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)
“How do you know it’s radiating? Measure the power consumption I’d say.
Look, if you want to turn this into a philosophical question”
It is quite a real problem, not just philosophical. You provided a good answer in terms of power consumption. That’s as philosophical as required.
Roy Spencer wrote:
“ALL
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.”
I find many things wrong with this statement.
1. One assumes that the 240 W/sq m “average” over a sphere is derived from 960 W/sq m over a disk – this is the fundamental basis of all the initial mathematical derivations of the greenhouse theory I have seen.
Using Stefan-Boltzmann – 240 W/sq m equates to a maximum temperature of about 255 K.
Using Stefan-Boltzmann – 960 W/sq m equates to a maximum temperature of about 360 K
Which temperature more accurately predicts the potential heating power of the Sun ?
Remember that one “disk shaped” hemisphere of the Earth is always illuminated – it is simply the portion of the Earth that changes form day to night.
2. Your statement appears to imply that you can heat something up beyond the power of the input radiation by reducing output. This is in contradiction of the results of Planck et all and is reflected in the Stefan-boltzmann equation and its application by climate science.
3. I remember that you once described a post I made where I stated my belief that the period of rotation of a planet was significant as :-
“This post is a good example of how some people have polluted the minds of others with pseudo-science. ”
I’m not sure I’ve polluted anyone’s mind – they are free to take it or leave it – as I am free to find little credibility in many of your arguments.
Are you now saying that the period of a planet is now crucial ?
The Moon and Mercury both represent heated surfaces radiating to space. Both have significant temperature swings from high to extremely high down to about 100 K – perhaps lower.
The Moon has been mapped and loses “heat” at about 1 K per hour average from noon to sunset – a time of about about 177 Earth hours to decrease from about 390 K to about 220 K and a further 120 K, maybe more, in the 354 Earth hour long night.
The Earth obviously loses less energy in 12 hours of night than the Moon in about 354 Earth hours or Mercury in its 2111 odd Earth hour long night.
Is not this the same as saying the Earth’s output is decreased as compared to the others ?
Venus is a totally different kettle of fish but I find it amazing that climate science can keep a straight face claiming Venus’ 730 K or 16,101 W/sq m – yes 16.1 KW/sq m – can be caused by the albedo adjusted solar input of 132 W /sq m – lecture 3 on the Greenhouse Effect UCLA.
4. When you say “So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.” I assume you refer to the “constant” input of solar radiation.
Even so the Stefan-Boltzmann determines that the temperature is determined by the radiation input so the question remains – can one increase the radiative input by adding the radiation from a cold atmosphere to the solar radiation to create a higher temperature ?
If so which radiative input – the more realistic 960 W/sq m during the day exclusively or the mythical averaged 240 W/sq m ignoring day and night ?
Remember a hemisphere of the Earth is always in day – under the 960 W/sq m (not my figures by the way but simply the albedo reduced solar constant over a disk as used in calculating the “effective radiating temperature of Earth) – it is simply rotating which part is under the Sun at its zenith.
I find “averaging” the solar input as meaningless for any real purpose – the Sun is either heating the surfaces of a part of the Earth during the day or not during the night.
Adding support to this is the real observation that my solar panels only produce during the day and they require an input of approximately 1000 W/sq metre to produce anything like their rated capacity.
At 240 W/sq m illumination at 15.6% efficiency (quoted for 1000 W/sq m – significantly less for lower power but why argue) and 1.2 sq m panels my 14 panel array would produce about 630 W per hour.
I know they produce a “feed in” of up to 7 KW hrs a day and supply up to 5 Kw hours of reduced energy from the grid.
I don’t have a 19 hour long day at 240 W/sq m solar input – we rarely have more than 14 hours sunlight a day being almost within the tropics.
Where I live the clear sky solar power at noon is calculated approximately to be Cos 27 degrees S x 1367 x 0.7 or about 852 W/sq m.
842 x 1.2sq m per panel x 14 panels results in a maximum of 2.2 Kw per hour and requires 6 to 8 hours of varying levels of solar radiation due to the angle of the Sun.
This result seems much more realistic to me.
Anyway – that’s my 2 cents worth.
By the way, if you shine your flashlight at the sun you won’t reach it. You need to lead. Slightly to the right. But your beam WILL hit the sun. And be just as effective as the rest of the light bouncing off the planet at increasing the sun’s overall temperature… unless you have a mighty bright flashlight.
