Heh. In response to a ridiculous claim making the rounds (I get comment bombed at WUWT daily with that nonsense) which I debunked here: A misinterpreted claim about a NASA press release, CO2, solar flares, and the thermosphere is making the rounds
Dr. Roy Spencer employs some power visual satire, that has truth in it. He writes:
How Can Home Insulation Keep Your House Warmer, When It Cools Your House?!
<sarc> There is an obvious conspiracy from the HVAC and home repair industry, who for years have been telling us to add more insulation to our homes to keep them warmer in winter.
But we all know, from basic thermodynamics, that since insulation conducts heat from the warm interior to the cold outside, it actually COOLS the house.
Go read his entire essay here. <Sarc> on, Roy!
UPDATE: Even Monckton thinks these ideas promoted by slayers/principia/O’Sullivan are ridiculous:
Reply to John O’Sullivan:
One John O’Sullivan has written me a confused and scientifically illiterate “open letter” in which he describes me as a “greenhouse gas promoter”. I do not promote greenhouse gases.
He says I have “carefully styled [my]self ‘science adviser’ to Margaret Thatcher. Others, not I, have used that term. For four years I advised the Prime Minister on various policy matters, including science.
He says I was wrong to say in 1986 that added CO2 in the air would cause some warming. Since 1986 there has been some warming. Some of it may have been caused by CO2.
He says a paper by me admits the “tell-tale greenhouse-effect ‘hot spot’ in the atmosphere isn’t there”. The “hot spot”, which I named, ought to be there whatever the cause of the warming. The IPCC was wrong to assert that it would only arise from greenhouse warming. Its absence indicates either that there has been no warming (confirming the past two decades’ temperature records) or that tropical surface temperatures are inadequately measured.
He misrepresents Professor Richard Lindzen and Dr. Roy Spencer by a series of crude over-simplifications. If he has concerns about their results, he should address his concerns to them, not to me.
He invites me to “throw out” my “shredded blanket effect” of greenhouse gases that “traps” heat. It is Al Gore, not I, who talks of a “blanket” that “traps” heat. Interaction of greenhouse gases with photons at certain absorption wavelengths induces a quantum resonance in the gas molecules, emitting heat directly. It is more like turning on a tiny radiator than trapping heat with a blanket. Therefore, he is wrong to describe CO2 as a “coolant” with respect to global temperature.
He invites me to explain why Al Gore faked a televised experiment. That is a question for Mr. Gore.
He says I am wrong to assert that blackbodies have albedo. Here, he confuses two distinct methods of radiative transfer at a surface: absorption/emission (in which the Earth is a near-blackbody, displacing incoming radiance to the near-infrared in accordance with Wien’s law), and reflection (by which clouds and ice reflect the Sun’s radiance without displacing its incoming wavelengths).
He implicitly attributes Margaret Thatcher’s 1988 speech to the Royal Society about global warming to me. I had ceased to work with her in 1986.
He says that if I checked my history I should discover that it was not until 1981 that scientists were seriously considering CO2’s impact on climate. However, Joseph Fourier had posited the greenhouse effect some 200 years previously; Tyndale had measured the greenhouse effect of various gases at the Royal Institution in London in 1859; Arrhenius had predicted in 1896 that a doubling of CO2 concentration would cause 4-8 K warming, and had revised this estimate to 1.6 K in 1906; Callender had sounded a strong note of alarm in 1938; and numerous scientists, including Manabe&Wetherald (1976) had attempted to determine climate sensitivity before Hansen’s 1981 paper.
He says, with characteristic snide offensiveness, that I “crassly” attribute the “heat-trapping properties of latent heat to a trace gas that is a perfect energy emitter”. On the contrary: in its absorption bands, CO2 absorbs the energy of a photon and emits heat by quantum resonance.
He says the American Meteorological Society found in 1951 that all the long-wave radiation that might otherwise have been absorbed by CO2 was “already absorbed by water vapor”. It is now known that, though that is largely true for the lower troposphere, it is often false for the upper.
The series of elementary errors he here perpetrates, delivered with an unbecoming, cranky arrogance, indicates the need for considerable elementary education on his part. I refer him to Dr. Spencer’s excellent plain-English account of how we know there is a greenhouse effect.
The Viscount Monckton of Brenchley (April 18, 2013)
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Matt in Houston, no it is not wrong. You are being misled about basic thermodynamic principles. I meant my statement exactly as stated: “ANYTHING that reduces heat loss from a heated object can increase its temperature”. Furthermore, that “thing” is almost always at a lower temperature than the heated object. The examples are literally all around you, even the clothes you wear.
tjfolkerts says (April 25, 2013 at 8:32 am): “Why use two different names for exactly the same process producing exactly the same spectrum of photons?”
=====================================================
a) because it causes confusion and b) because it makes possible to mislead via the trick “the same spectrum of photons”/”warming effect”, although the warming effect is physically impossible.
