emitting from one of its sides by placing another identical cube next to it, but not touching it and separated by a vacuum, then it is forced to emit heat from the remaining 5.
It is producing 1800 watts so 1800 watts divided by 5 square metres is 360 watts per square metre. According to the S-B law that body will have a temperature of +9.12C.
Very simple and basic secondary school physics.
It is accepted that there is interference and cancellation in intersecting energy fields. Why is this ignored in Climate Science?? If the field is cancelled it transfers no energy. If the field experiences interference the energy is reduced.
Again, why are these issues that have been extensively documented in radiative physics ignored and Climate Sciences bore on with the idea that the colder body heats the warmer body when interference and cancellation will greatly reduce or erase the flux???
Roy, it is very easy to quantify Willywolfes explanation of AGW (July 28th, 7.23am) which you endorse but which “they” do not accept.
Start with a bare earth, receiving W watts per square metre from the sun. Balance the incoming and outgoing radiative energy and you will obtain an average temperature of 255 K.
Now add an atmosphere which reduces the outgoing radiant energy by half, returning half to the surface – the “explanation” of AGW.
The atmosphere is mow emitting W/2 to space and receiving W. The surface is receiving W plus W/2, and emitting W.
To restore the balance, the surface must absorb the surplus energy, increasing its temperature until it emits 2W. The earth will then receive and emit 2W (W solar, W back from the atmosphere), and the atmosphere will emit W, as the bare earth did originally.
The ratio of atmospheric earth emission to bare earth emission is 2. From Stefan Bolzmann, the corresponding ratio of the temperatures is the fourth root of 2, or 1.19.
The atmospheric surface temperature will consequently increase to 1.19 x 255 = 303.5K, an increase of 48.5 degrees, which is rather more than we observe, but in the ball park.
That temperature increase is what Woods was looking for. He did not find it. Before you call this theory an explanation you need to say why not, or repeat the experiment.
The AGW theory does not stop there, however. Sadly, the absorption distance in the atmosphere mean that is has more than one layer, and every layer will absorb the incident energy and re-emit half downwards. Repeat the calculation and you will find:
One layer – Fourth root of 2 = 1.19. Tsurf = 1.19 x 255 = 303K
Two Layers – Fourth root of 3 =1.315 T surf = 1.315 x 255 = 335K
Three Layers – Fourth root of 4 = 1.415 Tsurf = 1.415 x 255 = 360K
Four Layers – Fourth root of 5 = 1.495 Tsurf = 1.495 x 255 = 381K and so on.
These results are absurd, but they are derived from the original AGW “explanation”.
It is less than half downward, but, who’s counting?
The earth would have to be an infinite plain to receive close to half the “downward” energy. The actual surface irradiated area below a molecule is the large end of a cone unless it is next to the surface or between irregularities close to the surface!!
Additionally the old hack earth with and without atmosphere completely ignores the “momentum” of cooling the mass of the atmosphere, the top layers of the surface itself (see the greenhouse moon, even if the math was flawed it simply needs to be computed correctly for a more reasonable estimate!!) not to mention the oceans. And where is the black body type radiation from the non-ghg atmosphere in all this?? Also, I really doubt with the oceans warming to 100 meters or more during the day that the simple thought experiment really captures the energy balance over them!!
Basically the earth energy balance is in now way amenable to a simplistic thought experiment to the degree that is being claimed. The error bars are TOO BIG!!
The field resulting from two EM waves progating in opposite directions through the same region of space has Poynting vectors that are in general neither constant in amplitude nor direction. It is not true to say that the energy flux is must always be the direction of the wave with the larger rms amplitude.
The resultant field cannot be expressed in terms of a single wave progating in the direction of the wave with the larger rms amplitude.
Try it, and also try deriving the Poynting vector as a function of space and time and show how this can be interpreted as corresponding to a single wave.
Further the energy density (a scalar) of the resultant field is the sum of the densities of the two seperate waves. It does not cancel out. If you have a high energy density and a low average flux, i.e. average Poynting vector you can be sure that this is not a description of energy being transported in just one direction.
Alex
Hmmm, “progating” I think I meant “propagating”, well at least I was consistent!
Oh the joys of growing old.
Alex
From Wikipedia
Poynting vector
“In physics, the Poynting vector can be thought of as representing the energy flux (in W/m2) of an electromagnetic field.”
The Poynting vector in plane waves
“In a propagating sinusoidal electromagnetic plane wave of a fixed frequency, the Poynting vector oscillates, always pointing in the direction of propagation. The time-averaged
magnitude of the Poynting vector is
(S) = (1/2uoC)Eo^2 = (eoc/2)Eo^2
where Eo is the maximum amplitude of the electric field and c is the speed of light in free space. This time-averaged value is also called the irradiance or intensity I.”
There is also a graphic showing a Dipole Antenna radiation pattern that has nulls and peaks.
———–
The Poynting Vector is a VECTOR (having a Magnitude and a Direction that ALWAYS points in the direction of Propagation) and has to be treated as such.
———–
Gord
TheLastMan says:
July 29, 2010 at 6:34 AM
“More accurately, we are talking about the heat emitted from the hotter object in the direction of the colder object being 300 and the heat emitted from the colder object in the direction of the hotter object being 200.”
You are aware, of course, that your 300 object can, in fact, be colder than 200 object. Right? How can you calculate *net* heat flux in this case? What will be direction of the heat flux?
I was assuming Dr Spencer’s original experiment, that is two identical objects. If one is radiating 300 units and the other is radiating 200 units then, by definition, the one emitting the larger amount of energy will have the higher temperature. To paraphrase the Stefan-Boltzmann law dreadfully
The total energy radiated per unit surface area of a black body per unit time (radiant flux) is directly proportional to the fourth power of the black body’s temperature
Our “units” of Radiant Flux are in fact energy per area per time period. Generally will generally be quoted in Joules per second per square metre. As a Watt is a Joule per second, the units usually quoted are Watts per square metre w/m2.
The “net” radiant flux is simply the difference between the two travelling in the direction of the colder object. On the assumption that the area concerned is one square metre, then in this case the formula is 300w/m2 – 200w/m2 = 100 w/m2. Energy flows from the hotter object to the colder one at a rate of 100 w/m2.
Remembering that both units are generating heat, this means the hotter object will lose less heat than if the object emitting at a rate of 200w/m2 were not there. Because it is generating heat at 300 watts and only able to lose 100 watts it will accumulate heat i.e. its temperature will rise.
When its temperature gets high enough it can once again lose all the heat that it is generating and it will settle at an equilibrium temperature.
I won’t go into the details of the Stefan-Boltzmann law that details what the temperatures might end up at, I will leave that to the reader’s further education. You could do worse than start with Wikipedia
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
Good luck!
I have already explained that an EM field is a vector quantity, provided the physics links as back-up and provided a link that explaines how to sum Vector fields.
This has been known and used by Physicists and Electrical Engineers for over a hundred years.
I also explained that an EM field is a Force field and how zero mass Photons can only move in the the direction of the stonger force.
It is impossible for the Photons to move in the direction of the weaker force, which means that heat Energy CANNOT flow from cold to hot.
“Second Law of Thermodynamics: It is NOT POSSIBLE for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.”
This is in complete compliance with the 2nd Law and The Law of Conservation of Energy.
If you violate the 2nd Law you will violate The Law of Conservation of Energy.
That’s why it is IMPOSSIBLE for you to compute an equilibrium temperature for your example when heat flows from cold to hot without violating the 2nd Law and The Law of Conservation of Energy.
The facts are that there has never been ANY measurement, ever done, where Heat energy flows from a colder body to a warmer body without any work having been done to accomplish this flow.
That’s why the 2nd Law of Thermodynamics and The Law of Conservation of Energy are called LAWS OF SCIENCE.
You just ignore all the facts, all the Science, all the measurements and continue to post your “opinions” without even a shread of scientific evidence.
This type of blind devotion to an irrational belief system is typically found in cults.
Gord
Impossible for photons to move in the direction of the weaker force? So I guess I can’t point a flashlight at the sun, or even a car headlight, huh?
“Photons moving in the direction of a weaker force” is a statement that is empty of any meaning at all.
If you bothered to read the established Laws of Science and EM Physics pertaining to your questions, you could answer them yourself.
Example: Pointing a flashlight at a more powerful Car headlight.
If the Photons could actually move from the weaker Flashlight to the more powerful Car headlight, they would be absorbed and heat up Car headlight filament, producing more light.
The now more powerful Car headlight would move Photons back to the Flashlight causing it to heat up and produce more light.
The now more powerful Flashlight would move even more Photons back to the Car Headlight causing it to heat up even more and produce even more more light.
The cycle would continue untill both the Flashlight and the Car Headlight were producing INFINITE energy!
A perpetual motion machine in a positive feedback loop.
——–
from wikipedia…
Perpetual motion
“The term perpetual motion, taken literally, refers to movement that goes on forever. However, the term more generally refers to any closed system that produces more energy than it consumes. Such a device or system would be in violation of the law of conservation of energy, which states that energy can never be created or destroyed.”
