The R. W. Wood Experiment

Guest Post by Willis Eschenbach

Pushed by a commenter on another thread, I thought I’d discuss the R. W. Wood experiment, done in 1909. Many people hold that this experiment shows that CO2 absorption and/or back-radiation doesn’t exist, or at least that the poorly named “greenhouse effect” is trivially small. I say it doesn’t show anything at all. Let me show you the manifold problems with the experiment.

To start with, let me give a curious example of the greenhouse effect, that of the Steel Greenhouse. Imagine a planet in the vacuum of space. A residue of nuclear material reacting in the core warms it to where it is radiating at say 235 watts per square metre (W/m2). Figure 1 shows the situation.

steel greenhouse 1Figure 1. Planet in outer space, heated from the interior. Drawing show equilibrium situation

This planet is at equilibrium. The natural reactor in the core of the planet is generating power that at the planet’s surface amounts to 235 W/m2. It is radiating the same amount, so it is neither warming nor cooling.

Now, imagine that without changing anything else, we put a steel shell around the planet. Figure 2 shows that situation, with one side of the shell temporarily removed so we can look inside.

steel greenhouse 2Figure 2. As in Figure 1, but with a solid steel shell surrounding the planet. Near side of the shell temporarily removed to view interior. Vertical distance of the shell from the surface is greatly exaggerated for clarity—in reality the shell and the shell have nearly the same surface area. (A shell 6 miles (10 km) above the Earth has an exterior area only 0.3% larger than the Earth’s surface area.)

[UPDATE: Misunderstandings revealed in the comments demonstrated that I  lacked clarity. To expand, let me note that because the difference in exterior surface area of the shell and the surface is only 0.3%,  I am making the simplifying assumption that they are equal. This clarifies the situation greatly. Yes, it introduces a whopping error of 0.3% in the calculations, which people have jumped all over in the comments as if it meant something … really, folks, 0.3%? If you like, you can do the calculations in total watts, which comes to the same answer. I am also making the simplifying assumption that both the planet and shell are “blackbodies”, meaning they absorb all of the infrared that hits them.]

Now, note what happens when we add a shell around the planet. The shell warms up and it begins to radiate as well … but it radiates the same amount inwards and outwards. The inwards radiation warms the surface of the planet, until it is radiating at 470 W/m2. At that point the system is back in equilibrium. The planet is receiving 235 W/m2 from the interior, plus 235 W/m2 from the shell, and it is radiating the total amount, 470 W/m2. The shell is receiving 470 W/m2 from the planet, and it is radiating the same amount, half inwards back to the planet and half outwards to outer space. Note also that despite the fact that the planetary surface ends up much warmer (radiating 470 W/m2), energy is conserved. The same 235 W/m2 of power is emitted to space as in Figure 1.

And that is all that there is to the poorly named greenhouse effect. It does not require CO2 or an atmosphere, it can be built out of steel. It depends entirely on the fact that a shell has two sides and a solid body only has one side.

Now, this magical system works because there is a vacuum between the planet and the shell. As a result, the planet and the shell can take up very different temperatures. If they could not do so, if for example the shell were held up by huge thick pillars that efficiently conducted the heat from the surface to the shell, then the two would always be at the same temperature, and that temperature would be such that the system radiated at 235 W/m2. There would be no differential heating of the surface, and there would be no greenhouse effect.

Another way to lower the efficiency of the system is to introduce an atmosphere. Each watt of power lost by atmospheric convection of heat from the surface to the shell reduces the radiation temperature of the surface by the same amount. If the atmosphere can conduct the surface temperature effectively enough to the shell, the surface ends up only slightly warmer than the shell.

Let me summarize. In order for the greenhouse effect to function, the shell has to be thermally isolated from the surface so that the temperatures of the two can differ substantially. If the atmosphere or other means efficiently transfers surface heat to the shell there will be very little difference in temperature between the two.

Now, remember that I started out to discuss the R. W. Wood experiment. Here is the report of that experiment, from the author. I have highlighted the experimental setup.

