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.
” It is the greenhouse properties of water vapor, CO2, and to a lessor extent, N2O and methane that hold all the IR, or as you call it, the “daily heat” in and keep it warmer at night. ”
BEEP! wrong!
jeez louise – this is not about radiation physics.
water vapor holds 50,000 times more HEAT than the CO2 on a parcel of atmosphere and it is perfectly capable of conduction without condensation. steam is used to cook things NONRADIATIVELY. it can also condense without radiating – dew you know?
nitrogen, oxygen, friday’s beans – all gasses of every description conduct heat.
you’re not heated by radiation when you swim in a sea of gasses and the fan on my heat sink isn’t moving IR rays, mmk?
and also, ffs – TEMPERATURE is not heat. it is also not a property of radiation physics.
temperature is kinetic energy. nitrogen and oxygen don’t need permission from radiation freaks to be hot or not. dip something in some lox and guess what happens. explain that with radiation – go ahead and make my day… lol
review avogadro, too – all molecules in a small volume have essentially the same temperature as the energy is completely shared by collisions.
co2 does not sit there glowering with special intensity. it gets slapped around just like all the other molecules so its temp is the same as the rest.
slow to follow says:
December 27, 2011 at 4:25
Try this:
“if the
temperature differential between the surface and the stratosphere increases the
tropopause must rise. If the differential decreases then the tropopause must fall.
“Suppose, for example, that the surface temperature and the tropospheric
temperature gradient are given and that the temperature of the stratosphere
varies. Then, a cold stratosphere will be associated with a high tropopause (low
tropopause pressure), and a warm stratosphere will correspond to a low
tropopause (high tropopause pressure).”
from here page 14:
http://journals.ametsoc.org/doi/pdf/10.1175/1520-
0442(2001)014%3C3117%3ATTITPR%3E2.0.CO%3B2pm
So if solar variability alters ozone quantities above the tropopause to warm or cool the stratosphere then the tropopause height will change accordingly.
In another thread it has been confirmed that (unexpectedly) an active sun cools the mesosphere by destroying ozone above 40/45km. It has been observed that the stratosphere also cooled when the sun was more active. I think the stratospheric temperature trend follows the mesosphere trend DESPITE an active sun creating MORE ozone below that height.
That makes my hypothesis set out here:
http://climaterealists.com/attachments/ftp/How%20The%20Sun%20Could%20Control%20Earths%20Temperature.pdf
likely to be correct and the only current climate description that fits the observations.
From the above, Lord Monckton used the term “Earth’s characteristic-emission temperature.” I believe that is the temperature of a body with a single approximate emission temperature that radiates the same power over its whole spherical surface as it receives as a disk from the sun, based on the albedo-reduced solar constant. This emission temperature should not be confused with any normal average temperature because, according to the Stefan-Boltzmann law, energy emission is proportional to the fourth power of the absolute temperature.
The solar constant is nominally 1368 W/m² and the ratio of the Earth’s surface area to its cross-sectional area is 4:1 so that the Earth only needs to radiate about 342 W/m² to be in stable equilibrium. The nominal reflectivity of the Earth is usually stated to be about 30 percent, which reduces the required IR energy flow to about 239 W/m². The Stefan-Boltzmann law says this is equivalent to a temperature of 254.9 degrees K. Once again, this is an equivalent uniform surface temperature, based on the simplifying assumption that the Earth radiates like a star.
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slow to follow says:
December 27, 2011 at 9:14 am
Think of getting in a submarine and descending in the sea – does the hull experience a change in pressure loading?…
****
Certainly. Just like the change in pressure loading that happens when it ascends, except the opposite. 🙂
I’m getting a better understanding of how some CAGW alarmists treat the Anthony Watts, Willis Eschenbachs, Tallblokes, Roy Spencers, etc. of the world. I chanced upon a critique of this post. The URL for that critique is:
http://wottsupwiththat.com/2011/12/26/a-controversial-look-at-blackbody-radiation-and-earth-minus-ghgs/
Although there are “nuts” on both side of the AGC/CAGW discussion, the case is sometimes made that on a per capita basis the AGC/CAGW side more than the skeptic side engages in name calling, ad hominem attacks, and shrill rhetoric. I believe it would be informative for the AGW and CAGW advocates who have commented on this thread to read that critique and assess its relevance to what I wrote. Everyone is entitled to his opinion; and since I entered the arena, criticism comes with the territory. That’s as it should be. However, if I were an AGC and/or CAGC advocate, I would try to suppress critiques such as the one given above. Then again, maybe the critique author knows more about my intentions, background, and beliefs than I do.