“Perpetual motion violates either the first law of thermodynamics, the second law of thermodynamics, or both”
“A perpetual motion machine of the first kind produces energy from nothing, giving the user unlimited ‘free’ energy. It thus violates the law of conservation of energy.”
————-
Why don’t you deal with reality instead of fairy tale irrational beliefs typically found in cults?
Gord
Dr. Spencer,
I didn’t read your experiment comments carefully because I didn’t have to. I was trying to explain your points to a blogger much the same way, but to no avail. I even used a variant of your colder object next to a hot object causing a rise in temperature in the hot object.
My first college physics professor asked the question: If you place water in a one foot cube container that is insulated except for the open top and place it in the desert, at what air temperature would it freeze. His answer was “about 40 degrees F.” If the sky were cloudy, the water would freeze at a lower air temperature. If one doesn’t get how this works, then one will never understand what you are describing.
I have never had to run an experiment because I have seen frost on the ground when the temperature remained above well above freezing and I noted this happens on clear nights.
yes, I agree with you Mark. But what seem obvious to most of us is not obvious to others. So, I’m trying to find different ways of demonstrating it…while hopefully learning something in the process.
We get it. I promise. I don’t doubt anyone reading your blog contests the sky’s ability to cool warmer objects exposed to it. Poor cube of water will freeze trying to warm the sky above. However it will take an awfully long time to freeze cubic foot of water at 40F. Freezing water releases a lot of energy.
On the other hand, the sky’s ability to warm warmer objects is not so obvious.
Quite so, Andrew. People have no problem seeing that clouds reduce visible light radiation from the earth to space even though they are “darker.” They may even see that the condition of the night sky affects the earth’s surface and lower atmosphere rate of cooling because they can observe it. Leaving out some events or certain colder objects, the night sky warms nothing. The condition of the sky affects the rate at which things cool. It’s about changing the equalibrium point. We all say with certainty that gas flows from high pressure to low. The reality is that a very large portion of the molecules at any given time are moving from low to high pressure. It’s only the net movement that is from high to low pressure.
I have been trying to think of a better way of phrasing this. How about:
“The ability of a cold sky to absorb less heat from a warmer Earth than an even colder sky”
In other words a “sky” (atmosphere) radiating energy at a higher level is less able to absorb heat than one radiating heat at a lower one, regardless of whether more or less radiation is coming from the Earth.
I think that makes the mechanism clearer.
BTW, as I am sure Dr Spencer is aware, this experiment was first carried out in 1954:
http://journals.ametsoc.org/doi/pdf/10.1175/1520-0469(1954)011%3C0121:AIDFMO%3E2.0.CO;2
They come to the conclusion that the night time back radiation from the atmosphere is (in rather quaint units) a range roughly between 0.4 and 0.45 cal cm-2 min-1 or “calories per square centimetre per minute”.
Which I convert to… (Gets out pocket calculator… tap, tap, tap) 279 to 314 w/m2.
And they measured the day time back radiation at the same sort of level. This proved that the radiation downwelling from the atmosphere is about the same during night and day and is not affected by sunlight to which (by definition) a “clear sky” is transparent.
Interesting stuff!
Dr. Spencer,
How do you know that the downwelling radiation you measure is not simply black body radiation – the radiation everything emits that is not at absolute zero? O2, N2, AR,… everything in the atmosphere emits this radiation.
It is radiation from everything in the atmosphere. That is all he is trying to prove. He is not saying it is only from CO2, in fact the biggest contribution is from water vapour. Your “black body” radiation is what is colloquially and misleadingly called the “greenhouse gas” effect.
His problem is that there are some deluded souls out there who think there is no “downwelling” infra-red radiation at all! Or, if there is, that there is no way it could have any effect on the rate of cooling of the atmosphere.
Those with even the scantiest secondary school physics will realise this is nonsense. The problem is a couple of theoretical physicists got there theoretical knickers in a twist trying to prove that this somehow violated the second law of thermodynamics. Every climate scientist with half an ounce of knowledge of the subject, whether they were skeptical of AGW or not, quickly disowned the paper saying it was just plain wrong.
The greenhouse gas theory is fully accepted by all serious climate scientists, skeptic and believer alike. The dispute is over the extent of feedbacks. The AGW proponents think that positive feedbacks dominate whereas the skeptics either think the feedbacks are negative or say that the feedbacks are unknown.
Pick your side, but there is no point in trying to deny the already well proven facts.
Here’s a crazy idea for another control scenario: Set up a “proxy deep space sink” over one of the jars. For instance, it might be possible to have a high-absorbance plate cooled with liquid nitrogen, placed a couple of metres over the jar. You’d need a clear barrier below the plate, with ambient air circulating below to ensure that heat doesn’t conduct through the air gap (I guess you could measure that air to make sure it wasn’t any colder than ambient air below the real sky).
That would give you an approximation of ‘non-GHG’ sky radiation, next to the jar that radiates through the GHG sky.
Or, as someone else suggested, just buy a slab of dry ice and hold that next to the opening. I’ve worked with LN2, and it’s a pain in the butt.
@ Gord
Apparently, the nesting level for comments is limited. But in response to:
If you bothered to read the established Laws of Science and EM Physics pertaining to your questions, you could answer them yourself.
Example: Pointing a flashlight at a more powerful Car headlight.
If the Photons could actually move from the weaker Flashlight to the more powerful Car headlight, they would be absorbed and heat up Car headlight filament, producing more light.
The now more powerful Car headlight would move Photons back to the Flashlight causing it to heat up and produce more light.
The now more powerful Flashlight would move even more Photons back to the Car Headlight causing it to heat up even more and produce even more more light.
The cycle would continue untill both the Flashlight and the Car Headlight were producing INFINITE energy!
A perpetual motion machine in a positive feedback loop.
——–
from wikipedia…
Perpetual motion
“The term perpetual motion, taken literally, refers to movement that goes on forever. However, the term more generally refers to any closed system that produces more energy than it consumes. Such a device or system would be in violation of the law of conservation of energy, which states that energy can never be created or destroyed.”
“Perpetual motion violates either the first law of thermodynamics, the second law of thermodynamics, or both”
“A perpetual motion machine of the first kind produces energy from nothing, giving the user unlimited ‘free’ energy. It thus violates the law of conservation of energy.”
————-
Why don’t you deal with reality instead of fairy tale irrational beliefs typically found in cults?
Gord
It’s impossible to have a rational discussion with someone so completely misinformed. The comment from Gord, above, is perhaps the best example of the Dunning–Kruger effect I’ve ever seen.
Photons carry no charge and have a rest mass of 0. So please explain to me what force it is that compels the photons from the flashlight beam to discern that the headlight is brighter and turn around. Is it the “stronger” photons from the headlight? Hint: all photons of a given frequency have identical energy, E=h*c, h is Planck’s constant and c is, as usual, the speed of light. Maybe it’s that the flashlight photons are overwhelmed by the shear number of headlight photons and beaten back? Photons do not interact in this fashion.
It’s true that the flashlight will heat the headlight which will emit radiation more energetically, etc. However, equilibrium is reached at a higher energy level and no “infinite feedback loop” is created.
Seriously, get a grip! This is like arguing with a three year old.
Rob, you are probably going to have to just let it go. If someone believes that the sun sets in the east, there’s not much you can do. I guess the headlight-flashlight scenario is used because it would be so hard to observe (without instraments) the effects. Shine a 1000 watt spotlight into a 2000 watt spotlight and, with some designs, the resulting fire would be quite observable.
“Shine a 1000 watt spotlight into a 2000 watt spotlight and, with some designs, the resulting fire would be quite observable.”
Why don’t you video the experiment of putting those spotlights together for us??
Rob,
It seems that you have never heard of the Electromagnetic Field that CARRIES Photon energy or the Electromagnetic Force that moves the zero mass Photons.
All EM fields are Vector fields that have a magnitude and a direction.
The EM fields are also Force fields, in fact, the Electromagnetic Force is one of the four fundamental Forces, and have a Vector magnitude and a direction.
Google “Vector addition of Fields” and learn how to use Vector Math.
The Headlight will produce the larger EM field and EM force.
The Flashlight will produce a smaller EM field and EM force.
The Resultant Vector Field will have a magnitude of [larger EM field – smaller EM field] and the Direction of propagation is from the Headlight to the Flashlight.
There is ZERO EM field propagation from the Flashlight to the Headlight and no Photons can propogate from the Flashlight to the Headlight.
If you use Electromagnetic Forces, the results are exactly the same.
There is ZERO EM field FORCE from the Flashlight to the Headlight and no zero mass Photons can move from the Flashlight to the Headlight.
Just like a block of wood with two opposing forces, the block of wood can only move in the direction of the larger force.
It is IMPOSSIBLE for the block of wood can only move in the direction of the weaker force.
And, It is IMPOSSIBLE for zero mass Photons to move in the direction of the weaker EM Force.
———
This is simple Electromagnetic Physics that has been known and used by Physicists and Electrical Engineers for over a hundred years.
The Vector addition of EM fields are used in all Propagation calculations for all EM fields ranging from long wave Radio through X-Ray frequencies.
Antenna radiation patterns for single or multiple arrays of antennas, use Vector addition of fields to detetermine the nulls and peaks of the resultant radiation pattern.