The notion of “photons” alone is misleading enough, because people fall into the trap and start counting them without any basis in real science. The idea about “all radiation being equal” has no basis in real science either and misleads people as well.
Most of science is merely the appearance of confidence. If you make something appear confident, people will follow it, totally independent of any rationality. In fact they’ll create sophistry, and will fully believe themselves, just to defend the apparition of confidence so that they can feel like they’re part of it or that the confidence is part of them. Then, they’ll get really upset about questions which threaten the foundation of the confidence. As we see. Just look at how people call it ridiculous to say that the Earth is heated on one side only by the Sun, and that this is the average solar input. People with PhD’s literally call this ludicrous. They call it ludicrous that the Sun heats the Earth on one side only, and that you can mathematically quantify this and that it physically makes a big difference to the usual assumptions. No one wants to even discuss it calmly, kindly, or openly, they just obfuscate and change goal posts, and all manner of other things. Mostly they get really upset and call people names. They expose themselves for the most part quite uncomfortable with what the scientific method is actually supposed to be about. But they have the appearance confidence, and the masses of number who follow confidence. It is not possible to have a civilized discussion with people who think it is ludicrous to talk about an Earth heated on one side only with a very high forcing temperature. There is no civility to be had between the idea that the Earth is heated on one side, and the idea that it is not. From there, the only thing that happens is that abuse gets thrown about from those who think it is ludicrous to speak of the Earth being heated on only one side. And then the people who say that the Earth is heated on one side, somehow, magically, become blamed for the abuse that other people created. This is all very fascinating.
I want to thank joeldshore (April 24, 2013 at 8:13 pm), LdB (April 24, 2013 at 8:19 pm) and Gary Hladik (April 24, 2013 at 10:34 pm) for responding to my comment. I’ll reply to each response. But before doing that, I want to make it clear– I am not trying to “slay” the “greenhouse effect dragon”!” For one thing, I’m not sure exactly what the “greenhouse effect dragon” is. If the “greenhouse effect” is the statement that some gases in the Earth’s atmosphere absorb/radiate electromagnetic energy in sub-bands of the IR, then I believe in the greenhouse effect. If the “greenhouse effect” is the statement that greenhouse gases in the Earth’s atmosphere will affect the temperature as a function of location on the Earth’s surface, then, again, I agree. If the “greenhouse effect” is the statement that greenhouse gases in the Earth’s atmosphere will increase the “average temperature” (whatever that is) of the Earth’s surface, then I don’t have sufficient knowledge to arrive at an informed opinion. My original comment was a response to Dr. Spencer’s statement: “Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, …”
Response to joeldshore.
I agree that increasing the amount of CO2 in the vacuum region of a thermos bottle won’t result in improved thermos bottle performance. However, based on the rest of your response I believe you agree with the statement: “adding any amount of CO2 gas to the region between the inner and outer walls of a vacuum thermos bottle will NOT improve thermos bottle performance.” If my belief is valid, then isn’t Dr. Spencer’s immediately above statement incorrect?
Response to LdB.
Who mentioned CLASSIC PHYSICS verses quantum mechanics? The experiment is independent of any particular theory. Injecting a fixed amount of greenhouse gas into the volume between the inner and outer walls of a vacuum thermos bottle will have one and only effect on thermos bottle performance–where thermos bottle performance is defined as the time interval required for the temperature of the coffee to reach the temperature of the heat sink into which the thermos bottle is placed. It will either (a) have no effect (equal time intervals), (b) improve performance (increase the time interval), or (c) degrade performance (decrease the time interval). If either (a) or (c) is true, then Dr. Spencer’s statement is wrong. If (b) then, like all physics’ statements/theories, the experiment doesn’t prove Dr. Spencer’s statement/theory is true under all conditions, but the experiment cannot be used to demonstrate that Dr. Spencer’s statement/theory is wrong.
Next, it seems to me that your introduction of “gain factors” is obfuscating the issue at hand–to wit, will adding CO2 to the vacuum space of a thermos bottle decrease the rate of cooling? And to your comment: “It is that simple and basic and no amount of stupid classic physics with thermos flasks is going to save you … simple question posed is CO2 a quantum gain medium answer yes …. end of story. No possibly way to talk around it unless you want to alternatively explain how a laser works” is both irrelevant and nonsensical. The issue of whether CO2 is a “quantum gain medium” may be relevant to a theoretical understanding of the phenomenon, but has no bearing on the experiment itself. And if the experiment shows that CO2 does NOT reduce the rate of cooling, then I have no need to explain “how a laser works”. The onus, if there is one, would be on you to show why the thermos bottle experiment doesn’t agree with your “quantum gain medium” position.
Response to Gary Hladik.