Note on the Theory of the Greenhouse

By Professor R. W. Wood (Communicated by the Author)

THERE appears to be a widespread belief that the comparatively high temperature produced within a closed space covered with glass, and exposed to solar radiation, results from a transformation of wave-length, that is, that the heat waves from the sun, which are able to penetrate the glass, fall upon the walls of the enclosure and raise its temperature: the heat energy is re-emitted by the walls in the form of much longer waves, which are unable to penetrate the glass, the greenhouse acting as a radiation trap.

I have always felt some doubt as to whether this action played any very large part in the elevation of temperature. It appeared much more probable that the part played by the glass was the prevention of the escape of the warm air heated by the ground within the enclosure. If we open the doors of a greenhouse on a cold and windy day, the trapping of radiation appears to lose much of its efficacy. As a matter of fact I am of the opinion that a greenhouse made of a glass transparent to waves of every possible length would show a temperature nearly, if not quite, as high as that observed in a glass house. The transparent screen allows the solar radiation to warm the ground, and the ground in turn warms the air, but only the limited amount within the enclosure. In the “open,” the ground is continually brought into contact with cold air by convection currents.

To test the matter I constructed two enclosures of dead black cardboard, one covered with a glass plate, the other with a plate of rock-salt of equal thickness. The bulb of a thermometer was inserted in each enclosure and the whole packed in cotton, with the exception of the transparent plates which were exposed. When exposed to sunlight the temperature rose gradually to 65 oC., the enclosure covered with the salt plate keeping a little ahead of the other, owing to the fact that it transmitted the longer waves from the sun, which were stopped by the glass. In order to eliminate this action the sunlight was first passed through a glass plate.

There was now scarcely a difference of one degree between the temperatures of the two enclosures. The maximum temperature reached was about 55 oC. From what we know about the distribution of energy in the spectrum of the radiation emitted by a body at 55 o, it is clear that the rock-salt plate is capable of transmitting practically all of it, while the glass plate stops it entirely. This shows us that the loss of temperature of the ground by radiation is very small in comparison to the loss by convection, in other words that we gain very little from the circumstance that the radiation is trapped.

Is it therefore necessary to pay attention to trapped radiation in deducing the temperature of a planet as affected by its atmosphere? The solar rays penetrate the atmosphere, warm the ground which in turn warms the atmosphere by contact and by convection currents. The heat received is thus stored up in the atmosphere, remaining there on account of the very low radiating power of a gas. It seems to me very doubtful if the atmosphere is warmed to any great extent by absorbing the radiation from the ground, even under the most favourable conditions.

I do not pretend to have gone very deeply into the matter, and publish this note merely to draw attention to the fact that trapped radiation appears to play but a very small part in the actual cases with which we are familiar.

Here would be my interpretation of his experimental setup:

r w wood experiment 2Figure 3. Cross section of the R. W. Wood experiment. The two cardboard boxes are painted black. One is covered with glass, which absorbs and re-emits infrared. The other is covered with rock salt, which is transparent to infrared. They are packed in cotton wool. Thermometers not shown.

Bearing in mind the discussion of the steel greenhouse above, I leave it as an exercise for the interested reader to work out why this is not a valid test of infrared back-radiation on a planetary scale … please consider the presence of the air in the boxes, the efficiency of the convective heat transfer through that air from the box to the cover plates, the vertical temperature profile of that air, the transfer of power from the “surface” to the “shell” through the walls of the box, and the relative temperatures of the air, the box, and the transparent cover.

Seems to me like with a few small changes it could indeed be a valid test, however.

Best regards,

w.

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tjfolkerts
February 8, 2013 5:45 am

Allen,
Suppose a warm metal block and a cool metal block are in contact. Do faster-than-average atoms in the cool block ever hit slower-than-average atoms in the warm block? Do such collisions ever transfer energy from the cool block to the warm block?
Is that ALSO “magical backward energy flow”?

