Guest Post by Reed Coray
On Dec. 6, 2011 12:12 am Lord Monckton posted a comment on a thread entitled Monckton on sensitivity training at Durban that appeared on this blog on Dec. 5, 2011. In that comment he wrote:
“First, it is not difficult to calculate that the Earth’s characteristic-emission temperature is 255 K. That is the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere. Since today’s surface temperature is 288 K, the presence as opposed to absence of all the greenhouse gases causes a warming of 33 K”.
Since I’m not sure what the definition of the “Earth’s characteristic-emission temperature” is, I can’t disagree with his claim that its value is 255 K. However, I can and do disagree with his claim that 255 K is “the temperature that would obtain at the surface in the absence of any greenhouse gases in the atmosphere”.
When computing the Earth’s surface temperature difference in “the presence as opposed to the absence of all greenhouse gases”, (i) two temperatures (A and B) must be measured/estimated and (ii) the difference in those temperatures computed. The first temperature, A, is the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases. The second temperature, B, is the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only–i.e., an atmosphere that contains non-greenhouse gases but is devoid of greenhouse gases.
For temperature A almost everyone uses a “measured average” of temperatures over the surface of the Earth. Although issues may exist regarding the algorithm used to compute a “measured average” Earth surface temperature, for the purposes of this discussion I’ll ignore all such issues and accept the value of 288 K as the value of temperature A (the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases).
Thus, we are left with coming up with a way to measure/estimate temperature B (the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only). We can’t directly measure B because we can’t remove greenhouse gases from the Earth’s atmosphere. This means we must use an algorithm (a model) to estimate B. I believe the algorithm most commonly used to compute the 255 K temperature estimate of B does NOT correspond to a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only”. As will be evident by my description (see below) of the commonly used algorithm, if anything that algorithm is more representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains both greenhouse gases and non-greenhouse gases” than it is representative of a model of “the temperature of the Earth’s surface in the presence of an atmosphere that contains non-greenhouse gases only.”
If I am correct, then the use of 255 K in the computation of the Earth surface temperature difference with and without greenhouse gases is invalid.
Although there are many algorithms that can potentially lead to a 255 K temperature estimate of B, I now present the algorithm that I believe is most commonly used, and discuss why that algorithm does NOT represent “the temperature of the Earth’s surface in the presence of an atmosphere that is devoid of greenhouse gases”. I believe the algorithm described below represents the fundamental equation of radiative transfer for the Earth/Sun system assuming (a) an Earth absorption albedo of 0.3, and (b) an Earth emissivity of 1.
(1) The “effective temperature” of the Sun [i.e., the temperature of a sun-size spherical blackbody for which the radiated electromagnetic power (a) is representative of the total solar radiated power, and (b) has a power spectral density similar to the solar power spectral density] is approximately 5,778 K.
(2) For a spherical blackbody of radius 6.96×10^8 meters (the approximate radius of the sun) at a uniform surface temperature of 5,778 K, (a) the total radiated power is approximately 3.85×10^26 Watts, and (b) the radiated power density at a distance of 1.5×10^11 meters from the center of the blackbody (the approximate distance between the center of the Sun and the center of the Earth) is approximately 1,367 Watts per square meter.
(3) If the center of a sphere of radius 6.44×10^6 meters (the approximate radius of the Earth) is placed at a distance of 1.5×10^11 meters from the center of the Sun, to a good approximation the “effective absorbing area” of that sphere for blackbody radiation from the Sun is 1.3×10^14 square meters; and hence the solar power incident on the effective absorbing area of the sphere of radius 6.44×10^6 meters is approximately 1.78×10^17 Watts (1.3×10^14 square meters x 1,367 Watts per square meter).
(4) If the sphere of radius 6.44×10^6 meters absorbs electromagnetic energy with an “effective absorption albedo” of 0.3, then the solar power absorbed by the sphere is 1.25×10^17 Watts [1.78×10^17 Watts x (1 – 0.3)].
(5) A spherical blackbody (i.e., a spherical body whose surface radiates like a surface having an emissivity of 1) of radius 6.44×10^6 meters and at a temperature 254.87 K (hereafter rounded to 255 K) will radiate energy at the approximate rate of 1.25×10^17 Watts.