If I understand what you wrote, you and I agree that adding more CO2 won’t improve thermos bottle performance. Then you mentioned thinking of the top of the atmosphere as “the inner layer of the vacuum thermos bottle”. By “inner layer” did you mean the wall of the chamber that holds the coffee, or the wall of the thermos that is in contact with the heat sink? I agree that with the exception of matter leaving the Earth (e.g., the velocities of many helium gas atoms at room temperature exceed “escape velocity”) radiation is the only mechanism for energy to leave the earth/earth atmosphere system. However, there is a difference between the “inner wall” of the thermos (i.e., the wall of the chamber that holds the coffee) and the top of the atmosphere. In particular, in the case of a vacuum between the walls, radiation escapes to the outer wall only from the inner wall. However, in the case of greenhouse gas in the region between the two walls, in addition to thermal conduction and convection between the inner and outer walls, there is no single “surface” from which radiation to the outer wall takes place. To various degrees, the greenhouse gases in the space between the walls will act like an “inner wall” in that a portion of the radiation from greenhouse gases at all locations between the walls will find its way to the outer wall. As I see it, the issue of an Earth atmosphere greenhouse effect is one of determining the Earth surface temperature for two scenarios: (a) a surface surrounded by a non-greenhouse gas atmosphere such that most if not all radiation that escapes to space originates from the surface, or (b) a surface surrounded by an atmosphere containing greenhouse gases such that radiation escapes to space from both the surface and the greenhouse gases in the atmosphere. As I mentioned at the outset, I have insufficient knowledge to determine these temperatures. I’m open to temperature (a) being both higher and lower than temperature (b).
Finally, if by “inner wall” you mean the thermos wall in contact with the heat sink, then my response is: “let the heat sink be the vacuum of cold space.” The nature of the heat sink will have no effect on whether the CO2 thermos outperforms the vacuum thermos.
Can someone point me to where the actual measured CO2 PPM for the upper troposphere or troposphere are stored?
Reed Coray:
I appreciate your challenging statements made. I write to answer a question you put because it comes between two people and – being neither of them – I may be able to help.
In your post at April 25, 2013 at 9:56 am you quote Roy Spencer having said
Spencer’s statement is correct.
(Any statement from him is likely to be correct, and if it were shown to be wrong then he would admit it.)
But it seems you have been confused about that correct statement by – in this case, also correct – statements by joeldshore.
(Your confusion should not dismay you because confusion is a normal outcome of interaction with joeldshore.)
You say
Your “belief” and Spencer’s comment are both correct. I explain as follows.
The “region between the inner and outer walls of a vacuum thermos bottle” contains a good vacuum. And a good vacuum is a very good insulator. Any gas will not be as effective an insulator as a good vacuum. Therefore, introducing any gas “between the inner and outer walls of a vacuum thermos bottle” reduces its insulation. The gas replaces a good insulator (vacuum) with a less good insulator (gas) and, thus, reduces the performance of the thermos bottle.
The atmosphere is a mixture of gases.
Most of the gases in the air (e.g. nitrogen and oxygen) do not “insulate” the Earth’s surface from flow of radiation from the surface to space. (They are similar to a gas being between the inner and outer walls the thermos bottle).
But greenhouse gases do “insulate” the Earth’s surface from flow of radiation from the surface to space. (They are like a good vacuum being between the inner and outer walls of the thermos bottle).
So, the Earth’s surface is better “insulated” when the gases of the atmosphere include greenhouse gases.
I hope this helps.
Richard
Reed Coray says (April 25, 2013 at 9:56 am): “By “inner layer” did you mean the wall of the chamber that holds the coffee…”
Yes. This wall touches the coffee on its inside surface and on the outside is surrounded by the vacuum.
“However, in the case of greenhouse gas in the region between the two walls, in addition to thermal conduction and convection between the inner and outer walls, there is no single “surface” from which radiation to the outer wall takes place.”
Correct. The earth’s upper atmosphere, however, does have something like a “surface” from which radiation finally escapes to space, i.e. the “effective radiating layer” which Steve Mosher mentioned above (April 24, 2013 at 3:42 pm).
“As I see it, the issue of an Earth atmosphere greenhouse effect is one of determining the Earth surface temperature for two scenarios: (a) a surface surrounded by a non-greenhouse gas atmosphere [snip] or (b) a surface surrounded by an atmosphere containing greenhouse gases”
Dr. Spencer covers that case. I thought it was a pretty good read:
http://www.drroyspencer.com/2009/12/what-if-there-was-no-greenhouse-effect/
Matt in Houston says (April 25, 2013 at 8:38 am): “…I hope Dr. Spencer and Mr. Watts will try to invite Dr. Pierre Latour to come in and write a response article on this matter. His writings on this subject are articulate, principled and clear.”