Your “three year old” denial of established Science is like denying that Antenna Radiation patterns have nulls in them, and that Radio, TV, Satellite, Microwave, IR, Optical…etc Communication links do not exist.
You are hilarious, totally ignorant of the obvious facts….but totally consistent with a Cult mentality.
————
If heat energy could flow from cold to hot objects without work being done to accomplish the flow, a perpetual motion machine will always result.
There is no “equilibrium” temperature that can be calculated because it will constantly create energy and result in an infinite cycle spiral to infinite energy.
—
The 2nd Law of Thermodynamics and The Law of Conservation of Energy are linked.
Violate the 2nd Law of Thermodynamics and The Law of Conservation of Energy will be Violated.
That’s why they are called Laws of Science and have NEVER been shown to be wrong.
If you think that Heat energy can flow from cold to hot objects without work being done to accomplish the flow, why don’t you post ANY measurement, EVER, where this has happened?
It simply DOES NOT EXIST.
All you will be able to do is babble your irrational cult beliefs without a shread of Science or Measurements to support your fairy tale claims.
Come on, Prove me wrong.
Gord.
Dr Spencer has done just that. Or don’t you understand the experiment?
TheLastMann,
What a HOOT!
Where did heat energy flow from cold to hot??
What I see is a duplication of the Solar Oven experiment done by the Physics Dept. of Brigham Young University.
The warmer Cavity DROPPED in temperature about -3 deg F to -5.3 deg F when exposed to the cold sky.
Dr. Spencer not only proved that heat energy can ONLY flow from hot to cold, he also dis-proved the fable that Back Radiation from a colder atmosphere can heat up a warmer Earth.
Your blind devotion to an irrational belief system when all the Science and Measurement facts are right in front of you is typically found in cults.
Gord.
Yes, that is the whole point! It only dropped to -3 and stayed there. Outer space is roughly 3k above absolute zero. The only reason it did not drop that low is because the atmosphere is warmer than outer space and the downwelling infra-red radiation prevented it getting any colder. The implication of the experiment is that it is this downwelling radiation from a cold atmosphere that stops a warmer Earth from getting as cold as outer space.
Clearly you did not understand the experiment then. Strange, I thought he explained it very clearly. Never mind, the rest of us understand it.
Does it ever worry you when everybody else seems to think one thing and it is only you that has the real answer?
A depressive or manic state can often give rise to something called a Grandiose Delusion
Ring any bells?
Gord:
I couldn’t possibly prove you wrong. You are so ludicrously misinformed that there’s no rational basis from which to start. Photons don’t “feel a force” from an EM vector field.
I wonder how you address this: spotlight, flashlight 20 meters apart pointed at one another. 0.1 meters from the flashlight, a light meter pointed at the flashlight will read much brighter light then when turned around and pointed at the spotlight. Somewhere further from the flashlight, the readings will be even. Do the photons emitted by the flashlight progress to that point? What do they do then?
I know it’s a waste of time and keystrokes to try to educate you and, it’s losing its fun factor as well. You apparently learned the words “vector” and “field” somewhere and heard that vectors have a magnitude and direction. You learned somewhere that there are four fundamental forces and that there are a couple of laws of thermodynamics and learned a simplistic way of stating them. At some junior high level you’ve absorbed it and are now misapplying it everywhere.
A correct and universal statement of the first law is: energy obeys a local conservation law (that is, change(energy inside boundary) = ? flow(energy, outward across boundary)). A correct and universal statement of the second law is: entropy obeys a local paraconservation law (that is, change(entropy inside boundary) ? ? flow(entropy, outward across boundary)). Nothing about a cooler radiator raising the temperature of a warmer radiator violates these laws. To show that it violates the second law you must define a system, an environment, and a boundary and show that the entropy decreases in the system without a consequent increase in the entropy of the environment (that is, you must show that a flow of entropy from the system to the environment across the boundary at least as large as the decrease in the system DOES NOT take place. Unfortunately, it does.
A fully correct statement would limit this to systems and environments where Newtonian mechanics provide sufficient accuracy. Where considerations of Special Relativity must be utilized, we must say the [energy, momentum] 4-vector is conserved and, considering General Relativity, we must say that conservation and paraconservation laws apply in regions of spacetime that are at least asymptotically flat.
None of this will, I’m sure, do anything to repair your complete misunderstanding of electromagnetics and thermodynamics.
Rob,
Electromagnetic Field Physics and their Vector nature have been understood and used by Physicists and Electrical Engineers for over a Century.
Simple EM Vector field analysis can be used to accurately predict the radiation pattern, including all nulls and peaks, produced by multiple Radiation sources.
Multiple Antenna arrays are used all the time in communication systems spanning the entire EM spectrum from long wave Radio, TV, Microwave, Satellite, IR and Optical frequencies.
By the way, I am a Professional Electrical Engineer that has been consulting and designing EM field communication systems for over 30 years.
Maybe you should consider dealing with REALITY instead of babbling your hilarious Cult rants, that are proven wrong everytime you turn on a Radio.
Ever consider Comedy as a career?
———————–
PS:
Like I said in my post to YOU on August 2, 2010 at 9:07 PM:
“The 2nd Law of Thermodynamics and The Law of Conservation of Energy are linked.
Violate the 2nd Law of Thermodynamics and The Law of Conservation of Energy will be Violated.
That’s why they are called Laws of Science and have NEVER been shown to be wrong.
If you think that Heat energy can flow from cold to hot objects without work being done to accomplish the flow, why don’t you post ANY measurement, EVER, where this has happened?
It simply DOES NOT EXIST.
All you will be able to do is babble your irrational cult beliefs without a shread of Science or Measurements to support your fairy tale claims.”
—
And, EXACTLY as predicted, all you have done is “babble your irrational cult beliefs without a shread of Science or Measurements to support your fairy tale claims.”
What a HOOT!
Gord.
Radiative heat transfer will always be from hotter to cooler at a rate proportional to deltaT^4. Earth’s gravity ensures that its tropospheric atmosphere will be warmer near the surface than at higher elevations. Therefore there is normally no radiative heat transfer from higher atmosphere to the surface. On low cloudy nights though, when water vapor is condensing, the water vapor releases its latent heat, and that warming effect may be felt on Earth’s surface.
Re: TheLastMan post August 3, 2010 at 6:49 PM
What a HOOT!
Like I said..
“Where did heat energy flow from cold to hot??”
OBVIOUSLY there was ZERO heat flow from COLD to HOT!
But, it did not prevent you from babbling that there was a heat flow from cold to hot!
TheLastMan says:
August 3, 2010 at 12:30 PM
“Dr Spencer has done just that. Or don’t you understand the experiment?”
That’s what I call a “Grandiose Delusion” that only an irrational CULT member could babble.
Ring any bells?
——————
The reason the Earth does not drop to 3K above absolute zero is because the SUN heats the Earth.
Take away the SUN and the Earth WILL drop to 3K above absolute zero.
The Earth has a THERMAL TIME CONSTANT, that means it CAN STORE HEAT ENERGY.
The Earth heats during the day and releases some of the STORED heat energy as it cools during the night.
THAT’S WHY THE EARTH REMAINS WARM AND THERE IS AN AMBIENT AIR TEMPERATURE AT NIGHT THAT IS WAY ABOVE 3K!!
Just like heating a heating a pot of water on a stove.
The larger the pot of water is, the longer it will take to heat up and the longer it will take to cool down, when the stove is turned off.
The water STORES HEAT ENERGY and will heat the air above the pot when it the stove is on AND for a period of time when the stove is turned off.
The heated air above the pot will ALWAYS have a LOWER TEMPERATURE than the water, just like the Earth’s atmosphere is colder than the Earth’s surface.
The stove energy heats the water and the heat energy in the water heats the air above the water on it’s way to the colder surroundings.
The heated water CAUSES the air temperature above the water to rise, not the other way around!
The heated air above the water DOES NOT CAUSE THE HOT WATER TO INCREASE IN TEMPERATURE, just like the Earth’s colder atmosphere does not cause the warmer Earth to increase in temperature.
Dr. Spencer’s experiment proved EXACTLY THAT!
———-
The Cavity COOLED by about -3 deg F to -5.3 deg F compared to the ambient air temperature, when it was exposed to the cold sky.
Dr. Spencer’s experiment proved that the Back Radiation from the colder atmosphere does NOT HEAT THE EARTH OR PREVENT THE EARTH FROM COOLING!
————————–
Your blind devotion to an irrational belief system when all the Science and Measurement facts are right in front of you is typical of CULT members.
Gord.
Gord, you are just not looking at the article you are commenting on. Look at the design of the experiment. Why do you think he has surrounded the sensor with a styrofoam box?
Yes, it is to isolate the sensor from the heat “stored” in the surrounding ground and atmosphere. The sensor he is using has no stored energy. Almost the first thing he says in the article above is
I am just gobsmacked that you still don’t get it. As I have said from the outset, there is no need for heat to flow directly from a colder sky to a warmer Earth. The second law of thermodynamics does not need to be broken.