A C Osborn
February 8, 2013 5:49 am

Isn’t it very odd that all the Practical Experiments (including those at Universities) show that rather than warming the object DWIR actually cools them in relation to their surrounding. But Dr Spencer with his box sees this as proving that DWIR is heating the surface because the air inside could theoretically get to the same temperature as outer space?
Only a true believer can say this demonstrates heating.

KR
February 8, 2013 6:51 am

richard verney – The “skin layer” measurements referred to in the RealClimate post (http://www.realclimate.org/index.php/archives/2006/09/why-greenhouse-gases-heat-the-ocean/) were taken in situ, on the oceans, not in a “mill pond”.
I’ll refer you to a description of heat loss through the thermal skin layer by Minnett (https://www.ghrsst.org/files/download.php?m=documents&f=120113121306-SSTDefinitionsDiscussion.pdf), primary research papers such as Minnett et al 2011 (http://www.sciencedirect.com/science/article/pii/S0967064510003024), Veron et al 2011 (http://journals.ametsoc.org/doi/abs/10.1175/2010JPO4491.1?journalCode=phoc), and for an overview Science of Doom (http://scienceofdoom.com/2011/01/18/the-cool-skin-of-the-ocean/) also discusses this in some detail.
Winds certainly do have an effect – the gradient between sub-skin temperatures and the skin layer is ~0.6K in low winds, dropping to perhaps ~0.13K in high winds. But the viscous surface water represents a thermal layer that energy must go through to leave the oceans, and energy transfer across that boundary is affected by the thermal gradient, by the difference between sub-skin water and the atmosphere.
A warmer atmosphere, with higher DLWR (even though _cooler_ than the sub-skin oceans), reduces the energy loss of the ocean through that boundary, meaning energy accumulation from incoming/penetrating sunlight, and hence ocean warming.
It may seem counter-intuitive, but those are the observations. If you disagree with the mechanism of outward energy transfer from the oceans, you will need to point to some supporting data.

Gary Hladik
February 8, 2013 7:09 am

Greg House says (February 7, 2013 at 10:01 pm): “It is not about whether colder objects can or can not radiate to warmer objects.
It is about whether radiation of colder objects affects temperature of warmer objects. That is the point, please, note that.”
Hi, Greg. I believe you’ve been referred to this experiment before, but just in case, check this out:
http://scienceofdoom.com/2010/07/31/the-amazing-case-of-back-radiation-part-three/

PJF
February 8, 2013 7:16 am

Greg House wrote:
“It is not about whether colder objects can or can not radiate to warmer objects.
It is about whether radiation of colder objects affects temperature of warmer objects. That is the point, please, note that.”

OK, I note that that is your point. So, if the radiation of colder objects doesn’t affect the temperature of warmer objects, what happens to the energy of that radiation?

Gary Hladik
February 8, 2013 7:24 am

Bart says (February 7, 2013 at 11:03 pm): “And, that is how I see your analogy potentially failing, because it places a vacuum between the planet and the shell which is not present for the Earth and its atmosphere.”
When teaching a new concept, it’s usually best to start with a simplified ideal model and work your way toward more complicated, more realistic cases. Willis’s model is step 1. The vacuum in the model is there to isolate the shell from conductive/convective contact with the surface. Of course the Earth’s upper atmosphere isn’t 100% isolated from the surface, but it’s still cooler than the surface and that’s the key factor illustrated in the model.

Ryan
February 8, 2013 7:28 am

@richardscourtney:”I put things in the apparatus, provide a power supply, and the apparatus heats the things I inserted to temperatures higher than the temperature of the apparatus.”
That’s all very well Richard, but this is an energy conversion process, not a heat energy transfer in a closed system.The argument we are having is whether a cooler object can make a hotter source even hotter, or if this would break the First or Second Law of Thermodynamics.
Personally I would say that if it were possible to make a hot object even hotter by getting it intimate with a cooler object, then making my bath water cooler by adding cold water would be a futile exercise, since the energy inherent in “cold” water (i.e. water way above 0K) would still be enough to heat up the hot water. Or maybe not…..
If you were able to do that, then you could take buckets of water at the same temperature, mix them together and get a bucket of water at double the temperature. Oh if only! You could make a steam turbine just by “concentrating” the heat in water at room temperature! Sadly this would contradict the 2nd Law of Thermodynamics (i.e. the Laws of Entropy).