His writings are handwaving. Might I suggest you contact him yourself and ask him why he has not performed the “Yes, Virginia” experiment to prove Dr. Spencer (and innumerable textbooks) wrong and win a Nobel Prize in the process?
http://www.drroyspencer.com/2010/07/yes-virginia-cooler-objects-can-make-warmer-objects-even-warmer-still/
“BOTH problems have a “star” at a fixed temperature, and then add a “heat shield”. BOTH calculate how much less power is required after adding the shield. The only real difference is that 1023 assumes the shell is “nearby” so the R ≈ r, while 1026 allows the outer shell to be any radius. 1023 is a limiting case of 1026, and both say the total power required is 1/2 as much when the shell is added nearby.” ~Tim
Here, let’s go ahead and see if this is the case, shall we?
“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings?
(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
Solution:
(a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ – T⁴/2, or T₁ = ∜T⁴/2 = T/∜2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.” ~Problem 1023
“A spherical black body of radius r at absolute temperature T is surrounded by a thin spherical and concentric shell of radius R, black on both sides. Show that the factor by which this radiation shield reduces the rate of cooling of the body (consider space between spheres evacuated, with no thermal conduction losses) is given by the following expression: aR²/(R² + br²), and find the numerical coefficients a and b.
Solution:
Let the surrounding temperature be T₀. The rate of energy loss of the black body before being surrounded by the spherical shell is
Q = 4πr²σ(T⁴ – T₀⁴).
The energy loss per unit time by the black body after being surrounded by the shell is
Q’ = 4πr²σ(T⁴ – T₁⁴), where T₁ is the temperature of the shell.
The energy loss per unit time by the shell is
Q” = 4πR²σ(T₁⁴ – T₀⁴).
Since Q” = Q’, we obtain
T₁⁴ = (r²T⁴ + R²T₀⁴)/(R² + r²).
Hence Q’/Q = R²/(R² + r²), i.e., a = 1 and b = 1.” ~Problem 1026
Anthony
Might I suggest yet one more resource page on WUWT? This topic simply brings out a level of misinformation that I for one find aggravating (which I promptly make worse by getting involved in the discussion).
Might it make sense to have a resource page with a small set of seminal articles on the topic? Willis’ Steel Greenhouse, Ira Glickstein’s series, perhaps one or two more. Putting the credible articles all in one place would be an excellent resource for those new to the discussion and would limit (I’d hope) the confusion and argument by circular reasoning that dominates threads like this one (and I think does more harm than good.)
REPLY: Yes, I’ll consider it. – Anthony
Max™ says (April 25, 2013 at 3:54 am): “First off, 1023 isn’t about the cooling rate.”
Ah, I think I see the source of confusion here. Actually problem 1023 is all about the cooling rate. Based on its temp, the sphere radiates (cools) at rate J, but thanks to the shield the system radiates (cools) at rate J1, where J1 = J/2.
In fact problem 1023 is much like problem 1026, which mentions “cooling” explicitly. The only difference is that in 1026 the radius of the shell is variable, but it reduces to problem 1023 when R is very close to r.
I notice you didn’t answer my question, but I’ll be happy to answer yours.
“To see what would happen though, consider the case where the shell is in contact with the sphere. If the gap was filled with the same material as sphere and shell are made of, then the volume of material goes up while the energy supply remains the same as it was originally, right?
Two spheres with the same power input/composition which only differ in size won’t reach the same temperature, will they?”
At equilibrium, the surface of the larger sphere will be cooler than the smaller, because it’s radiating the same power from a larger surface. To calculate temp, use the Stefan-Boltzmann calculator
http://calculator.tutorvista.com/stefan-boltzmann-law-calculator.html
Set emissivity to 1, choose arbitrary constant power input, vary the surface area, and solve for temp.
“So now you remove a section from the larger sphere, such that it is the same size as the other sphere, what happens?”
Surface temp goes up, matches the second identical sphere.
“Now remove the same amount of material from the larger sphere, except for a thin shell at the original surface, such that the inner sphere is the same size as the smaller sphere, what happens?”
Surface temp of the sphere goes up, because according to problem 1026 (your reference), “this radiation shield reduces the rate of cooling of the body”. Same power input, reduced cooling, the sphere’s surface temperature must go up until at equilibrium the shell is radiating as much power as the inner sphere receives.
Congratulation, Max! You’ve used a standard text to derive the Steel Greenhouse! The wizard is slain! Willis would be proud of you! I know I am. Well done.
Gary Hladik says:
“Handwaving” is a rather nice euphemism. “Complete and utter nonsense” is a more accurate summary of what Pierre Latour has to say on the subject.