(6) If independent of the direction of energy incident on a sphere, the surface temperature of the sphere at any instant in time is everywhere the same, then the sphere possesses the property of perfect-thermal-conduction. Thus, for (a) an inert (no internal thermal energy source) perfect-thermal-conduction spherical body of radius 6.44×10^6 meters and uniform surface temperature 255 K whose center is placed at a distance of 1.5×10^11 meters from the center of an active (internal thermal energy source) spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K, and (b) the inert perfect-thermal-conduction spherical body (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates electromagnetic energy with an emissivity of 1 then the perfect-thermal-conduction inert spherical body at temperature 255 K will be in radiation rate equilibrium with the active spherical blackbody at temperature 5,778 K. If the phrase “inert perfect-thermal-conduction spherical body of radius 6.44×10^6 meters” is replaced with the word “Earth,” and the phrase ” active spherical blackbody of radius 6.96×10^8 meters and uniform surface temperature 5,778 K” is replaced with the word “Sun”, it can be concluded that: If (a) an “Earth” at temperature 255 K is placed at a distance of 1.5×10^11 meters from the “Sun” and (b) the “Earth” (i) absorbs electromagnetic energy with an effective absorption albedo of 0.3, and (ii) radiates energy with an emissivity of 1, then the “Earth” will be in radiation rate equilibrium with the “Sun.” For the above conditions, the temperature of the “Earth” in radiation rate equilibrium with the “Sun” will be 255 K.
This completes the algorithm that I believe is commonly used to arrive at an “Earth’s characteristic-emission temperature” of 255 K, and hence is used to compute the 33 K temperature difference.
Even ignoring the facts that (1) it is incorrect to use the “average surface temperature” when computing radiation energy loss from a surface, and (2) in the presence of an atmosphere, (a) the blackbody radiation formula may not apply, and (b) blackbody radiation from the surface of the Earth is not the only mechanism for Earth energy loss to space (the atmosphere even without greenhouse gases will be heated by conduction from the Earth surface and both conduction and convection will cause that thermal energy to be distributed throughout the atmosphere, and the heated atmosphere will also radiate energy to space), the problem with using the 255 K temperature computed above to determine the difference between the Earth’s temperature with and without greenhouse gases is that the effective Earth absorption albedo of 0.3 used to generate the 255 K temperature is in part (mainly?) due to clouds in the atmosphere, and atmospheric clouds are created from water vapor, which is a greenhouse gas.
Thus an effective absorption albedo of 0.3 is based on the presence of a greenhouse gas–water vapor. It is illogical to compute a difference between two temperatures both of whose values are based on the presence of greenhouse gases and then claim that temperature difference represents the temperature difference with and without greenhouse gases. Without water vapor, there won’t be any clouds as we know them. Without clouds, the effective absorption albedo of the Earth will likely not be 0.3, and hence without the greenhouse gas water vapor, the Earth’s surface temperature in the absence of greenhouse gases is likely to be something other than 255 K. Thus, the 255 K “Earth characteristic-emission temperature” as computed using the algorithm above is NOT relevant to a discussion of the Earth surface temperature difference for an atmosphere that does and an atmosphere that does not contain greenhouse gases. Only if 0.3 is the effective absorption albedo of the Earth in the presence of an atmosphere devoid of all greenhouse gases is it fair to claim the presence of greenhouse gases increases the temperature of the Earth by 33 K.
Because clouds reflect a significant amount of incoming solar power, without water vapor I believe the effective absorption albedo of the Earth will be less than 0.3. If true, then more of the Sun’s energy will be absorbed by an Earth whose atmosphere is devoid of greenhouse gases than by an Earth whose atmosphere contains clouds formed from the greenhouse gas water vapor. This implies a higher Earth surface temperature in the absence of water vapor than the “Earth’s characteristic-emission temperature of 255 K”.
For an effective absorption albedo of 0, the temperature of the Earth in radiation rate equilibrium with the Sun will be approximately 278.64 K (hereafter rounded to 279 K). If this value is used as the Earth temperature in the presence of an atmosphere devoid of greenhouse gases, then it can be argued that the presence of greenhouse gases introduces a warming of approximately 9 K (288 Kelvin minus 279 K).
In summation, using the simplified arguments that I believe are also used to arrive at the 33 K temperature difference (i.e., assumed perfect-thermal-conduction Earth, blackbody Earth emission, greybody Earth absorption with an effective absorption albedo between 0 and 0.3, and ignoring atmospheric radiation to space for an Earth atmosphere devoid of greenhouse gases), I conclude the presence of greenhouse gases in the Earth’s atmosphere increases the Earth’s temperature by somewhere between 9 K and 33 K. Thus, I believe the claim that the presence of atmospheric greenhouse gases increases the temperature of the Earth by 33 K is based on an argument that has little relevance to the Earth’s temperature in the presence of an atmosphere devoid of greenhouse gases; and hence at best is misleading and at worst incorrect.