Gail Combs
February 8, 2013 7:40 am

Mervyn says:
February 8, 2013 at 5:32 am
I have extensively checked out the greenhouse effect supposition (GHE). At the end of the day, however, I side with individuals such as astrophysicist, Joseph E. Postma, in relation to this issue. His work is sufficiently authoritative to dismiss the GHE supposition….
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>
For those interested Dr. Postma posts a counter rebuttal to the “Skeptical Science” blog critique of the recent paper by astrophysicist Joseph E. Postma, ‘Copernicus meets the greenhouse effect,’

Gary Hladik
February 8, 2013 7:41 am

Ryan says (February 8, 2013 at 7:28 am): “The argument we are having is whether a cooler object can make a hotter source even hotter, or if this would break the First or Second Law of Thermodynamics.”
What’s your take on MikeB’s three-body reference above?
http://scienceofdoom.com/2010/11/05/the-three-body-problem/

tjfolkerts
February 8, 2013 7:51 am

Ryan (and many others),
You are making a false analogy when you say things like “Personally I would say that if it were possible to make a hot object even hotter by getting it intimate with a cooler object, then making my bath water cooler by adding cold water would be a futile exercise,”
The analogy is more like this. Fill a bath tub with water, put a small electric heater in the water, and set the whole thing outside on a very cold (for example -30 C) winter day (ie in contact with with the very cold surroundings). Measure the temperature of the water … maybe the heater can only heat the water to 5 C.
Then try it again on a sort of cold day (for example, 0C). Now maybe the heater (with the same power as before) will warm the water to 20 C. Putting the bathtub & water in contact with COLD air (0 C) instead of in contact with VERY COLD air is part of the cause allowing the water to warm from 5 C to 20 C. Even though the air is still COLD, it reduces the heat loss, thereby allowing the heater to be more effective.
THAT is what GHGs do. They let the earth be “in contact” (via thermal radiation) with COLD surroundings (the atmosphere) rather than in contact with VERY COLD surroundings (outer space). This, in conjunction with a continuous input of energy from the sun, makes the earth warmer.

tjfolkerts
February 8, 2013 7:55 am

To be more specific, that last paragraph should read “They let the earth‘s surface be “in contact” (via thermal radiation) … ” This allows the surface level to be warmer than it would be otherwise.

richardscourtney
February 8, 2013 8:14 am

Ryan:
Greg house had claimed there was no “experiment” which shows a cooler object warming a hotter object. I pointed out that I own an apparatus – a microwave oven – which does exactly that by the action of a flow of energy providing electromagnetic radiation to the hotter object.
At February 8, 2013 at 7:28 am you have replied saying in total

@richardscourtney:

”I put things in the apparatus, provide a power supply, and the apparatus heats the things I inserted to temperatures higher than the temperature of the apparatus.”

That’s all very well Richard, but this is an energy conversion process, not a heat energy transfer in a closed system.The argument we are having is whether a cooler object can make a hotter source even hotter, or if this would break the First or Second Law of Thermodynamics.
Personally I would say that if it were possible to make a hot object even hotter by getting it intimate with a cooler object, then making my bath water cooler by adding cold water would be a futile exercise, since the energy inherent in “cold” water (i.e. water way above 0K) would still be enough to heat up the hot water. Or maybe not…..
If you were able to do that, then you could take buckets of water at the same temperature, mix them together and get a bucket of water at double the temperature. Oh if only! You could make a steam turbine just by “concentrating” the heat in water at room temperature! Sadly this would contradict the 2nd Law of Thermodynamics (i.e. the Laws of Entropy).