Joseph Postma says:
No…We don’t deny that it heats the Earth from one side only. What we deny is that you can ignore half the Earth in your average. And, if this concept is too confusing for you, then don’t work with averages at all…Look at the total energy in and out as I explained in this post: http://wattsupwiththat.com/2013/04/24/spencer-slays-with-sarcasm/#comment-1286615
The reason that we are not so calm and kind is we have been round and round on this before (e.g., a couple years ago on Judith Curry’s blog) and I really think at this point, you should know better. You either have one heck of a monstrous mental block or you really do understand but have, for some bizarre reason, decided that you would rather engage in sophistry. Either way, it is not a pretty sight.
Kristian says (April 25, 2013 at 4:06 am): “The system is radiating out power to its surroundings twice as fast as soon as the surrounding shell is removed, Gary. The sphere isn’t.”
When the shell is snatched away (along with its insulating effect), the sphere is the system.
“Q (or J if you like) remains unchanged.”
No, but let’s go on.
“The central sphere has been putting out the same power density flux all along, because that is what its emission temperature dictates, the emission temperature in turn being dictated by the heat supplied to the surface of the sphere from its heat source, and only that.”
No. As Dr. Spencer points out, you need to know both the heating rate and the cooling rate. According to problem 1023, the system was in equilibrium with a cooling rate of J1, which is half of the sphere’s cooling rate J. Snatch away the shell and instantly the cooling rate doubles. Same power input, double cooling rate, the temp must go down. This is basic thermodynamics.
In fact the temp will go down until at equilibrium the sphere’s cooling rate is J1, which was (surprise!) the original (constant) power input.
“joeldshore says:
April 25, 2013 at 4:40 am
gbaikie says:
What is expected by some is that CO2 will cause there to less loss of heat. So generally since it doesn’t cause the day to be warmer, it could cause the nights to become less cold.
The notion that CO2 can’t cause the days to become hotter is nonsense.
There is “cooling” going on even during the daytime in the sense that the Earth is always radiating heat. It is just that the warming due to the sun is larger than the cooling due to the Earth radiating. If you increase GHGs, you can change that balance and cause it to become warmer during the day too. There are reasons why nighttime temperatures are expected to warm more than daytime temperatures under rising GHGs, but both are expected to rise (and have been rising).”
As I said there people who believe this, but there is no evidence which supports it.
The lunar surface during the day is much hotter than Earth is during the day, and the
Moon has no greenhouse effect.
According to greenhouse theory, the greenhouse effect adds 33 C to *average temperature*.
And Earth’s greenhouse effect can’t be making it warmer during the day, as Earth is colder
than the Moon [which has no greenhouse effect] during the day.
During the night Earth is far warmer than the Moon during it’s night. So the only large
effect possible of Earth’s greenhouse effect, or this +33 C which is suppose to added
from the greenhouse effect can only be in keeping the night times warmer.
So there is no evidence of it.
And the theory which believer believe in, does not support the idea that
greenhouse gases increase daytime temperature, yet still many believers
hold this view. [due to being brainwashed].
Again, I just posted the actual text from the problem at 12:10 PM, unaltered, ctrl+F > cooling:
“Consider a black sphere of radius R at temperature T which radiates to distant black surroundings at T = 0K.
(a) Surround the sphere with a nearby heat shield in the form of a black shell whose temperature is determined by radiative equilibrium. What is the temperature of the shell and what is the effect of the shell on the total power radiated to the surroundings?
(b) How is the total power radiated affected by additional heat shields?
(Note that this is a crude model of a star surrounded by a dust cloud.)
Solution:
(a) At radiative equilibrium, J – J₁ = J₁ or J₁ = J/2. Therefore T₁⁴ – T⁴/2, or T₁ = ∜T⁴/2 = T/∜2
(b) The heat shield reduces the total power radiated to half of the initial value. This is because the shield radiates a part of the energy it absorbs back to the black sphere.” ~Problem 1023
Note: I missed your question I’ll check and see what it was real quick… hmmm, not sure which one you meant.
“Surface temp of the sphere goes up, because according to problem 1026 (your reference), “this radiation shield reduces the rate of cooling of the body”. Same power input, reduced cooling, the sphere’s surface temperature must go up until at equilibrium the shell is radiating as much power as the inner sphere receives. ~Gary
Ah, so you like 1023 when it agrees with you, but now it isn’t important?
According to 1023 the shell radiates half as much power as the inner sphere receives.
The problem specifically states that it is a simple model of a star in a dust cloud, hence it is roughly analogous to an internally heated sphere and shell.
Perhaps you can find me a Hubble image of a dust cloud glowing as bright as the surface of the stars embedded within, to support this “the sphere’s surface temperature must go up until the shell is radiating as much power as the inner sphere receives” conjecture?
A C Osborn says:
April 25, 2013 at 10:44 am
Can someone point me to where the actual measured CO2 PPM for the upper troposphere or troposphere are stored?
AIRS has it. But it will not release the measurements for the lower or upper troposphere.