Joe Postma says: Well this isn’t really relevant to the planet Earth …
Well, the discussion is specifically relevant to the situation described above (introduced April 24, 2013 at 2:39 pm and refine April 24, 2013 at 4:58 pm). Quite clearly this is not the earth. But to the extent that correct physics is relevant everywhere, it is relevant to earth.
” … the input is not 240 W/m^2 but 610 W/m^2, as an integrated average over the projection factor on the hemisphere. “
No, the input from the sun (accounting for albedo) is ~ 0.7 * 1370 W/m^2 = ~ 960 W/m^2
Averaged over a hemisphere, that is 480 W/m^2
Averaged over a whole sphere that is 240 W/m^2
Your 610 W/m^2 is averaged over a CYLINDER, not a hemisphere (or equivalently, it only applies at the equator, but no where else). You are too big by a factor of 4/π for a hemisphere.
“The error originates when we assume that sunshine is -18C instead of +49C.
Sunshine has (approximately) the spectrum of a ~ 5800 K blackbody. As such, the only truly meaningful “temperature” of the photons is 5800 K. Assuming either -18C or 49C is wrong.
* Focus sunlight onto an object from all sides and that object will be 5800 K (but sunlight can never be focused to warm the object above 5800 K because that would violate the 2nd Law of Thermodynamics).
* If that sunlight comes from only a small part of the sky (ie a disk ~ 0.5 degrees across as seen at 1 AU, with the rest of the sky @ur momisugly 2.7 K) ) and is spread over a hemisphere with albedo 0.7, the object will be ~ 88 C
* if is is spread out over a cylinder, the object will be ~ 49 C
* if is is spread out over a sphere, the the object will be ~ -18 C
Sunlight is 5800 K. Geometry and albedo combine to limit its warming ability. (But if you make the rest of the sky warmer than 2.7 K, the combined effect of this sunlight and “atmospheric light” will combine to make the new temperature -18 C < T < 5800 K, ie somewhere between the temperature of the cool atmosphere and the warm sun).
“Again, we have seen the text book examples which show that the interior sphere does not become heated.”
Here is one “textbook example” (Sorry I don;t have the reference for the original).
http://i341.photobucket.com/albums/o396/maxarutaru/bunnytaru/lol%20censorship/heatsourceinshellatequilibrium_zps1c9d662a.png
This will get a little subtle, so hopefully people will not get confused — the situation in this textbook and the situation Greg introduced are SIMILAR, but not the same.
The “black sphere” (eg a star) in this example DOES stay the same temperature BUT that was part of the assumptions for this example. What DOES change is the power input. This star maintains the same temperature but with HALF the power when the shell is added.
That means that if the power was “turned back up” to the original value (corresponding to Greg;’s example), the interior would necessarily warm up.
Sorry, but this particular textbook supports me and not you. Do you have a “textbook example” you can cite that supports your claim?
Joseph Postma says:
Cut the sophistry, Joe. The Earth absorbs an amount of energy each second equal to (pi*R^2)*L_solar*(1-albedo)*(1 second) where the albedo is about 0.3, the solar constant L_solar = 1361 W/m^2, and the radius of the Earth is R = 6371 km. That makes for 1.21 x 10^17 Joules.
If the Earth were at an average temperature of 49 C, its emission of energy each second would be [at least because of Holder’s Inequality] (4*pi*R^2)*sigma*T^4*(1 second) where T = 322 K and the Stefan-Boltzmann constant sigma = 5.67 x 10^-8 W/[m^2*K^4]. That makes for 3.11 x 10^17 Joules. [If instead we use the correct average temperature of 255 K (-18 deg C), we get an emission of 1.22 x 10^17 Joules, which within the accuracy of the calculation, gives energy balance.]
In other words, you are spewing nonsense in an apparent attempt to confuse people who don’t know any better.
“That makes for 1.21 x 10^17 Joules.”
And these Joules are absorbed over a hemisphere with the intensity distribution going as the cosine of the solar zenith angle. This is reality.
“If the Earth were at an average temperature of 49 C”
I did not ever say that the Earth was at +49C. I said this is the equivalent temperature value of the average input. This too is reality. This is very different from the usual assumption of a -18C or 240 W/m2 input. It makes a big difference in what the input can and can not do. The total energy is conserved, the input is equal to the output in terms of total energy, but not in terms of power, not in terms of what the input and output power flux can do. The input can do a lot more work than the output. The input can generate very high temperatures on the surface. The atmosphere is then heated by the surface conductively/convectively and radiatively. That’s all that needs to happen because the input is so hot. Hot heats cold. You can go outside and feel the hot input on a sunny day. Latent heat carries a lot of this input around the planet via water vapor and liquid water.