I agree that I described an “energy conversion process”: the energy of electromagnetic radiation is converted to heat in the contents of the microwave oven and is powered by a source of energy flow (i.e. mains electricity in the microwave oven).
Similarly, the temperature rise of Willis’ hypothetical planet’s surface is an “energy conversion process”: the energy of electromagnetic radiation is converted to heat in the planet’s surface and is powered by a source of energy flow (i.e. the nuclear reactions in the planet)..
I fail to understand what you mean by it is “not a heat energy transfer in a closed system”: in each case electromagnetic energy is emitted from one object and is converted to heat in a hotter object such that the temperature of the hotter object is raised.
And neither case violates the First or Second Law of Thermodynamics.
Your statements about mixing water are irrelevant. They say nothing about how electromagnetic radiation interacts with matter.
Richard

Ryan
February 8, 2013 8:16 am

“THAT is what GHGs do. They let the earth be “in contact” (via thermal radiation) with COLD surroundings (the atmosphere) rather than in contact with VERY COLD surroundings (outer space). This, in conjunction with a continuous input of energy from the sun, makes the earth warmer.”
No, the planet’s surface would be warm even if the planet were directly in contact with space. Furthermore the presence of an atmosphere means you have a temperature gradient developed across that atmosphere which will make the Earth’s surface even warmer as it will slow the rate of cooling – you don’t need GHGs for that. Thirdly, the top of the stratosphere is actually hotter than the Earth’s surface due to UV hitting the outer layer of the atmosphere, so the temperature gradient is actually inverted for cooling and inherently stable. Forthly, if we consider the cooling scenario separately then the Earth’s surface cannot itself get any hotter since it is effectively the source of the heating effect and the hottest part of the system – in your example you are only decreasing the temperature gradient across the system by raising the temperature of the outside of the system (i.e. you would be in effect increasing the temperature of outer space), which is NOT what GHGs do.
Go back to Willis’ shell system, but instead of a thermo-nuclear reaction consider a massive ball of water at a smidgen under 100Celsius. In the first diagram the water is cooling into space quite freely. But what about the second diagram. You are now reflecting a lot of IR back to the water, i.e. it seems to be receiving a lot more energy. Does that mean the water will now boil, or will it just stay a smidgen under 100Celsius?

February 8, 2013 8:22 am

: The argument we are having is whether a cooler object can make a hotter source even hotter,
Cooler than what? Cooler than the hotter source? Or cooler than the 4 deg K cosmic background?
Look at it from the point of view of the hotter object. Without the steel shell, it is staring into a 4 deg K “shell” of space. The hotter object gets wrapped in shell of 200 deg K. There will be a drop in Heat loss from the hotter object, and if the hotter object has an internal sourse of heat, its temperature will rise at equilibrium. “Come out of the cold and into my igloo.”
In Perpetuum Mobile, tjfolkerts (2/14/12 10:13am)

I think we can agree that:
* it is NOT a violation of thermodynamics to run a machine perpetually using the hot sun (5770 K) and cold outer space (3 K) as the two thermal reservoirs.
* it IS a violation of thermodynamics to run a machine perpetually using the warm shell (279 K) and warm surface (279 K) as the two thermal reservoirs.
The “million tiny suns” earth is using the hot sun (albeit inefficiently) and cold outer space (albeit inefficiently) as the two thermal reservoirs.
Rasey:Let’s cut the sun up into 200,000 stars of 6000 deg K. Each will then be 12.6 square arc seconds or less than 4 arc seconds across. Jupiter, viewed from earth is 32 arc seconds or over 64 times bigger in area (though much dimmer!) than one of our 200,000 starry suns. One brilliant 4 arc-second spec for every spot in the sky the size of the sun..

This exchange taught me the danger of taking an average energy flux in W/m2 and backing into a temperature. The tiny million sun example of 6000 deg K specks in a field of 4 deg K void can give the same flux as a uniform 279K field. (Thanks TJ).