All that is needed is for the rate of heat flow from hot to cold to be reduced or inhibited. We all understand that the atmosphere is colder than the Earth. But it matters how much colder, the smaller the difference in temperature the lower the rate of heat loss. If the difference is 275 kelvin then the Earth will lose heat quicker than if the difference is 1 kelvin. To me it is common sense and really does not need a scientific education to understand.
…why am I bothering?
The CAVITY STORED HEAT ENERGY FROM THE SUN!
“When I first closed up the box with the thermometer placed in the cavity, I was surprised how hot the cavity became. The maximum temperature recorded yesterday afternoon was 158 deg. F, and that must have been the limit for the sensor, because the temperature then flatlined for about an hour.
The reason for the high temperature was some direct sunlight reflecting off of one wall of the airspace, above the cavity. Even though the cavity was painted white, it still absorbed enough energy to make the air very hot.”
And then the Cavity COOLED by by about -3 deg F to -5.3 deg F compared to the ambient air temperature, when it was exposed to the cold sky.
Dr. Spencer’s experiment proved that the Back Radiation from the colder atmosphere does NOT HEAT THE EARTH OR PREVENT THE EARTH FROM COOLING!
————-
The atmosphere does not affect the Earth’s temperature any more than heated air above a pot of warmer water affects the water temperature.
Example:
– The SUN, the ONLY energy source, heats the Earth to +15 deg C (288K), so using the Stefan-Boltzmanm Law, the Earth receives 390 w/m^2.
– The average surface temperature of the Earth is +15 deg C (288K) so using the Stefan-Boltzmanm Law, it will radiate 390.10 w/m^2.
– With no atmosphere the Earth will radiate 390 w/m^2 to cold space.
– This complies with The Law of Conservation of Energy (Energy In = Energy Out)
—
– Now include an atmosphere and assume the average atmospheric temperature is -20 deg C (253K) and it will radiate 232.32 w/m^2 in all directions.
– Between the Earth and Atmosphere the Vector EM fields are opposing and the Resultant EM field will have a Magnitude of 390.10 – 232.32 = 157.78 w/m^2 and the Direction of propagation is from the warmer Earth to the colder atmosphere.
There is ZERO w/m^2 propagating from the Cold atmosphere to the Warmer Earth.
– Past the atmosphere, the the Vector EM fields are in the same direction (towards Cold Space), so they ADD.
– The Resultant Vector EM field will have a Magnitude of 157.78 + 232.32 = 390.10 w/m^2 and a Direction of propagation towards Cold Space.
This complies with The Law of Conservation of Energy (Energy In = Energy Out)
ALL the Earth’s energy received from the Sun, is still flowing to Cold Space, EXACTLY the same as with no atmosphere!
Like I said…
“The atmosphere does not affect the Earth’s temperature any more than heated air above a pot of warmer water affects the water temperature.”
———-
Maybe it’s time you started using established Laws of Science instead of babbling like an irrational Cult member.
Gord.
Wrong in so, so, many, many ways…
As you say, the earth is at about +15 deg C in this warmly atmosphere blanketed Earth.
It is a well established fact (Google it) that at the top of the atmosphere the Earth recieves an average 342w per m2 of surface area (averaging for night/day and lattitude etc).
However…
6% is reflected straight back by the atmosphere itself
20% is reflected straight back by clouds
4% is reflected straight back by the surface.
This means that the actual amount of the Sun’s radiation reaching the Earth and absorbed by it is just 70% of the radiation received from the Sun at the top of the atmosphere. In other words only around 240 w/m2 actually hits the ground and is absorbed.
But, as you so rightly say, the temperature is +15 deg C. So it must be emitting at a rate of 390 w/m2. The S-B law cannot be wrong.
But… if it is only getting 240 w/m2 from the Sun, where can those extra 150 w/m2 be coming from?
You have 3 hours for this paper. Please write on both sides of the paper. Calculators are permitted in the exam room.
LastMan illustrates the fallacy of using averages of solar input, temperatures, etc. over a diurnal cycle.
Daytime solar input 1370 Wm-2 – 30% albedo = 959 Wm-2 input to earth surface which according to SB should raise the surface temp to 87.5C or 360K if a day was long enough to reach equilibrium, but it only has time to reach 15C or so.
At night, the earth loses some of this accumulated heat primarily due to latent heat transfer from the surface to atmosphere and ultimately to space.
The earth radiates ~390 Wm-2 at night because it accumulated 959 Wm-2 during the day, not because GHGs add back an “extra 150 Wm-2.”
hey Hockey Schtick, I don’t know whether you realize it, but you just assumed the world is flat. 🙂
I don’t assume the world is flat, but Kiehl & Trenberth et al do, and I am simply pointing out the fallacy.
You divide the solar constant by 4 to spread solar energy across the spherical earth, before accounting for albedo…when you didn’t do that, you assumed the Earth was flat. 🙂
Ok, hope I’m getting there…
1370/4=342-30%=239.4 W/m2 average insolation during the day.
0 W/m2 “average insolation” at night
Average of day & “night” insolation = 120 W/m2
K-T show the average of day & night insolation on their “flat earth with continuous solar input” to be 198 W/m2
Why?
Ok, 959/2=479.5 Wm-2 average daytime solar input, so I correct my statement on the fallacy to:
The earth radiates ~390 Wm-2 at night because it accumulated 480 Wm-2 during the day, not because GHGs add back an “extra 150 Wm-2.”
You DO realize that even though the sun shines only during the day, that IR is emitted day and night, right?
Correction: K-T show 168 W/m2 average day & night insolation
Why not 120?
This was still awating moderation since August 6, 2010 at 7:33 PM, so I will post it without the full links.
RE:TheLastMan says:
August 6, 2010 at 3:49 AM
Your post is just more typical Cult babbling.
The Earth’s atmosphere is NOT an energy source.
The Earth’s colder atmosphere can only COOL the Earth.
The Sun is the ONLY energy source and is, therefore, the ONLY cause of the Earth’s heating.
(See The Law of Conservation of Energy)
The AGW QUACKS have Deliberately and Fraudulently Lowered the Sun’s temperature to 5778K!
——-
PROOF:
Temperature on the Surface of the Sun
There are five sources for the surface temp of the Sun (6000,5500,5700,6000 and 5600 deg C).
The average is 5800 deg C or 6073 K and a max of 6000 deg C or 6273 K.
See hypertextbook dot com/facts/1997/GlyniseFinney dot shtml
I have seen a number of AGW’er papers and other sources that correctly state that the actual Sun temperature is much higher than 5778 K.
Here is one example:
HEATING THE EARTH
“3. If the Sun were a blackbody, this emissivity would correspond to a surface
temperature of 5798°K. However, the wavelength of maximum intensity is at 0.475 microns (green light). By Wien’s Law, this is the maximum that would be produced by a blackbody at a temperature of 6101°K.”
See www dot climates dot com/SPECIAL%20TOPICS/GW/heatingtheearth dot pdf
—-
Here is another… Sun’s emissivity curve as a function of wavelength:
File:EffectiveTemperature 300dpi e.png
See en dot wikipedia dot org/wiki/File:EffectiveTemperature_300dpi_e dot png
See how the Solar spectral irradiance exceeds the black body approximation in the visible light range (400 to 700 nm).
The visble light spectrum from 400 nm to 700 nm (that will absolutely heat the Earth) has been FRAUDULENTLY under valued with a Sun Temp = 5777 K!!
If you look closely at the Effective Temperature graph above you will see that the Green Light real peak is at about 462 nm and using Wein’s Displacement Law will produce a Sun temperature of 6273 K (thats 6000 deg C) and agrees with the link I quoted above!
——-
This equation relates Sun Temperature to Earth Temperature
See en dot wikipedia dot org/wiki/Black_body
TE = TS (((1-a)^0.5 * Rs)/(2*D)))^0.5)
Where TE is blackbody temp of the Earth in K
TS is the surface temp of the SUN in K
Rs is radius of the Sun (6.96 X 10^8 meters)
D is distance between the Sun and Earth in meters (1.5 X 10^11)
a is albedo of the Earth
Results: (Albedo = 0, Black Body Earth WITHOUT AN ATMOSPHERE)
TS = 5778 K (Trenberth’s FRAUDULENT “lowered” Sun Temperature)
TE = 278.68 Kelvin
TE = 5.53 Celsius
Note: This is very close to the value obtained in this link:
“If an ideal thermally conductive blackbody was the same distance from the Sun as the Earth, it would have an expected blackbody temperature of 5.3 °C.”
See en dot wikipedia dot org/wiki/Greenhouse_effect
TS = 6073 K (average Sun temperature)
TE = 292.91 Kelvin
TE = 19.76 Celsius
TS = 6101 K (as per the AGW’er Heating The Earth link above)
TE = 294.26 Kelvin
TE = 21.11 Celsius
TS = 6273 K (max Sun temperature)
TE = 302.55 Kelvin
TE = 29.40 Celsius
All the Sun temperatures (except Trenberth’s) produce Earth temperatures that exceed the Earth’s average temp of +15 deg C.
This means that the ADDITION OF AN ATMOSPHERE AND ALL OTHER FACTORS ACTUALLY COOLED THE EARTH !!!!