Note: Upon first publication – the guest author Reed Coray was accidentally and unintentionally omitted.
@ur momisugly Dave Springer
“No, you’re wrong. The ocean has an effective albedo near zero and it radiates very little. Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation.”
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You surprise me. It may be true that the ocean loses most of its heat by evaporation, however water has an emissivity of about the same as black oil. Look it up. Write it down.
Water radiates heat very efficiently, certainly better than concrete or brickwork which are both about 0.93. Water is about 0.99. You can show this by covering a boiling pot with a lid. The water will boil at a much higher rate not only because of the insulating layer of air+lid and the limiting of evaporation, but because of the retention of a couple of hundred watts of energy that normally radiates from the water surface of the open pot onto the ceiling above.
Painting the oceans black, for example, would barely increase the radiation of heat at all.
Crispin in Waterloo says:
December 29, 2011 at 12:08 am
@ur momisugly Dave Springer
“No, you’re wrong. The ocean has an effective albedo near zero and it radiates very little. Look up any ocean heat budget study in the literature and you will discover that the ocean loses most of its heat by evaporation not radiation.”
+++++++++
You surprise me. It may be true that the ocean loses most of its heat by evaporation, however water has an emissivity of about the same as black oil. Look it up. Write it down.
Water radiates heat very efficiently, certainly better than concrete or brickwork which are both about 0.93. Water is about 0.99. You can show this by covering a boiling pot with a lid. The water will boil at a much higher rate not only because of the insulating layer of air+lid and the limiting of evaporation, but because of the retention of a couple of hundred watts of energy that normally radiates from the water surface of the open pot onto the ceiling above.
Painting the oceans black, for example, would barely increase the radiation of heat at all.
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And so also an efficient absorber of thermal infrared radiation, heat, as direct from the Sun..?
“And so also an efficient absorber of thermal infrared radiation, heat, as direct from the Sun..?”
Yes, it all get’s absorbed in the first centimeters, but there’s not much infrared from the sun, most is from the atmosphere. As it does not mix that well, it’s not very plausible that the “missing heat” is in the deep ocean. If you go swimming in the summer, you’ll notice the first centimeters are much warmer than even 1m below the surface.
*****
slow to follow says:
December 28, 2011 at 7:36 am
<i.Apologies if you are on top of this and I have misunderstood your pov.
****
slow to follow, I was following Patrick Michaels’ (who at the time was the VA State Climatologist before they found a way to get rid of him) objections to AGW back in the early 80’s when the AGW industry was just getting started. I was following John Daly’s site, then Climate Audit when it first appeared (~2000 IIRC). I’m a mechanical engineer w/a solid foundation in thermo. And I’ve heard the argument about “compression heating of the atmosphere” for years.
Instead of replying w/a long post, I recommend a site I was reading extensively some months ago:
http://scienceofdoom.wordpress.com/
and search for “Venus”.
He’s listed here as “Pro AGW”, but I found him accurate and w/o an apparent agenda (unbelievable). Eventually, the specific discussion there about Venus & GH effects reached an inconclusive end when some aspects of Venus’ lapse rate didn’t make sense according to standard theory. So I’m quite open to the idea that there is indeed something we don’t understand about Venus (& Earth too). But it ISN’T simple adiabatic compression/decompression — that’s straightforward thermo.
Read away — it’s a good discussion.
iya says:
December 29, 2011 at 6:47 am
“And so also an efficient absorber of thermal infrared radiation, heat, as direct from the Sun..?”
Yes, it all get’s absorbed in the first centimeters, but there’s not much infrared from the sun, most is from the atmosphere. As it does not mix that well, it’s not very plausible that the “missing heat” is in the deep ocean. If you go swimming in the summer, you’ll notice the first centimeters are much warmer than even 1m below the surface.
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Thermal infrared direct from the Sun is invisible, it is the heat we feel from the Sun, the Sun’s thermal energy on the move to us. It arrives on Earth at the same time as non-thermal visible light, you can feel thermal infrared heat is direct from the Sun for example when a cloud temporarily blocks the Sun on a bright coldish spring day. It is readily absorbed by the body, being mostly water, it moves water molecules into vibrational states which heats up water.
Water has also a very high heat capacity, unlike say carbon dioxide with its low heat capacity, (which means it heats up very quickly and as quickly, practically instantly, gives up that heat), water actually stores the energy. It takes a long time to heat up but so keeps it longer, it takes longer to cool down – why we have water in our radiators in central heating systems. The temperature of the ocean isn’t a complete picture of the amount of heat energy it is absorbing and storing because heat capacity is the amount of heat energy required to raise temp one degree per given volume and water readily absorbs vast amounts of it before you’d notice any temperature change. Water has to absorb about four times as much heat energy as the same amount of air to raise its temperature 1°C. That’s how lakes and oceans have little temperature change even while the air around them is changing.