February 8, 2013 8:25 am

[mods: please delete my 8:22 am with an unclosed Link ref. and use this one]
: The argument we are having is whether a cooler object can make a hotter source even hotter,
Cooler than what? Cooler than the hotter source? Or cooler than the 4 deg K cosmic background?
Look at it from the point of view of the hotter object. Without the steel shell, it is staring into a 4 deg K “shell” of space. The hotter object gets wrapped in shell of 200 deg K. There will be a drop in Heat loss from the hotter object, and if the hotter object has an internal sourse of heat, its temperature will rise at equilibrium. “Come out of the cold and into my igloo.”
In Perpetuum Mobile, tjfolkerts (2/14/12 10:13am)

I think we can agree that:
* it is NOT a violation of thermodynamics to run a machine perpetually using the hot sun (5770 K) and cold outer space (3 K) as the two thermal reservoirs.
* it IS a violation of thermodynamics to run a machine perpetually using the warm shell (279 K) and warm surface (279 K) as the two thermal reservoirs.
The “million tiny suns” earth is using the hot sun (albeit inefficiently) and cold outer space (albeit inefficiently) as the two thermal reservoirs.
Rasey:Let’s cut the sun up into 200,000 stars of 6000 deg K. Each will then be 12.6 square arc seconds or less than 4 arc seconds across. Jupiter, viewed from earth is 32 arc seconds or over 64 times bigger in area (though much dimmer!) than one of our 200,000 starry suns. One brilliant 4 arc-second spec for every spot in the sky the size of the sun..

This exchange taught me the danger of taking an average energy flux in W/m2 and backing into a temperature — there is more than one solution. The tiny million sun example of 6000 deg K specks in a field of 4 deg K void can give the same flux as a uniform 279K field. (Thanks TJ).

jae
February 8, 2013 8:40 am

I suspect that no page numbers will be forthcoming to support the shell-game, because there is no support for the shell-game and its derivatives. I cannot weld steel with a propane torch. But according to the shell game, all I have to do is use 2 or 3 propane torches, adding their heat together, to do the trick. Sorry, no can do!!

davidmhoffer
February 8, 2013 8:51 am

tjfolkerts says:
February 8, 2013 at 7:51 am
>>>>>>>>>>>>>
An excellent explanation tj. Of course it will be refuted by some (snip) claiming an experiment in 1909 with cardboard boxes proves your explanation impossible followed by yet (snip) another shouting something about the second law being violated, and yet another (snip) complaining that he should be able to heat his house with two slabs of iron at the same temperature…
This thread reminds me of someone who sneers down their nose at the notion that the earth is flat and proceeds to prove that it isn’t by explaining how the sun orbits the earth.

Ryan
February 8, 2013 9:51 am

If you want to improve Willis model to look like the real GHG system you should have a heat source outside the shell only, that turns on and off every 12hours and has a temperature of say 1000Celsius. Imagine the planet as a ball of water. This heat source will heat the ball of water so that when the heat source is on and there is no shell present the temp will reach 100Celsius, but when the heat source is off it will normally cool to space reaching 50Celsius.
Now re-introduce the shell during the periods when the heat source is off only. What will happen next is the ball of water inside the shell will not be able to cool either by conduction or radiation so it’s temp will remain at 100Celsius. Now remove the shell and re-apply the heat. Since the ball of water has now started at 50Celsius the temperature with the heat source on will reach 150Celsius (well, it will boil).
This is the greenhouse theory at its simplest. You notice it requires no detailed knowledge of DWIR or any other details to work – it only requires you to get the model right.
I did not say, however, that this would work on planet Earth.

Gary Hladik
February 8, 2013 10:12 am

jae says (February 8, 2013 at 8:40 am): “I suspect that no page numbers will be forthcoming to support the shell-game, because there is no support for the shell-game and its derivatives.”
Willis quoted chapter and verse (February 7, 2013 at 9:11 pm).
I pointed to a site that quotes several textbooks (February 7, 2013 at 7:32 pm).
Your claim is demonstrably wrong.