—————–
Maybe it’s time you started using established Laws of Science instead of babbling like an irrational Cult member.
Gord.
To: Dr.Spencer and Hockey Schtick
This equation relates Sun Temperature to Earth Temperature:
TE = TS (((1-a)^0.5 * Rs)/(2*D)))^0.5)
Where TE is blackbody temp of the Earth in K
TS is the surface temp of the SUN in K
Rs is radius of the Sun (6.96 X 10^8 meters)
D is distance between the Sun and Earth in meters (1.5 X 10^11)
a is albedo of the Earth
See en dot wikipedia dot org/wiki/Black_body
It uses the Earth as a “flat disk” (Area = pi*R^2) for radiation received from Solar Energy.
This is the projection of a spherical Earth onto a two dimensional plane, the same result as if a Surface Integral were used to determine the Average Solar Flux density of the Spherical Earth surface facing the Sun.
That is why the, so-called, Solar Constant is divided by 4.
“flat disk” (Area = pi*R^2) divided by Spherical Earth Area 4*pi*R^2 = 1/4.
This is included in the development of the Equation above.
The Equation assumes Solar Energy is received as a “flat disk” during the day and is equated to the IR re-radiation by the Earth, as a Sphere, day and night.
There is no need to divide the Solar insolation by 2 since it is already an average and equated to the day and night IR re-radiation by the Spherical Earth.
——
The problem is that Trenberth and most AGW’ers have fraudulently lowered the Sun’s surface temperature to 5778 K, when it is actually much higher.
(see my Proof of this in my Post Gord says: August 8, 2010 at 7:42 PM above)
“If you look closely at the Effective Temperature graph above you will see that the Green Light real peak is at about 462 nm and using Wein’s Displacement Law will produce a Sun
temperature of 6273 K (thats 6000 deg C) and agrees with the link I quoted above!”
And, a 6273 K Sun Surface temperature will produce a (albedo = 0, no atmosphere) Black Body Earth temperature of 29.40 deg C!
So, the addition of an Atmosphere and all other factors, produces a COOLING of the Earth, down to +15 deg C!
——–
PS:
The equation above can be modified to have a Spherical Earth receiving the Solar Energy, which will produce a much higher Earth temperature that is equivalent to the Earth’s
temperature at the Equator, where the Solar Energy received is normal to the Spherical Earth surface.
The new Equation becomes:
TE = TS (((1-a)^0.5 * Rs)/(D)))^0.5)
The strong heating of the Oceans at the equator and the Earth’s rotation (called the Coriolis Acceleration Force) will ensure that water heated at the Equator will transfer heat to the Colder regions of the Oceans, including the Polar Regions.
Gord.
Gord,
Thanks for your very interesting observations (and I have also appreciated your comments elsewhere on solar cookers disproving “back-radiation”)!
Here’s what someone emailed me in response to your comment above: “I have also suspected that the sun’s temperature might be underestimated. Reason: When fitted to a Planck profile, parts of the sun’s emission exceed a blackbody’s profile at the estimated temperature. If the sun is a not-quite blackbody, then, yet the emission profile is accurate, then it must be hotter. Gord could be correct.”
If you don’t mind, I will be creating a new post containing some of your insights giving credit to you, whoever you are.
Also, what do you think about this:
1370/4=342-30% albedo=239.4 W/m2 average insolation during the day.
0 W/m2 “average insolation” at night
Average of day & “night” insolation = 120 W/m2 (K-T say 168)
Average day & night global temperature: ~15C
Average atmospheric temperature: ~-15C
Using these avg temps in the SB equation, Earth should be radiating a diurnal average of 120 W/m2 to the atmosphere, which is exactly the “average insolation” above.
Agree?
when you divided 1370 by 4, you already averaged the sunlight over 24 hours (the day and night sides of the Earth)…can’t divide by 2 again
Ok, my bad, should have looked at Claes Johnson’s site first where he replied “The day-night compensation is already included in the factor 4 dividing 1370, so 120 does not come from 240/2. Instead it comes out from a balance between thermodynamics setting the lapse rate and radiation operating on the lapse rate.”
But I still don’t understand how K-T come up with an average solar insolation of 168 W/m2 if anyone wants to explain to me.
Its because part of the absorbed solar (67) is absorbed directly by the atmosphere. There is direct absorption by water vapor, and some absorption by clouds (the size of that is controversial). The number (67) has always seemed high to me, but all I have is gut instinct to go on.
Sorry to keep beating this into the ground, but if K-T are anywhere near correct that the average solar insolation is 168 W/m2, by SB that means the surface temp should be -40C.
Almost all explanations of the “GHE” state that the earth is 33C warmer (15C) than it would otherwise be (-18C), since they use 240 W/m2 as the average solar insolation at the surface.
I see you said you think 67 is high, but how do you explain such a huge discrepancy between K-T (168) and most descriptions of the GHE (240)?
They don’t use 240 for the surface. They use 240 for what the top-of-atmosphere emitted IR is. Since that corresponds to blackbody emission temperature that is 33 deg. C less than the OBSERVED surface temperature, the greenhouse effect is quantified at 33 deg. C surface warming.
If there were no IR absorbers, the surface IR emission and top of atmosphere emission would be the same, so no greenhouse effect.
Dr. Spencer,
You claim the “Greenhouse Effect” has caused the the Earth’s surface temperature to rise from -18 deg C to +15 deg C….a 33 deg C rise in temperature.
If the only energy source, the Sun, only provides 240 w/m^2 to heat the Earth to -18 deg C and a +15 deg C Earth radiates 390 w/m^2…then there must be 150 w/m^2 CONSTANTLY flowing from the Colder atmosphere to a Warmer Earth Surface to maintain the +15 deg C temperature.
WHERE DID THE EXTRA 150 WATTS/M^2 COME FROM?…WAS IT CREATED?
WHY CAN’T WE MEASURE OR HARNESS THIS PHANTOM “DAY AND NIGHT” COLD “BACK RADIATION” HEATING OF THE WARMER EARTH, THAT EXCEEDS THE SOLAR ENERGY BY 150 WATTS/M^2 ?
——
The facts are:
– Heat energy can only flow from hot to cold objects, unless work is done to make the reverse flow possible…just like The 2nd Law states.
– Spontaneous heat energy flow from cold to hot is not possible and would result in a Violation of The Law of Conservation of Energy (creation of energy and perpetual motion).
– There is no Law of Science that supports the fantasy “Greenhouse Effect”.
– There is not even ONE measurement, ever done, that shows that a colder atmosphere can heat up a warmer Earth.
– ALL actual measurements PROVE that a colder atmosphere CANNOT heat up a warmer Earth.
The “Greenhouse Effect” is pure fiction.
Gord.
Gord, by your reasoning, I could put a 240 Watt heat source in a heavily insulated 1 cubic meter enclosure, and the temperature would never get above -18 deg. C. You are confusing the rate of energy flow with accumulated energy (temperature).
No, I’m NOT confusing rate of energy flow with temperature.
The Stefan-Boltzmann Law relates Watts/m^2 to Temperature!
—–
– Without any heat energy being received by the Earth from the Sun, the Earth would be at a temperature near absolute zero.
– If the Earth receives and absorbs 240 Watts/m^2 from the Sun, the temperature will rise to 255.07 K or -17.93 deg C. (See the Stefan-Boltzmann Law)
– If the Earth is to rise in temperature to +15 deg C (288 K) it has to absorb 390.10 Watts/m^2. (See the Stefan-Boltzmann Law)
The Earth needs to absorb an ADDITIONAL 390.10 – 240 = 150.10 Watts/m^2 in order to rise in temperature from -17.93 deg C to +15 deg C…a 32.93 deg C rise in temperature.
But, the only energy source (the Sun) can only provide 240 Watts/m^2, so this temperature rise is NOT POSSIBLE because it requires CREATION OF ENERGY.
Your “Greenhouse Effect” CREATES the ADDITIONAL 150.10 Watts/m^2 from NOTHING….a Perpetual Motion Machine.
———
The same applies to your example.
– If a heavily insulated 1 cubic meter enclosure is suspended in cold space, it would also be at a temperature near absolute zero.
– If a Heating source supplies 240 Watts to a body that has a radiating surface area of 1 m^2 inside the enclosure, it will radiate 240 Watts/m^2 and reach a temperature of -17.93 deg C.
The body CANNOT increase in temperature past -17.93 deg C unless the 240 Watts is INCREASED.
This is simple Conservation of Energy….Energy CANNOT BE CREATED.
—
PS:
Add all the “Greenhouse Gasses” you want to the enclosure, they are NOT ENERGY SOURCES and temperature of the Body will DROP because “Greenhouse Gasses” require energy to heat up and will also Increase the Radiating Surface Area of the Body.
Just like a “Heat Sink” works to COOL the microprocessor in your Computer.
The Watts disipated by the microprocessor remains the same but the Watts are distributed over a larger radiating surface area DROPPING the watts/m^2 and therefore the TEMPERATURE of the microprocessor DROPS!
———-
The “Greenhouse Effect” requires that Energy be CREATED and is pure fiction.
Gord.