Water is transparent to visible light, visible light can’t heat the oceans at all.
Heat radiation from the Sun heats oceans, light radiation from the Sun is simply transmitted through.
The ‘energy budget’ as it is depicted with visible light claimed to heat land and oceans, is junk physics. Regardless it is now taught everywhere, visible light isn’t able to do this.
iya says:
December 28, 2011 at 9:35 am
The question is, would the (surface air-) temperature change, if earth had a much denser atmosphere, of equal composition?
Yes because of broadening of the spectral lines of the GHG spectra, this is why the CO2 isn’t as effective on Mars.
See here for part of the CO2 spectra on the two planets:
http://i302.photobucket.com/albums/nn107/Sprintstar400/Mars-Earth.gif
richard verney says:
December 27, 2011 at 3:57 pm
@R. Gates says:
December 27, 2011 at 2:40 pm
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Mr Gates
You are right to point out the effectiveness of water vapour. In pointing this out, one can see how little warming/heat retention CO2 plays. However, the position is very complicated and we do not know precisely how everything interplays.
Let us assume the following: (i) that N2 and O2 cannot radiate heat and therefore can only give up heat by way of convection or conduction; (ii) that there is no CO2 nor water vapour in the atmosphere and the atmosphere consists only of N2 and O2; (iii) at the end of the day the N2 and O2 molecules are at a temperature of about 40degC (this being the air temperature measured in the shade in the usual way say 1.5m above the ground); (iv) it is a windless night; and (v) there is very thick cloud cover at say 50ft extending miles high which has a temperature of say 39 deg C on the desert side and below freezing on the space side (an unlikely scenario over a desert but since we are looking at theoretical compositions of the atmosphere a legitimate exercise).
Without water vapor in the atmosphere where does the cloud cover come from?
John Marshall says:
December 27, 2011 at 3:39 am
I am sorry to disagree with the famous Lord but the theory of GHG’s violates the laws of thermodynamics.
Certainly doesn’t, which suggests that you don’t understand the thermodynamics
A moments thought will show that if a molecule of CO2 is preferentially heated it will immediately share this heat with its surroundings. It cannot do anything else. It cannot ‘store it’ because the 2nd law must hold true. Any heated volume of gas in the atmosphere will convect due to density changes and convecting air in the atmosphere must expand and loose heat due to adiabatic expansion.
Indeed you don’t understand thermo, ‘adiabatic’ means with no heat transfer!
Adiabatic cooling involves a drop in temperature as the gas expands, not a loss in heat.
The gas will very soon be cooler than the surface so cannot re-radiate heat to the surface, heat can only flow by whatever means from hot to cold as the laws of thermodynamics dictate.
Not true, radiation occurs in all directions irrespective of what the temperature of its ultimate target, how would we receive the 3K background radiation from the ‘Big Bang’?
Rosco says:
December 26, 2011 at 2:07 pm
Nitrogen and Oxygen may not absorb much IR but they MUST become heated and MUST radiate.
They not only mustn’t they radiate but it is experimentally observed that they don’t.
The amount of IR from water vapour is at best a trace amount and the amount of IR from CO2 is tiny – unless CO2 possess some hidden magical power that science has not detected yet.
On the contrary, the emission of H2O and CO2 is substantial and is experimentally observed.
steveta_uk says:
December 26, 2011 at 5:15 am
John Brookes (December 26, 2011 at 4:18 am)
(245+265) / 2 = 255
sqrt( sqrt( ((245^4)+(265^4)) / 2 ) ) = 255.5864
Not what I was expecting.
Why ever not, it’s high school math?
If the mean temperature is T and the range is a then what you want is
∜( (T+a)^4+(T-a)^4))/2
all the odd terms in a cancel so you’re left with:
∜( T^4+6(Ta)^2+a^4)
in the case you quoted T = 255 and a =10 you get the result above, if a=20 it’s around 256, for small a the deviation from T will be small.
Well I don’t understand just where all of this methane flap comes from.
For starters, at a black body Temperature of 288K which Trenberth et al use for their earth surface LWIR emission, the spectral peak wavelength is 10.1 microns. elementary high school BB radiation theory says that only 1% of the total BB spectral energy falls below one half of the peak wavelength or 5.05 microns, so the shorter wavelength methane and nitrous oxide absorption peaks, are quite innocuous as is the CO2 4 micron band. Only 25% is emitted below the peak wavelength of 10.1 microns (obtainable from Wien’s Law), and only 1% is emitted above 8 times the peak wavelength or 80.8 microns.