Here’s the original, Tim: that up there is just a snapshot I took to embed.
There was nothing about the power input being changed in that example, go to problem 1026 to see another examination of this problem.
As for the integration over a sphere, you’re treating his 610 value as a projection, it is integrated from the peak value at the subsolar point (1366~ before albedo) to the value at the edges (0~) of the terminator.
After albedo it works out to around 476 W/m^2, or as I prefer to express it, 1.22×10^17 Joules absorbed by one hemisphere, and ~5.11×10^16 Joules emitted by each hemisphere.
@ur momisugly Max April 24, 2013 at 6:21 pm
Your “textbook example” is similar to the one I just discussed. It says that to maintain a constant temperature, an object needs less power by a factor of
R^2 / (R^2 + r^2)
If the radius of the shell (R) is about the same as the radius of the sphere (r), this becomes 1/2. So only 1/2 as much power is needed to keep the inner sphere warm with the shell in place as was required without the shell. This example has the advantage of also showing what would happen is the shell was significantly larger that the sphere inside.
And as I just said, if the power was “turned back up” to the original value (corresponding to Greg’s example), the interior would necessarily warm up above its original temperature.
Max, fair enough, I took the “TM” as a slight bit of “argument from authority”, my apologies, I did not realize it was a simple contraction from a common name.
And my generic observation about using the right units was not “aimed” at anyone specific, just a “look out that you don’t step in that same mud puddle I just soiled my pants with” kind of observation. I have been tripped up many times when I mismatched the correct units for a calculation. One of the most valuable lessons back in college was to “reconcile your units” always, always, always…..
So, in summary, I believe that the proponents of the “GHE” have some valid points; for one, a photon leaves its source without regard to where it’s going. And yes a colder object (or gas cloud) can affect the temperature of a warmer object (ONLY by slowing the velocity of energy flowing away from the warmer object, that’s what the fiberglass in the walls of your house accomplishes)).
But the “deniers” of the GHE also have some valid points; the proponents of the GHE hypothesis have not demonstrated that the “effect” actually reduces the velocity of heat/energy (they alternate between IR EMR and molecular rate of vibration (i.e. temperature)) through the system. In fact I maintain that the “GHE” only causes heat/energy to make multiple passes through the system at the speed of light (there are several flavors of the “speed of light”, the fastest is in a vacuum, and the slowest is in the “flint” glasses) which is known to be MUCH faster than the speed of heat through any gases (GHG or otherwise) or the Oceans or Icesheets or soil in the complex Earth/Atmosphere system.
The much ballyhooed “GHE” only delays the transit of heat/IR by a few tens of milliseconds. That is why you can measure it with an IR sensor with no time reference (Dr. Spencer’s “look I can measure it” posting) and not realize that it is simply “passing through” at the speed of light and causes no ”Higher Equilibrium Temperature”.
Cheers, Kevin.
“Do you have a textbook example?”
Here are some textbooks examples of how a continuous source of heat as a sphere trapped inside a shell doesn’t cause an increase of temperature of there sphere (with continuous input):
http://books.google.co.in/books?id=dQGC0ifkE34C&pg=PA24&lpg=PA24&dq=concentric+sphere+black+body&source=bl&ots=Zh6N1e35jc&sig=m-7nVWch4_zv-l3ISR5k7bluSUQ&hl=en&sa=X&ei=_ldYUd7EFZLU9ATVzYHoDw&redir_esc=y#v=onepage&q&f=false
Also problem *1023* above it. Oh, I see that’s the one you referred to. The problem says nothing about halving the input to the sphere, nor does the other one. You’re making that up. The problem specifically refers to a continuous input which means the sphere is heated to a constant temperature. You could think of it as an infinite heat sink. There is nothing in the two examples about the input being reduced – the input is constant.
“Averaged over a whole sphere that is 240 W/m^2”
You can not average the input over a sphere because the input does not occur over a sphere. The input does not occur over a sphere, it only ever occurs over a hemisphere. The 610 W/m^2 average input is the integrated average intensity over the hemisphere. The energy comes in over a hemisphere, and so this is the correct and only meaningful way to average it, as an integrated average over that geometry.