I would not bother with Gord, Dr Spencer. He just does not get it and never will. I have tried multiple lines of reasoning but he still seems to think that an increase in temperature requires the “creation of energy”.
The fact that we can measure the Greenhouse Effect empirically, and have been able to do so for more than 50 years seems to have passed him by too.
The fact that somebody so clearly ill-educated is trying to argue that the Greenhouse Effect does not exist with somebody who measures it for a living beggars belief!
A classic case of Dunning-Kruger.
I have been suffering from SIWOTI (Somebody Is Wrong on the Internet) compulsion. That is the tendency to try and correct people who make blatantly wrong statements.
I am beginning to learn that such people are usually deluded or set in their views so the effort is usually wasted.
…and I’m beginning to regret I ever started this little exercise. 🙂
TheLastMan says:
“I have tried multiple lines of reasoning but he still seems to think that an increase in temperature requires the “creation of energy”.”
My post to TheLastMan July 29, 2010 at 9:22 PM:
“I completed the calculation of the two bodies, showed that they would both be radiating 300 units at equilibrium and that it complied with The Law of Conservation of Energy.
I suggest that do the same, with units of energy flowing from the 200 body to the 300 body.
Compute the equilibrium units of energy for both bodies and show that it complies with The Law of Conservation of Energy.
Good Luck.”
Any LUCK computing the equilibrium units of energy for both bodies and show that it complies with The Law of Conservation of Energy???
No?….Why not?
———
TheLastMan says:
“The fact that we can measure the Greenhouse Effect empirically, and have been able to do so for more than 50 years seems to have passed him by too.”
Why don’t you POST these Phantom “Greenhouse Effect” Measurements???
———–
I look forward to your next babbling post.
Gord.
Gord,
I created a post based upon your observations:
hockeyschtick.blogspot . com/2010/08/is-greenhouse-effect-is-based-on-cool . html
and invite you to comment on it. Thanks
Hockey Schtick,
This equation relates Sun Temperature to Earth Temperature:
TE = TS (((1-a)^0.5 * Rs)/(2*D)))^0.5)
Where TE is blackbody temp of the Earth in K
TS is the surface temp of the SUN in K
Rs is radius of the Sun (6.96 X 10^8 meters)
D is distance between the Sun and Earth in meters (1.5 X 10^11)
a is albedo of the Earth
The equation you posted on your website is not correct.
TE = TS ((((1-a)/e)^0.5 * Rs)/(2*D)))^0.5)
This equation has e = emissivity of the Earth included again.
(1-a) is equal to e and it already includes the emissivity of the Earth as well as the Clouds.
Division by e is obviously wrong since if e = 0 (a perfect reflecting Earth), it will produce an INFINITE Earth temperature (TE)!
Someone has changed the correct equation on Wikipedia to this hilarious and obviously wrong form.
Gord.
Hockey Schtick,
Re: Division by 2 to get 120 w/m^2
I covered this in my previous post:
“The Equation assumes Solar Energy is received as a “flat disk” during the day and is equated to the IR re-radiation by the Earth, as a Sphere, day and night.
There is no need to divide the Solar insolation by 2 since it is already an average and equated to the day and night IR re-radiation by the Earth.”
Dr. Spencer also made a similiar comment, which I agree with.
————–
Some points regarding the, so called, Solar Constant being 1370 w/m^2.
– The Solar Constant can be be calculated from this equation:
SC = BC*TS^4 * Rs^2/D^2 in w/m^2
Where BC = Boltzmann Constant = 5.67 X 10^-8
TS is the surface temp of the SUN in K
Rs is radius of the Sun (6.96 X 10^8 meters)
D is distance between the Sun and Earth in meters (1.5 X 10^11)
If a fraudulent Sun temperature of 5778 K is used, the Solar Constant is 1367.88 w/m^2.
This is the Solar Constant value used in the K-T energy budget diagram.
If a true Sun temperature of 6273 K is used the Solar Constant is 1900.38 w/m^2.
This equation relates the Earth Temperature to the Solar Constant.
TE = (SC * (1-a) / (4*BC))^0.25
A 1900.38 w/m^2 Solar Constant will produce the same (a = Albedo = 0, no atmosphere) Black Body Earth temperature = 29.40 deg C that I calculated before.
———-
Some comments regarding K-T energy budget diagram.
The very first and most basic point is that, the Sun is the ONLY Energy Source.
The atmosphere and Earth are NOT Energy sources, however K-T treats the atmosphere and Earth as energy sources that EXCEED the energy that the Sun provides.
This is a blatent Violation of The Law of
Conservation of Energy as well as a Violation of The 2nd Law of Thermodynamics.
PROOF:
K-T shows the Solar Energy at the top of the atmosphere (using a fraudulent low Solar Constant) as being 1367.88 w/m^2 /4 = 342 w/m^2
Reflection by clouds and the Earth’s surface = 107 w/m^2 leaving 342-107 = 235 w/m^2 to heat the atmosphere and the Earth.
The K-T energy budget diagram balances the 235 w/m^2 (that heats the atmosphere and the Earth) with 235 w/m^2 out-going top of atmosphere long wave radiation and claims that this satisfies The Law of Conservation of Energy.
This is a SCAM, because the Earth surface Radiation (390 w/m^2) and the Atmospheric “Back Radiation” (324 w/m^2), below the top of the atmosphere, BOTH EXCEED the 235 w/m^2 value!
This is a clear Violation of The Law of Conservation of Energy.
Further, of the 235 w/m^2 Solar Energy, 67 w/m^2 is absorbed by the Atmosphere leaving only 168 w/m^2 to heat the Earth’s surface.
Yet, the Earth is shown to be radiating 390 w/m^2 of longwave IR energy!…a physical IMPOSIBILITY.
(The 168 w/m^2 Solar Energy being absorbed by the Earth’s Surface could only heat the Earth’s Surface to 233.31 K or -39.69 deg C!)
———————–
If you total all the “In-Coming Energy” at the Earth’s Surface you get:
– Solar Energy = 168 w/m^2 (based on a fraudulent lower Sun temperature)
– Back Radiation from the colder atmosphere being absorbed by and heating the warmer Earth’s surface = 324 w/m^2 (a violation of the 2nd Law and not supported by ANY Measurement)
(Counting the Back Radiation as an energy source is like adding the watts consumed by a Resistor (atmosphere) to the Watts provided by the Battery (Sun), in Battery-Resistor
Circuit!)
Total In-Coming Energy = 168 + 324 = 492 w/m^2
—
If you total all the “Out-Going Energy” at the Earth’s Surface you get:
– Earth Surface Radiation = 390 w/m^2
– Thermals = 24 w/m^2
– Evapo-transpiration = 78 w/m^2
Total Out-Going Energy = 390 + 24 + 78 = 492 w/m^2
Total In-Coming Energy at the Earth’s surface = Total Out-Going Energy at the Earth’s surface = 492 w/m^2!
So, What should the temperature of the Earth’s surface be according K-T and the Stefan-Boltzmann Law?
Assuming an emissivity of 1, 492 w/m^2 will produce an Earth Surface temperature of 305.2 K or +32.2 deg C!!
————
The K-T diagram is full of obvious Violations of Laws of Science, contains a Fraudulent Sun temperature, is completely dis-proven by actual measurements and is filled with hilarious contradictions.
The K-T Earth Energy Budget Diagram really is a CARTOON.
Gord.
Aw come on Lastman & Dr. Spencer, what wrong with a little lively debate?
Gord & Dr. Spencer: thanks I am now thoroughly straitened out on dividing the solar constant by 4, except that the latest Chilingar et al paper I just posted says we should be dividing by 3.5 due to the current Earth precession angle of 23.44 degrees:
hockeyschtick . blogspot . com/2010/08/paper-increasing-co2-cools-climate . html
agree or disagree?
Secondly, Lastman himself states above “around 240 w/m2 actually hits the ground and is absorbed,” which is also what just about every GHE description I have seen says, yet when I asked Dr. Spencer why K-T says it’s only 168 absorbed by the surface he says GHE descriptions “don’t use 240 [for the surface]. They use 240 for what the top-of-atmosphere emitted IR is. Since that corresponds to blackbody emission temperature that is 33 deg. C less than the OBSERVED surface temperature, the greenhouse effect is quantified at 33 deg. C surface warming.”
So, which is it guys, 168 or 240 absorbed by the surface?
BTW that 67 “absorbed by the atmosphere” seems to be added back to the surface as the 66 difference between surface radiation & “back-radiation”
Hi Hockey Schtick,
Thanks for your link to the paper:
Cooling of Atmosphere Due to CO2 Emission
The Paper attempts to tie adiabatic heating and Radiation heating together.
While this commendable, I just have a few comments regarding Radiation.
The Earth modelled as a disk for Solar Insolation is, of course, a gross simplification.
Example:
– There is no accounting for the Earth’s rotation or speed of rotation, variation of Solar flux effective heating based on latitude and/or longitude, seasonal axis tilt or change of orbital distance from the Sun.
– The Earth’s albedo is also affected by some of these parameters, as well as the varying quantity and distribution of clouds etc.
These are just a few of the variables (I’m sure there are many, many more) that are all continuously changing at the same time.