And the peak spectral irradiance varies as the fifth power of the Temperature, so if the equivalent BB temperature is only 255K and not 288K then the peak spectral irradiance is only 54% as large, and that peak is moved to an even longer wavelength of about 11.4 microns moving it further into the clear atmospheric window, and even away from the 9.6 micron Ozone band.
Note that the Planck formula BB spectral calculations represent the limiting envelope of any real thermal radiation source emissions, so those numbers above are worst (or best) case.
I simply cannot see how GHGs sans H2O can possibly absorb the majority of the surface LWIR emissions.
And any radiation shorter than about 4.5 microns has to be from incoming solar spectrum energy; it can’t be earth generated thermal emissions, so any atmospheric absorption below that wavelength has to be solar energy trapping in the atmosphere, which ALWAYS results in LESS solar energy being absorbed by the earth, so that leads to cooling of the surface.
RE: Main Article
“Since I’m not sure what the definition of the ‘Earth’s characteristic-emission temperature’ is, I can’t disagree with his claim that its value is 255 K.”
That is the temperature indicated by the Stefan-Boltzmann law to be characteristic of the average rate of energy flow escaping to outer space per square meter of the Earth’s surface.
“For temperature A almost everyone uses a ‘measured average’ of temperatures over the surface of the Earth.”
Since this calculation is based on average energy and the Stefan-Boltzmann law says this energy is proportional to the fourth power of the *absolute* temperature, what we really have is a fourth root of the mean fourth powers average temperature in degrees K. This applies to any calculation based on average radiant energy.
The one potential weak link in this equation is the assumption of a 30 percent average optical reflection factor that reduces the incoming 1368 W/m² nominal solar radiation to 958 W/m² over the disk of solar radiation intercepted by the earth. This hypothesis assumes that the atmosphere has absolutely no effect on determining the temperature of the Earth. That, of course, is unlikely in reality but this is a hypothetical value, not a real prediction, except, perhaps, on an airless planet having an average surface optical reflection factor of 30 percent.
As atmospheres without greenhouse gases cannot cool themselves except by contact with the ground, I think such an atmosphere subject to solar heating might expand indefinitely as it continued to heat and eventually escape to outer space.
Spector, I don’t understand your comment:
“As atmospheres without greenhouse gases cannot cool themselves except by contact with the ground, I think such an atmosphere subject to solar heating might expand indefinitely as it continued to heat and eventually escape to outer space.”
The ground has a non-zero emissivity, so it will radiate the heat out. In Earth’s position, it would tend towards a global heat distribution that has a quartic-root-mean-to-the-power-fourth temperature of 255 K.
The atmosphere can still convect and conduct, if it got hotter than the surface then heat would flow throug convection and conduction back into the ground. The ground would cool it off and prevent it expanding indefinitely.
R. Gates says:
December 28, 2011 at 8:29 pm
Your numbers are way off. The temperature of the moon’s darkened surface has been measured during a total lunar eclipse, with a maximum cooling of about 30K per hour. The first few centimeters of rock and dust on the moon cool rapidly during a lunar eclipse.
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Hmm. I was using a model – no, not mocking models. Seen many odd numbers for Moon’s temperature. Example: a cable on top of the surface heats from ca. 150K to ca. 310K in a very short period of time (graph appears almost vertical – don’t have table) after sunrise. Then it takes many hundreds of hours to go from 310K to 390K near noon. And drops from 390K down to ca. 310K at sunset. Then plummets to near 165K (again, graph near vertical) in a short period of time, and after that cools very slowly to ca. 150K near sunrise.
I can explain the rapid rise in temperature due to the cable’s cross section at sunrise of 2 * r * length * sin theta, where theta is the angle of the cable to the sun. So the cable acts as a massless object. The cable would cool based on its surface of 2 * pi * r * length. Some radiation is exchanged between the cable and the surface of the moon.
The moon should heat up much slower at sunrise than a cable, as the surface of the moon is at a 90 degree angle to the sun at sunrise. But obviously, I would need to know specific heat, conductivity, and other such fun numbers – like how bumpy the surface of the moon is – in order to have my model match observations. Flipside, a standard Stevenson screen on the moon might give noon temperature as low as 45K or as high as 280K.
What I don’t get is the ‘knee’ in the lunar temperature data. Perhaps it is when conduction from subsurface slows further drop in temperature. That isn’t in my model, and needs to be.