“the only truly meaningful “temperature” of the photons is 5800 K. Assuming either -18C or 49C is wrong.”
Tell that to the GHE crowd. Of course, we are speaking about the heating potential of sunlight at the distance of the Earth and given its input distribution on the Earth. Which results in the integrated average as discussed. Of course the raw heating potential from sunlight for a flat blackbody is +121C.
The only way to get higher temperatures than this from sunlight at the Earth is to re-condense it using a magnifying glass. GHG’s do not re-condense incoming sunlight, hence they can not create 5000K out of sunlight. GHG’s only interact with the outgoing IR, and as you have pointed out they can not create a higher temperature than their source – the surface of the Earth.
Of course, the actual input is +49C as a temperature and this is higher than the surface temperature. Total energy balances out but the input itself is much stronger than the output – the input can heat the Earth to high temperature, melt ice etc etc.
Well that’s awfully convenient. Greenhouse
Theoryhypothesis cannot be experimentally reproduced so we’ll just have to take the word of climate kooks instead. Experiments replaced by models and slideshows. Yep. The natural conclusion of the pop-science movement ever since Sagan, Ehrlich, Hansen and Mann lowered the bar to unimaginable levels.P.S. some C02 related experiments can be reproduced Steven …
Roy I am so glad that scientists on the AGW side of climate science is taking on the pseudoscience junkies and there is no greenhouse effect garbage it had been worrying me for sometime.
The fact all these pseudoscience types ignore is the ability to challenge it is way out the league of climate science stupidity because it meshed in Planck’s law which forms part of the foundations of Quantum Mechanics.
Planck already worked out classic physics of Boltzmann, Wein and Kirchoff were simplifications based on classic physics. Using quantum mechanics he showed you could derived the classic physics law it is an important piece of science history like E=MC2.
A reasonable short version link on the subject:
http://csep10.phys.utk.edu/astr162/lect/light/radiation.html
Wikipedia carries the full version of derivations of the classic laws from Planck’s laws
http://en.wikipedia.org/wiki/Planck%27s_law_of_black-body_radiation
The dragon slayers like all pseudoscience lunatics miss the fact we know QM is right we have tested it extensively and continue to test it everyday of every year and to date it has never been shown wrong.
I have stated this before we also know that the greenhouse gases are quantum active because all of the gases involved can also be used as the medium in a standard laser tube (CO2, N2O, Water Vapour). If you want absolute proof that you can optically pump the gases involved it doesn’t get more basic than a laser because there are only 3 requirements for a laser tube
A laser is constructed from three principal parts: (http://en.wikipedia.org/wiki/Laser_construction)
An energy source (usually referred to as the pump or pump source),
A gain medium or laser medium, and
Two or more mirrors that form an optical resonator.
So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.
All of this is way beyond doubt and argument especially when your argument omits the fact that your alternative explaination is falsified by Quantum mechanics.
Whoops, I inadvertently typed the “D” word, apparently my latest comment is “trapped”…………….. ha ha ha
I still don’t see much “slaying” going on here, Dr. Spencer’s “The Sun/Earth system is just like your House/Furnace” analogy is (for lack of a better match) still a severely “urine deprived” analogy.
Cheers, Kevin
tjfolkerts says (April 24, 2013 at 7:10 pm): ‘And as I just said, if the power was “turned back up” to the original value (corresponding to Greg’s example), the interior would necessarily warm up above its original temperature.’
Aaaaaand we’re back to the steel greenhouse:
http://wattsupwiththat.com/2009/11/17/the-steel-greenhouse/
Roy Spencer says:
April 24, 2013 at 1:37 pm
ALL:
It is important to remember that there is NO WAY to determine the temperature of anything based upon the rate of energy input alone, for example the Earth absorbing an average of ~240 W/m2 from the Sun. Temperature is a function of BOTH energy input (typically not temperature dependent) AND energy loss (typically VERY temperature dependent), neglecting issues related to heat capacity which mainly affect the time required for the system to equilibrate. The temperature of anything heated will increase until the rate of energy *loss* equals the rate of energy *gain*. So, temperature can be increased by increasing INPUT, or decreasing OUTPUT.
———
Kirchhoffs law says that emission = absorption. Absorption at the TOA equals apx 341 W/M ^2 which if we plug into the S-B equation gives us 5˚ C for any albedo value given Kirchhoff’s law.