——–
That being said…
I did not see any reason or explaination why the Solar Constant should be divided by 3.5 rather than 4 to account for the Earth’s angle of 23.44 degrees.
But, maybe I missed something?
Gord.
maybe this has already been answered, but 240 is absorbed by the system….most at the surface, some directly by the atmosphere.
Dr. Spencer,
Dr Hertzberg Ph.D. has these Videos on Youtube where he describes the “Greenhouse Effect” as non-existent.
“Climategate” and Scientific Inquiry, part 6 of 7 (No Greenhouse Effect)
www.
youtube.com/watch?v=wsePItb-Anw&NR=1
“Climategate” and Scientific Inquiry, part 5 of 7 (Venus temp vs pressure, No Greenhouse Effect)
www.
youtube.com/watch?v=8t5vXiQzos4&NR=1
Are you aware of his views?
Have you ever had any discussion with Dr Hertzberg on the subject of the “Greenhouse Effect”?
I already have in my post of:
August 6, 2010 at 3:49 AM
..but you just chose to ignore it.
The amount of solar radiation received at the surface, after deducting direct reflection from the atmosphere and surface is roughly at a rate of 240w/m2. I won’t post a link, because links cause the post to be lost in moderation, but just Google it – the measurements have been carried out by various satellites over the years and the results put on the net. It is a well established fact.
Despite this, the temperature of the Earth means that it is emitting radiation at 390 w/m2. This can also be measured directly simply by calculation from the surface temperature.
From this we have to deduce that the surface of the Earth is receiving 150 w/m2 of energy from somewhere, because otherwise its temperature would be 33c lower.
It is also possible to measure back-radiation from the atmosphere using the various devices described by Dr Spencer here including his ingenious “box” and hand-held IR sensors.
These measurements show that the IR energy back-radiated from the atmosphere is sufficient to account for the higher temperature at the surface.
The IR energy being emitted from the top of the atmosphere has also been measured at 240 w/m2 – so no energy is created and no energy is lost. All that happens is that the rate of energy loss from the surface of the Earth is slowed down by back-radiation from the atmosphere causing it to rise in temperature relative to the upper levels of the atmosphere.
I repeat, these emission figures have been measured empirically in the real world. Conjecture that “it just isn’t possible” is pointless.
Now I understand that you have a problem with this because of your belief that no energy emitted or reflected from a lower temperature body can be absorbed by a higher temperature body because this somehow violates the 2nd law of Thermodynamics and/or some misunderstanding of the way EM radiation is emitted and received between bodies at different temperatures. I am afraid you are just plain wrong about that which is the source of your whole difficulty with this issue.
I already have in my post of:
August 6, 2010 at 3:49 AM
..but you just chose to ignore it.
The amount of solar radiation received at the surface, after deducting direct reflection from the atmosphere and surface is roughly at a rate of 240w/m2. I won’t post a link, because links cause the post to be lost in moderation, but just Google it – the measurements have been carried out by various satellites over the years and the results put on the net. It is a well established fact.
Despite this, the temperature of the Earth means that it is emitting radiation at 390 w/m2. This can also be measured directly simply by calculation from the surface temperature.
From this we have to deduce that the surface of the Earth is receiving 150 w/m2 of energy from somewhere, because otherwise its temperature would be 33c lower.
It is also possible to measure back-radiation from the atmosphere using the various devices described by Dr Spencer here including his ingenious “box” and hand-held IR sensors.
These measurements show that the IR energy back-radiated from the atmosphere is sufficient to account for the higher temperature at the surface.
The IR energy being emitted from the top of the atmosphere has also been measured at 240 w/m2 – so no energy is created and no energy is lost. All that happens is that the rate of energy loss from the surface of the Earth is slowed down by back-radiation from the atmosphere causing it to rise in temperature relative to the upper levels of the atmosphere.
I repeat, these emission figures have been measured empirically in the real world. Conjecture that “it just isn’t possible” is pointless.
Now I understand that you have a problem with this because of your belief that no energy emitted from, or reflected by, a lower temperature body can be absorbed by a higher temperature body because this somehow violates the 2nd law of Thermodynamics and/or some misunderstanding of the way electromagnetic radiation is emitted and received between bodies at different temperatures. I am afraid you are just plain wrong about all of that which is the source of your whole difficulty with this issue.
TheLastMan said:
“The fact that we can measure the Greenhouse Effect empirically, and have been able to do so for more than 50 years seems to have passed him by too.”
My response was:
“Why don’t you POST these Phantom “Greenhouse Effect” Measurements???
I look forward to your next babbling post.”
And, the RESULT is YOU PRODUCED ANOTHER BABBLING POST WITHOUT POSTING “ANY MEASUREMENT”!!!
PS: I am STILL waiting for your answer on this too:
Any LUCK computing the equilibrium units of energy for both bodies and show that it complies with The Law of Conservation of Energy???
No?….Why not?
————–
You did NOT post ANY link to ANY Phantom “Greenhouse Effect” Measurements on August 6, 2010 at 3:49 AM or any other time.
Do you even know what the fantasy “Greenhouse Effect” is?
The “Greenhouse Effect” is the Colder atmosphere heating up a warmer Earth Surface!
There has never been ANY measurement, EVER done, that shows that a Colder Atmosphere can heat up a Warmer Earth Surface!
That’s why all you can do is BABBLE about this PHANTOM Measurement when, in fact, NO SUCH MEASUREMENT EXISTS!
YOU CAN’T POST ANY MEASUREMENT BECAUSE IT SIMPLY DOES NOT EXIST!
———
I have already covered these items in my previous posts:
– ALL direct measurements of Back Radiation uses instruments that have CRYOGENICALLY COOLED IR DETECTORS, far below the -20 deg C average atmospheric temperature, to make direct measurements possible.
– Dr. Spencer’s “box” proved that Back Radiation CANNOT heat the Earth.
The Cavity DROPPED in temperature between about -3 deg F and -5.3 deg F, proving that heat from the Cavity ONLY moves to the Colder Atmosphere.
– Infrared Thermometers work exactly the same way, the Thermistor in these units drop in resistance when pointed at the cold atmosphere.
All the above confirm exactly what the 2nd Law of Thermodynamics states:
“Second Law of Thermodynamics: It is NOT POSSIBLE for heat to flow from a colder body to a warmer body without any work having been done to accomplish this flow. Energy will not flow spontaneously from a low temperature object to a higher temperature object.”
——
Regarding the Sun, I have also posted this:
– I posted Proof that the AGW QUACKS have Deliberately and Fraudulently Lowered the Sun’s temperature to 5778K!
– Using PROOF of a measured Sun temperature of 6273 K, the Earth’s Black Body Surface temperature (albedo = 0, no atmosphere) temperature is 29.40 deg C!
– So, the addition of an Atmosphere and all other factors, produces a COOLING of the Earth, down to +15 deg C!
———
What I have shown is that the Fantasy “Greenhouse Effect” does NOT EXIST, confirmed with ACTUAL MEASUREMENTS.
In fact, EVERY MEASUREMENT EVER DONE, confirms that a Colder Atmosphere CANNOT heat up a Warmer Earth Surface!
———
So all YOU are left with is:
– ZERO measurements to support your TOTALLY FALSE CLAIM:
TheLastMan said:
“The fact that we can measure the Greenhouse Effect empirically, and have been able to do so for more than 50 years seems to have passed him by too.”
– Another irrational babbling post totally devoid of any science or measurements, that ignores all established Laws of Science, all proven facts and all actual measurements.
What a HOOT!
It really would be funny, if wasn’t so Pathetic.
Gord.
So how do you think insulation works?
Such as putting cold reflective foil behind my radiators which make the radiators hotter.
Would a cold body that reflected 100% of the energy from a hot body back to that hot body warm it up?
If so, why is radiation reflected by a colder body any different from radiation emitted by a colder body?
We have discussed this endlessly before in this thread but all that is needed to comply with the 2nd law of thermodynamics is that the net movement of energy is from warmer to colder. It is still possible for a hot body to be both absorbing and emitting energy at the same time. After all, do you deny that the earth absorbs UV while simultaneously emitting IR?
If so, why can it not absorb IR while also emitting IR? They are after all just the same form of energy but at different wavelengths.
As for this:
Now it is my turn to say “what a hoot!”.
– Insulation is not an energy source, so it cannot heat ANYTHING up. That’s why Coffee in a Thermos Bottle will always cool and not HEAT UP.
– Reflective foil does NOT make your radiator HOTTER, it just concentrates more radiation in one direction while reducing it in another direction. (See Antenna radiation pattern nulls and peaks)
– 100% reflection does increase energy. That’s why you don’t burst into flames when you look in a mirror. (See The Law of Conservation of Energy).
– If any energy could flow from cold to hot, you would burst into flames when you look in a mirror. (See 2nd Law of Thermodynamics and Perpetual Motion)
– Cold objects can absorb heat from Warm Objects, not the other way, that’s why you don’t burst into flames if you stand on a glacier. (See 2nd Law of Thermodynamics and Perpetual Motion)
– I see you deny Weins Displacement Law in addition to The 2nd Law, The Law of Conservation of Energy, Heat Transfer Physics, Electromagnetic Physics, Vector Mathematics, Cause & Effect and common sense.