However, the scientists working on the project concluded, from all the measurements, that Carbon Dioxide in the atmosphere was lumpy and not at all well-mixed, to their astonishment as they had been brainwashed by this AGW fictional fisics meme, and so insignificant compared with water vapour in the atmosphere.
They released this conclusion at the same time as releasing cherry picked mid troposphere graphics which didn’t show the lumpiness and insignificance of carbon dioxide which they had found to arrive at their conclusion.
Their elementary sleight of hand worked and a flurry of discussions followed as if the conclusion related to the graphics released, which it didn’t. Although to some extent these showed it not well mixed, it didn’t show how lumpy it was.. Instead showing a spread within a few ppm, lumpy suggest areas with zilch.
Carbon dioxide is a real gas, not an “ideal gas” as claimed for the AGW’s Greenhouse Effect.
It has weight because of mass subject to gravity, it is one and a half times heavier than air, this means it will always sink displacing air unless work is being done to change that. Heavier than air gases sink, lighter than air gases rise. It will not readily rise from the ground into the atmosphere, just like the dust on your desk. If it is heated it will as all gases do, expand becoming less dense and so lighter than air will rise, but on releasing this heat it will again condense, real gases condense, and so becoming again heavier than air will sink back to the surface.
Winds, which are volumes of air on the move, that is, convection currents created by differential heating, can move carbon dioxide from one place to another, but, when the wind stops real carbon dioxide will not defy gravity but will sink to the surface. If large amounts of carbon dioxide get into the the great trade winds it can get carried great distances, over the sea to Hawaii for example, but most winds are local and short lived and will move any carbon dioxide around only at a local level. Our great wind systems do not cross hemispheres, so even if carried by these it will not become “well mixed in the atmosphere”. This is what makes carbon dioxide lumpy together with the fact that some areas just don’t produce much and some use all or most of what they produce locally.
And that’s the carbon dioxide which doesn’t come down as rain, all rain is carbonic acid.
AGWSF doesn’t have rain in its “carbon cycle”, firstly because it has eliminated the Water Cycle to create its “Greenhouse Effect”, and secondly because the AGWSF “carbon dioxide” is not real, it is an “ideal gas with no mass, volume, weight, attraction not subject to gravity”, and so not the real carbon dioxide and real water vapour which are greatly attracted to each other in the atmosphere. Real carbon dioxide is in this fully part of the Water Cycle with the residence time of water in the atmosphere of 8-10 days.
There was some Japanese study a short while back which confirmed the lumpiness of carbon dioxide, I think over China, and I recall something about the graphic being pulled and some ambiguous one put in its place.
To Gary Hladik,
Why not actually try this for yourself, as an experiment? That way you would know what happens. I have attempted a similar experiment. My result was that the inner black body did not heat up when it’s outgoing. Radiation was reflected back apron it’s self.
My experimental result may be challenged as simplistic and flawed in some respects but in my view it is better than arguing from theory. Maybe some can do a better experiment. Until they do it is my view that cooler objects DO NOT WARM hotter objects, even acting as insulators.
tjfolkerts says, April 25, 2013 at 4:49 am:
“You are ignoring the “J1″ = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..”
In what way am I ignoring it? Did you see the J = (J – J1) + J1?
“The NET loss from the planet is J/2
The NET loss from the shell is J/2.
The power required from a heater to maintain this situation is J/2.”
Nope. The heater will still have to provide J. J = J/2 + J/2. It has to maintain both losses. They both originally trace back to the surface of the sphere, after all … as J.
“You are ignoring the “J1″ = J/2 that is clearly shown coming back to the inner sphere from the shell due to the temperature of the shell..”
Ok Tim, so let’s see what this entails …
At first you have the sphere alone in space, powered by an internal heat source providing the surface of the sphere with a constant and evenly distributed power density flux (the heat input) of say 400 W/m^2. This flux brings the surface temperature to a steady 290K and this temperature in turn sets the outgoing flux from the surface of the sphere (the heat loss) to an equal 400 W/m^2. We have energy balance and a steady surface temperature corresponding to the input/output – Stefan-Boltzmann.
Then we surround the sphere concentrically with the shell, a narrow gap of vacuum between the two bodies (R ~ r). What happens?
Well, the internal heat source of the sphere still supplies the surface with its constant flux of 400 W/m^2 which in itself could still only bring the surface temperature to the same 290K as before … and no further. These 290K would in turn by necessity (dictated by the laws of physics) also still produce the same emission flux of 400 W/m^2, output equalling input, as before.
In other words, nothing relating specifically to the sphere and its internal heat source has changed – same input (400 W/m^2), same corresponding emission temperature (290K), same corresponding output (400 W/m^2).
So what has changed with the shell in place? The surface of the sphere now gets 200 W/m^2 in return as ‘back radiation’ from the shell as it’s heated to a temperature of 244K by the outgoing flux from the sphere. That is, it now receives an additional 200 W/m^2 which weren’t there before.