It just so happens that the average ocean temperature by mass is apx equal to 5˚C and the atmospheric temperature by mass is apx equal to 5˚C. Which clearly demonstrates the accuracy of Kirchhoff’s law and the S-B equation. 341 W/M^2 in = 341 W/M’2 out, which essentially never changes! Which means that the average temperature never changes either!
But we aren’t really talking about average temperatures here are we? We are talking about transient boundary conditions where the surface of the oceans temp varies from 0 to 30˚ and the temperature of the air immediately over the ocean has the same temperature.
But we aren’t really talking about that either are we? In fact the whole AGW meme is over the response time, the rate at which temperature lags radiation inputs.. And that is the question isn’t it? Because the lag time can range from a few seconds for a photon at 15 microns to travel out of the atmosphere impeded by CO2 ( and yes warming the atmosphere temporarily) to millenia if the photon is headed into the dark recesses of the ocean.
While we don’t know the exact response times, we do have clues don’t we? We know that CO2 levels have been much higher in the past and there were no hockey sticks, so we can falsify Hansen’s AGW theory out of hand. That doesn’t mean the GHG effect isn’t real, it is, we are just a little uncertain about its limits.
Gary Haldik wrote;
KevinK says (April 24, 2013 at 5:58 pm): “Different thought experiment, different expected outcome.”
“So if it’s radioactively heated, you wouldn’t expect plate 1 to get warmer when plate 2 is introduced?”
Frankly I have not thought about that one yet, few of us folks out in the real world have access to radioactive heat sources, although I do scan ebay once in a while to see if I can get rid of this d—n utility company that wants my money in exchange for their energy.
But if it is radioactively heated (an effective “unlimited” heat source, anybody else remember back in the 50s and 60s when electricity was going to be “too cheap to meter” ?) I would expect that the “cycle time” concerns would go away. And then plate 1 and plate 2 (and all the unicorns in the neighborhood) would all be warmer and cozier. So as soon as the Sun stays UP all day the GHE should show up. I could of course model that for say about $2 million US (oh wait I forgot about all the travel to warm exotic places to WARN everybody about the terrible dangers presented by warm cozy unicorns) Better make that $1 billion.
Cheers, Kevin.
Dr. Spencer, whom I greatly admire, wrote:
Roy Spencer says: April 24, 2013 at 12:12 pm
Greenhouse gases (thermodynamically like insulation in your house) reduce the rate at which heat flows from higher temperatures to lower temperatures, thus making the warm side warmer, and the cool side cooler.”
I disagree with the above statement. I am going to apply what you wrote to a vacuum thermos bottle. Place coffee in a thermos bottle and place the thermos bottle in a heat sink at a fixed temperature lower than the temperature of the coffee. The thermos bottle has a “warm side” (the wall of the chamber that holds the coffee–i.e., the inside wall) and a “cool side” (the wall in contact with the heat sink–i.e., the outside wall). In a vacuum thermos bottle, the volume between these two “walls” is empty–i.e., is a vacuum. According to what you wrote, relative to a vacuum thermos bottle inserting a greenhouse gas into the vacuum space will “reduce the rate at which heat flows from the higher temperature to the lower temperature.” If true, the performance of the thermos bottle (i.e., the time required for the temperature of the coffee to reach the temperature of the heat sink) is enhanced (lengthened) by the presence of the GHG. I don’t believe it. If this were true, then why does anyone manufacture vacuum thermos bottles? Wouldn’t CO2 (a greenhouse gas) thermos bottles be preferred?
Although you didn’t explicitly say so, if what you say is true, shouldn’t increasing the amount of GHG placed in this space result in improved thermos bottle efficiency? Do you really believe that relative to a vacuum thermos bottle, a thermos bottle with high pressure CO2 gas (or for that matter, CO2 gas at any pressure) in the region between the warmer (inside) wall and the cooler (outside) wall will reduce the rate thermal energy leaves the coffee, and therefore reduce the rate at which the temperature of the coffee declines?
Genghis says:
(1) No Kirchhoff’s Law says emissivity = absorptivity, not emission = absorption.
(2) Kirchhoff”s Law is true only at a fixed wavelength. An object can have a completely different absorptivity for radiation from the sun, which is primarily at visible and near-IR and UV wavelengths, as it has emissivity for radiation at terrestrial temperatures, which is primarily at far-IR wavelengths.