February 8, 2013 10:18 am

Bart says:
February 7, 2013 at 11:03 pm
Willis Eschenbach says:
February 7, 2013 at 9:11 pm
“The first term reflects the flow in one direction, and the second term, the flow in the opposite direction.”
In large part, this is an argument of semantics, or of mental constructs which produce a particular perspective. There are not really individual flows of energy which bounce back and forth between the planet’s surface and the inside of the shell. Electromagnetic radiation is a wave phenomenon. A some points, the fields will interfere constructively, and at others destructively. So, there will be peaks of electric field and nulls at various places along the way, as there will of the magnetic fields. At equilibrium, there will be some dc level of electromagnetic energy filling the gap, as well as fluctuations from the non-stationary interference patterns which could appear to move in either direction on a finite time scale.

Not correct there are indeed counter flowing flows of electromagnetic radiation, interference does not prevent the propagation of radiation in either direction, if it did a laser cavity wouldn’t work!

Shawnhet
February 8, 2013 10:21 am

jae says:
February 8, 2013 at 8:40 am
“I suspect that no page numbers will be forthcoming to support the shell-game, because there is no support for the shell-game and its derivatives. I cannot weld steel with a propane torch. But according to the shell game, all I have to do is use 2 or 3 propane torches, adding their heat together, to do the trick. Sorry, no can do!!”
Oh, come on. You ask for specific pages of textbooks to give you an idea of where others are coming from and then once you are given them, you respond by arguing against a straw man position that has nothing to do with what you have been pointed to. I guess you can point someone to dozens of textbooks but you can’t make them understand any of them.
BTW, where does the extra heat come from that warms someone after they put on a parka when it is cold outside. Does that extra heat become any less real if you call it “backradiation”?
Cheers, 🙂

February 8, 2013 10:23 am

Greg House says:
February 7, 2013 at 3:08 pm
Phil. says, February 7, 2013 at 12:08 pm: “Nothing to do with ‘climate science’ it’s the basic physics of radiation heat transfer.
Arises from S-B law, light emitted by hot bb object = const.A.Th^4
light emitted by cool bb object = const.A.Tc^4
Net heat transferred between the two objects = const.A.(Th^4-Tc^4)”
==========================================================
This “net” thing does not “arise” from S-B law, and it has apparently never been proven experimentally. As long as it has not been proven experimentally, it remains a fiction. A sort of “radiation arithmetic” without any basis in science.

Just because you’re not prepared to read about something doesn’t mean it doesn’t exist. Feel free to continue in ignorance but don’t expect others to accept your nonsense. Either read some basic texts on radiation heat transfer (I suggest Hottell) and learn about the subject or shut up.

February 8, 2013 10:36 am

mkelly says:
February 7, 2013 at 11:43 am
q = ε σ (Th4 – Tc4) Ac
A typical radiative heat transfer equation from Engineering Tool Box. Where in this formula can the radiation from the shell cause an increase in the temperture of the globe? Again it seems to me that when Th = Tc then q is zero.

Increase Tc and the heat loss from the globe goes down, since heat input to the globe is constant the temperature will go up (Th) thus q will increase until q equals the input to the globe. If Th=Tc there will be no heat loss from the globe so there will be an increase in the globe temperature until q = input.

February 8, 2013 10:56 am

MiCro says:
February 7, 2013 at 1:27 pm
So, let me ask this, if a 15u wavelength photon is captured and thermalize by a molecule of Co2, what would it’s temperature be?

Its translational temperature will stay the same its rotational and vibrational temperatures will increase depending on which particular energy levels have just been populated. The excited state will now attempt to lose that excess energy by a combination of collisional deactivation and radiation. Which one is favored will depend on conditions, near the earth’s surface collisions are the major route, up near the tropopause it’s radiation.

February 8, 2013 11:04 am

Willis Eschenbach says:
February 7, 2013 at 9:11 pm
q = A C s T1^4 – A C s T2^4 (5.10)
Thanks for proving my point. When T1 and T2 are equal q goes to zero and heat transfer stops. So the shell cannot warm the globe.

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