Now, you want to add this flux of 200 W/m^2 to the already supplied flux to the surface of the sphere of 400 W/m^2 from the internal heat source, according to you in the end bringing the new system to a steady state where the shell is at a temperature of 290K and the surface of the sphere is at 345K. At this point the sphere radiates 800 W/m^2 to the shell but gets 400 W/m^2 in return (to a net of 400 W/m^2) and the shell radiates 400 W/m^2 to space, equal to the still constant internal input to the surface of the sphere.
So what’s wrong with this picture?
Raising the temperature of an object or a surface takes absorption of heat. Objects are heated by positive heat transfer. Heat only spontaneously transfers from a warmer to a cooler object.
So how is the 200 W/m^2 flux from the cool shell to the warm surface of the sphere ever going to be an inward positive transfer of heat?
If you want to claim that the 290K sphere warms up to a higher steady-state temperature (by 19%) in the presence of the 244K shell, then you are in effect saying that the shell operates as a second, independent heat source for the sphere.
Why? Because nothing else has changed. The internal heat source still provides but the 400 W/m^2 to the surface of the sphere. This flux will in itself not warm it past the 290K it did before. The sphere also still emits its corresponding 400 W/m^2 flux from its surface based on this temperature.
The 400 W/m^2 from the internal heat source still goes straight out from the surface. It doesn’t in any way linger.
So if the temperature then still rises, then this must somehow be caused by something else. It must be caused by extra absorbed heat from somewhere. A positive transfer of (a gain in) thermal energy.
Well, the only other body in this system is the 244K shell. The only thing that’s different is the additional 200 W/m^2 flux from the cool shell to the warm sphere.
You want to claim that this radiative flux rather somehow slows the cooling rate of the sphere, thus heating it indirectly? Fine, but then you should also explain and show how specifically this is accomplished without the 200 W/m^2 flux of 244K spectrum radiation itself transferring HEAT to it and thereby raising the warmer sphere’s kinetic energy level beyond what it would be otherwise – and hence its temperature? Does it somehow disallow half of the 400 W/m^2 of 290K BB spectrum radiation from leaving the surface of the warm body? No. You’re just adding the 200 to the 400, aren’t you? Effectively as extra transfer of heat to the sphere.
Max™ says (April 25, 2013 at 1:08 pm): “Note: I missed your question I’ll check and see what it was real quick… hmmm, not sure which one you meant.”
Not important. You asked essentially the same question in your own way when you derived the Steel Greenhouse.
“Ah, so you like 1023 when it agrees with you, but now it isn’t important?”
1023 and 1026 are essentially the same problem. In fact, as tjfolkerts agrees (April 25, 2013 at 6:18 am) 1026 reduces to 1023 when R is close to r. Now 1023 is cuter, but I like both 1023 and 1026 equally. 🙂
“The problem specifically states that it is a simple model of a star in a dust cloud, hence it is roughly analogous to an internally heated sphere and shell.”
Right, problem 1026. That’s why they made R larger than r, since you don’t see many dust clouds a kilometer or so from a star’s “surface”. In fact in the case of Zeta Ophiuchi
http://www.spitzer.caltech.edu/images/5517-sig12-014-Massive-Star-Makes-Waves
R would be about half a light year, and even then the dust isn’t a perfectly opaque shell.
“Perhaps you can find me a Hubble image of a dust cloud glowing as bright as the surface of the stars embedded within, to support this “the sphere’s surface temperature must go up until the shell is radiating as much power as the inner sphere receives” conjecture?”
Conjecture? Since when are the First Law of Thermodynamics and the Stefan-Boltzmann Law “conjecture”? You don’t believe your own reference? 1023 says the shell cools at a rate J1, half the rate J of the inner sphere. If the system is in equilibrium (no temp change), J1 must be the same as the power input to the sphere. The sphere cools at rate J = 2*J1, but it receives J1 from its power source plus J1 from the shell (look at the diagram), so it’s in equilibrium, too.
Now take away the shell. The sphere (temporarily) radiates/cools at rate J, but now receives only J1 = J/2 from its power source. It must cool until it radiates/cools at rate J1 (plug your own figures for J and J1 into the Stefan-Boltzmann calculator).
Add the shell back, and the sphere’s temp goes back up because, as problem 1023 says, at equilibrium the shell cools/radiates to the surroundings at half the radiation/cooling rate of the sphere, and from Stefan-Boltzmann a surface radiating at J = 2*J1 is hotter than a surface radiating at J1. Easy peasy.
Myrrh says:
April 25, 2013 at 1:24 pm
Myrrh,
The language used by the AIRS people is quite confusing: calling a variation of a few ppm’s (+/- 2% of the full range) over the seasons as “not well mixed”, while about 20% of all CO2 in the atmosphere is exchanged with CO2 from other reservoirs over the same seasons. That only shows that CO2 in the atmosphere is well mixed and quite rapidely. Any huge change in CO2 at ground level is mixed within days to weeks for the same altitude and latitude band, it takes weeks to months to mix it over a full hemisphere and 1-2 years between the hemispheres.