(3) The 341 W/m^2 value is the amount that is incident at the TOA, not the amount that is absorbed.
And, no, we are not talking about “lag times”. We are talking about the ***rate*** at which energy is emitted and absorbed.
It is really painful to read all of these misconceptions. It is fine to not understand things, but please don’t trumpet your misunderstandings as wisdom.
“explaination is falsified by Quantum mechanics”
A laser is not a demonstration of the GHE. QM in fact shows that radiation trapped inside a cavity with a continuous source simply creates a blackbody spectrum. This is the very origin of QM, of Planck’s Law. Trapped radiation does not change its own frequency distribution and hence its temperature.
“So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.” ~LdB
So… your argument is that because you can excite CO2 molecules and then stimulate them to emit photons… therefore greenhouse effect?
>.>
LdB;
The dragon slayers like all pseudoscience lunatics miss the fact we know QM is right we have tested it extensively and continue to test it everyday of every year and to date it has never been shown wrong.
>>>>>>>>>>>>>>>>
Well said. Every day engineers design and build everything from mundane kitchen appliances to nuclear reactors using the exact laws of physics applied in exactly the same way that Spencer and tjfolkerts and Joel Shore and others are explaining. The things they design work. When they don’t, it is always because there was a design error. There has never been a case otherwise. P*stma and company would have us believe otherwise.
Reed Coray says:
No. Different effects dominate in different circumstances. Heat transfer is a process that can involve conduction, convection, and radiation. For the Earth, the only significant communication of energy out into space is via radiation. For a thermos bottle, conduction and convection of heat are more important (and CO2 probably doesn’t have much effect on radiation because the effect on reducing radiation from the Earth is a subtle interplay between the absorption and re-emission of radiation by greenhouse gases and the temperature structure of the atmosphere which controls how much they emit). Hence, in a thermos, conduction and convection would be increased by putting CO2 in and radiation would hardly be affected at all.
LdB;
OH GOD LORD, not the atmosphere acts like a laser argument AGAIN…..
So dragonslayers there is your absolute proof that the substances must be a gain medium because they can be used as a laser medium.
Absolute proof that the substances must be a gain medium………
Where to start;
1) The “gain medium” in a laser DOES NOT, EVER, NEVER, NO WAY, NO HOW provide energy gain, PERIOD, END OF SENTENCE… (moderator please add lots of periods if you have nothing else to do)
2) EVERY laser known to mankind CONSUMES FAR MORE ENERGY INPUT than it produces in coherent EMR (light) output, typically with a ratio of about 5%.
3) For example, a typical semiconductor laser diode produces say 5 milliwatts of coherent optical power and consumes something like 50 milliamps times 2.5 volts or 125 milliwatts of electrical power, for an efficiency of 5/125 = 4%. Go to DigiKey™ or RadioShack™ and check the specs for an “off the shelf” laser diode, these are typical values.
SHOW US THE OPTICAL RESONANT CAVITY IN THE ATMOSPHERE……… (again lots more periods needed here)
It turns out that making optical resonant cavities is in fact VERY DIFFICULT (hence there are very few “garage shops” that make lasers).
A gas CAN be a gain medium (power gain, NOT energy gain) when properly confined and excited, but this is HARD to do.
Cheers, Kevin.
@Reed Coray
You can’t use classic physics to try and understand any of this CLASSIC PHYSICS is wrong or at best a gross simplification you are dealing with a quantum effect.
The anti-greenhouse position is basically saying a non gain medium does not exist it was an argument that the advent of the laser disproved. There are mediums that have a gain factor because of quantum mechanics.
You can’t talk your way around this with classic physics garbage its a simple test can a medium have a gain factor QM said yes it can and the laser was invented.
The fact CO2 laser exist tells you that CO2 has a medium gain there is no other option or the CO2 laser wouldn’t work.
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.
If there was a gas layer where the major energy inputs were ozone absorbing shortwave (solar radiation, especially UV) and convective inputs from below, and the energy outputs were primarily CO2 molecules radiating long wave to outer space, then extra CO2 could well have a cooling effect, because more CO2 molecules would cause more radiation. This effect would probably be in the nature of a negative feedback rather than an overall cooling.
“The fact CO2 laser exist tells you that CO2 has a medium gain there is no other option or the CO2 laser wouldn’t work.” ~LdB
Ok, are you really equating the process of exciting CO2 molecules and stimulating emission of photons… to the description of the greenhouse effect?