————-
So, all YOU are left with, AGAIN, is:
– ZERO measurements to support your TOTALLY FALSE CLAIM:
TheLastMan said:
“The fact that we can measure the Greenhouse Effect empirically, and have been able to do so for more than 50 years seems to have passed him by too.”
– Another irrational babbling post totally devoid of any science or measurements, that ignores all established Laws of Science, all proven facts and all actual measurements.
What a HOOT!
You would be really funny, if you were not so Pathetic.
Gord.
Right from the very beginning of this entire thread you have completely misunderstood the whole point! My house is a heat source, as is the surface of the Earth. I have a furnace pumping out heat. The insulation, which is cold, causes the temperature to rise (i.e. “heat up”) because it reduces the rate of heat loss from a body that is generating heat. The Earth absorbs energy from the Sun (call it a furnace), and if that energy cannot escape, or escapes less fast, then the Earth will rise in temperature (i.e. “heat up”).
Oh yes it does. I have measured it. I have an electric storage heater in a conservatory. Putting foil behind it increases the surface temperature by about 2c.
Yes, back towards the radiator… while reducing it in another direction. (See Antenna radiation pattern nulls and peaks)
Yes, out of the wall into the cold night air. And what do you think happens to the energy reflected back by the foil towards the radiator?
More rubbish. The 2nd law of thermodynamics only requires that in net terms that energy flow from hot to cold.
I do not burst into flames when I look in a mirror because the net gain in temperature is absolutely tiny because I am such a poor heat source.
Shine an arc lamp into a mirror from 10cm and see if it bursts into flames.
You are just plain wrong. A glacier is giving out IR because it has a temperature above absolute zero. I am warmer standing on a glacier than I would be standing on dry ice.
The rate of flow of heat from hot to cold increases when the difference in temperature is greater. Dry ice is colder than water ice, hence I am warmer standing on a Glacier than dry ice.
Now that is common sense.
From the forum:
First Results from THE BOX: Investigating the Effects of Infrared Sky Radiation on Air Temperature
Roy W. Spencer, Ph. D. says:
July 29, 2010 at 2:38 PM
you cannot focus an extended source of thermal radiation.
—————
Wrong!
Remotely operated infrared radiometer for the measurement of atmospheric water vapor
David A Naylor?, Brad G. Gom, Ian S. Schofield, Gregory J. Tompkins, Ian M. Chapman
Department of Physics, University of Lethbridge, Lethbridge, Alberta, Canada
3.2 Optical
“The optical components of the radiometer are shown in Figure 2. Light enters the radiometer through a 13?m thick
polypropylene window, which acts as dust/moisture protector yet has negligible absorption at 20?m. The 100mm, f/1,
90º off-axis parabolic mirror brings the beam to a focus, behind a 1mm thick ZnSe optical window, on a 1mm2 Mercury-
Cadmium-Telluride (MCT) photoconductive detector (Kolmar Technologies; http://www.kolmartech.com) designed to
have a long wavelength cutoff at ~22 ?m and for operation at 77 K.”
http:
//www
.uleth.ca/phy/naylor/documents/pdf/SPIE_IRMAIII.pdf
This instrument uses a COOLED IR detector at the focal point of a Parabolic Mirror to measure atmospheric IR!
Gord.
sorry, I meant magnify…focus was too general a term.
Wrong again!
Focusing produces magnification.
The Magnifying Properties of a Magnifying Mirror
“Because of the focusing of the light caused by the parabolic shape of magnifying mirrors, the light is concentrated more and more as it approaches the focal point. This concentration of the beams of light results in a magnification of the image reflected by the mirror. As the beams of light get closer and closer to the focal point, they become more and more magnified, which accounts for the increase in magnification as you step away from the mirror.”
http:
//www
.ehow.com/how-does_5016150_magnifying-mirrors-work.html
Gord.
(I missed a breaking up a link.
2nd try.)
From the forum:
First Results from THE BOX: Investigating the Effects of Infrared Sky Radiation on Air Temperature
Roy W. Spencer, Ph. D. says:
July 29, 2010 at 2:38 PM
you cannot focus an extended source of thermal radiation.
—————
Wrong!
Remotely operated infrared radiometer for the measurement of atmospheric water vapor
David A Naylor?, Brad G. Gom, Ian S. Schofield, Gregory J. Tompkins, Ian M. Chapman
Department of Physics, University of Lethbridge, Lethbridge, Alberta, Canada
3.2 Optical
“The optical components of the radiometer are shown in Figure 2. Light enters the radiometer through a 13?m thick
polypropylene window, which acts as dust/moisture protector yet has negligible absorption at 20?m. The 100mm, f/1,
90º off-axis parabolic mirror brings the beam to a focus, behind a 1mm thick ZnSe optical window, on a 1mm2 Mercury-
Cadmium-Telluride (MCT) photoconductive detector (Kolmar Technologies) designed to have a long wavelength cutoff at ~22 ?m and for operation at 77 K.”
http:
//www
.uleth.ca/phy/naylor/documents/pdf/SPIE_IRMAIII.pdf
This instrument uses a COOLED IR detector at the focal point of a Parabolic Mirror to measure atmospheric IR!
Gord.
RE: Anonymous says:
August 13, 2010 at 3:07 PM
You said…
“The insulation, which is cold, causes the temperature to rise (i.e. “heat up”) because it reduces the rate of heat loss from a body that is generating heat. The Earth absorbs
energy from the Sun (call it a furnace), and if that energy cannot escape, or escapes less fast, then the Earth will rise in temperature (i.e. “heat up”).”
Wrong!
An increase in the temperature of the Earth past what the Sun (the only energy source source) provides energy for, requires CREATION of Energy.
This is simple Conservation of Energy that even Grade School Kids understand.
Just like a blanket cannot CREATE energy and CANNOT HEAT UP a human body.
PROOF:
Radiation emitted by a human body
“The total surface area of an adult is about 2 m^2, and the mid- and far-infrared emissivity of skin and most clothing is near unity, as it is for most nonmetallic surfaces.Skin
temperature is about 33 deg C, but clothing reduces the surface temperature to about 28 deg C when the ambient temperature is 20 deg C. Hence, the net radiative heat loss is about Pnet = 100 W.”
http:
//en.wikipedia.org/wiki/Black_body
Did you get that? The human body’s surface temperature DROPPED from +33 deg C to +28 deg C!
I will repeat it AGAIN, The human body’s surface temperature DROPPED from +33 deg C to +28 deg C!
Blankets can’t increase a human body’s surface temperature of +33 deg C because the body has to supply heat energy to the colder blanket to increase it’s temperature.
The result is a DROP in the human body’s surface temperature down from +33 deg C to +28 deg C!
It is IMPOSSIBLE for the blanket to increase the Body surface temperature of +33 deg C since that would require CREATION OF ENERGY.
That’s why it is IMPOSSIBLE for the colder atmosphere to heat a -18 deg C Earth up to +15 deg C as the fantasy “Greenhouse Effect” claims.
The colder atmosphere CAN ONLY COOL THE EARTH just like a blanket CAN ONLY COOL THE HUMAN BODY.
———————–
If ANY, I repeat ANY, heat energy from your +33 deg C body were reflected back by a mirror and absorbed by your body, your body would increase in energy and it’s temperature would rise above +33 deg C.
This already VIOLATES the Law of Conservation of Energy since ENERGY HAS BEEN CREATED and a perpetual motion machine has been created.
But it does not stop there, it is just the beginning of an upward spiral of energy Creation and temperature increases.
Your now warmer body that exceeds +33 deg C will radiate more energy to the mirror and that would be reflected back to your Body further increasing it’s energy and temperature.
This is a perpetual motion machine in a positive feedback loop.
The cycles of energy increase and temperature increase would continue until your Body burst into flames!
(See 2nd Law of Thermodynamics, Perpetual Motion and The Law of Conservation of Energy)
I suggest that you immediately patent your Radiator/Tin Foil Perpetual Motion Machine, you will be RICH and FAMOUS.
——————
The Earth not only absorbs UV from the hotter SUN, it absorbs heat energy from the ENTIRE EM spectrum produced by the hotter SUN.
And, the Earth will radiate the same amount watts/m^2 it absorbs from the Sun in the IR spectrum.
The Earth’s radiated heat energy can only flow to COLDER objects just like the 2nd Law states and The Law of Conservation of Energy requires.
——————
If heat energy could flow from cold to hot objects, every colder object around you would transfer heat energy to your body and increase your body temperature.
You would simply increase in temperature until you burst into flames.
That’s why Dry Ice, a Glacier or any other colder object does not cause your body temperature to increase.
——————-
So, all YOU are left with, AGAIN, is:
– ZERO measurements to support your TOTALLY FALSE CLAIM:
TheLastMan said:
“The fact that we can measure the Greenhouse Effect empirically, and have been able to do so for more than 50 years seems to have passed him by too.”
– Another irrational babbling post totally devoid of any science or measurements, that ignores all established Laws of Science, all proven facts and all actual measurements.
What a HOOT!
You would be really funny, if you were not so Pathetic.
Gord.