Anyway, over the full globe, except for the first few hundred meters over land, the measured CO2 levels are within +/- 2% of the full range. That is for over 95% of all air…
richardscourtney says: April 25, 2013 at 10:45 am
Thanks for trying to remove any confusion that might exist. Maybe I can’t see the forest for the trees; but Dr. Spencer did not qualify his statement by caveating that it applies to the Earth and its atmosphere. In fact, he parenthetically included a reference to a house. Taken at face value, Dr. Spencer’s statement:
“Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, …”
imply that everything else being equal, if two objects exist at different temperatures the presence of a greenhouse gas between the objects will, relative to the absence of that greenhouse gas, reduce the rate heat flows from the high-temperature object to the low-temperature wall. If this statement is true, then a CO2 thermos should outperform a vacuum thermos. I haven’t conducted the experiment; and it may turn out that, yes, a CO2 thermos outperforms a vacuum thermos. I don’t believe that is the case–i.e., I believe the vacuum thermos will outperform the CO2 thermos. If the vacuum thermos outperforms the CO2 thermos, then Dr. Spencer’s statement without caveats is not correct. The presence of a greenhouse gas does NOT reduce the rate heat flows from the thermos hotter inner wall to the thermos cooler outer wall.
I’m probably nitpicking here, but from my perspective many people on both sides of the AGW fence arrive at conclusions based on statements made by “experts”. I agree that Dr. Spencer is an expert. But if Dr. Spencer makes a statement that applies only under certain conditions but when making that statement does not identify those conditions, then I’m concerned that people will inappropriately apply that statement to conditions not covered by the caveats. Occasionally the process snowballs out of control leading to incorrect conclusions with societal impact.
Put into yes/no questions, in your opinion
(a) For hot coffee inside the chamber of a vacuum thermos bottle located in a room at a lower temperature than the coffee, is there one or more hot objects and one or more cold objects?
(b) Relative to the outer wall, is the thermos bottle chamber wall a hot object?
(c) Relative to the chamber wall, is the thermos bottle outer wall a cool object?
(d) Will heat flow from the chamber wall to the outside wall?
(e) Will the rate of heat flow from the chamber wall to the outside wall be reduced if CO2 gas is injected into the space (a vacuum prior to CO2 gas injection) between the inside wall and the outside wall?
If the answers to these questions are (a) yes, (b) yes, (c) yes, (d) yes, and (e) no, then do those answers constitute a counterexample to the statement: “Greenhouse gases reduce the rate at which heat flows from higher temperatures to lower temperatures?”
If your answer to the last questions is “No”, then I’d appreciate hearing your rationale.
Thank you for your time.
gbaikie says:
No…They believe it because they understand physics.
Look, you can only get so far by comparing the Earth and the moon. The fact that one has a greenhouse effect and the other doesn’t is only one difference. There are other important differences: The moon has very little atmosphere at all and does not have water. This means that the effective heat capacity or “thermal inertia” is much smaller so temperatures vary more dramatically between day and night. Furthermore, the moon has a much longer day than the Earth…about 30 times longer, so this also favors larger temperature variations. Neither of these differences should affect the average temperature (or, more precisely, the average of T^4), which is determined by radiative balance, but they do affect the range of temperatures.
For heaven’s sake, think things through before making statements about people being brainwashed. You might find that you can actually learn from people who have thought about these things much more than you apparently have.
Myrrh says:
April 25, 2013 at 1:24 pm
It has weight because of mass subject to gravity, it is one and a half times heavier than air, this means it will always sink displacing air unless work is being done to change that. Heavier than air gases sink, lighter than air gases rise. It will not readily rise from the ground into the atmosphere, just like the dust on your desk.
Think one moment further: why should the dust on your desk or the sand of the Sahara travel thousands of km through the air, with a density some 100 times that of air and CO2 at only 1.5 times heavier would sink out? Lookup “Brownian motion” works as good for dust and feathers or sand as for CO2 molecules (or even for far heavier CFC molecules reaching the stratosphere…).
Kristian says (April 25, 2013 at 1:52 pm): “So what’s wrong with this picture?
Raising the temperature of an object or a surface takes absorption of heat. Objects are heated by positive heat transfer. Heat only spontaneously transfers from a warmer to a cooler object.
So how is the 200 W/m^2 flux from the cool shell to the warm surface of the sphere ever going to be an inward positive transfer of heat?”
Aaaaaand the imaginary second law of thermodynamics strikes again!
http://scienceofdoom.com/2010/10/07/amazing-things-we-find-in-textbooks-the-real-second-law-of-thermodynamics/