R. Gates says:
December 27, 2011 at 2:40 pm
“It is the greenhouse properties of water vapor, CO2, and to a lessor extent, N2O and methane that hold all the IR, or as you call it, the “daily heat” in and keep it warmer at night.”
If so, why diurnal temp difference in Mars is so big despite having about 9 times more CO2 in much smaller surface area in its atmosphere?
Beng thinks that the adiabatic heating is a one time only event and has long gone.
Wrong. The compression due to mass is there all the time with changes, which are happening all the time both locally and between night and day hemispheres, only replacing losses due to radiation to space.
How does Jupiter maintain its temperature of 280K at the 1000mb level without any greenhouse gasses?
The theory of GHG’s has yet to be shown to be true within the bounds laid down by the laws of thermodynamics. Sorry Beng.
As has been known since 1966,[58] Jupiter radiates much more heat than it receives from the Sun. It is estimated that the ratio between the power emitted by the planet and that absorbed from the Sun is 1.67 ± 0.09. The internal heat flux from Jupiter is 5.44 ± 0.43 W/m2, whereas the total emitted power is 335 ± 26 petawatts. The latter value is approximately equal to one billionth of the total power radiated by the Sun. This excess heat is mainly the primordial heat from the early phases of Jupiter’s formation, but may result in part from the precipitation of helium into the core.[59]
Dennis Ray Wingo says:
December 27, 2011 at 10:54 am
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Dennis
Very interesting papers.
In one of my earlier posts, I suggested a rotation of about an hour was necessary to maintain temps. Fig 3 shows that the temps drop off like a cliff (the rock has little stored heat capacity at the depths of the probes) and it may be that an even faster rotation would be required. One would need more probes covering a different range of depths and a different scale to Fig 3 to make a more realistic estimate of the required speed of rotation.
SGW says:
December 28, 2011 at 12:58 am
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The argument that the AGW proponents raise is that the atmosphere is so thin. Obviously, it is true that the Marsian atmosphere is thin, but whether this is the reason why the GHE cannot be seen on Mars is moot.
*****
John Marshall says:
December 28, 2011 at 2:36 am
Beng thinks that the adiabatic heating is a one time only event and has long gone.
Funny how I think, huh? No, adiabatic compression-heating is a constant process in a dynamic atmosphere. But so is adiabatic decompression-cooling.
Wrong. The compression due to mass is there all the time with changes, which are happening all the time both locally and between night and day hemispheres, only replacing losses due to radiation to space.
The act of air compression is work, so produces heat. An equal amount of air elsewhere, tho, HAS to move upward to replace the air that moved downward (nature abhors a vacuum). That air decompresses and cools by essentially the same amount. The net result is zero.
How does Jupiter maintain its temperature of 280K at the 1000mb level without any greenhouse gasses?
Leftover internal heat! Jupiter’s mass is huge, and preserves much of the initial compression heat from formation. Earth’s residual heat is tiny in comparison (& lost much more quickly), & the insulating crust slows heat loss down to a minimum (guessing a couple of W/m2). Insignificant compared to solar input (except at the mouth of a volcano).
The theory of GHG’s has yet to be shown to be true within the bounds laid down by the laws of thermodynamics. Sorry Beng.
Don’t be sorry, John. I’m not the one lacking a basic understanding of thermo. 🙂
R. Gates says:
December 26, 2011 at 11:00 am
A shocking revelation isn’t it. At the moon’s equator on the sunlit side the temperature is around 120C whereas on the opposite dark side, the temperature is about -230C. Amazing what an ocean and greenhouse atmosphere can do!
———————————————————————
That has almost nothing to do with oceans and atmosphere, and everything to do with rotation speed. The surface of the Earth near the equator with cloudy skies heats up at a typical maximum rate of just under 1 C per hour in the late morning. Luna near the equator also heats up at a maximum rate of around 1 C per hour in the late morning. The difference is that earth rotates about 29.5 times faster than Luna, so Luna receives 29.5 times as much gross energy per unit area per day cycle.
Plugging in 393K, 288K and 43K (Note: -230C = 43K is only correct for those parts of Luna in permanent shadow, so not germane to the current thread), one gets, using 288K as unitary, (393/288)^4 or 3.4673 times as much energy in Luna late afternoon equilibrium, and 0.0005 times as much energy in Luna permanent shadow. All this tells us, is that, absent atmosphere, Earth would be a pretty good insulator.
For typical daily ranges on Earth, we get a ~22C range in the desert, and a ~11C range where people typically live. The minimum to maximum time is around 8-9 hours. For Luna near the equator, the daily range is typically from 165K to 375K, a range of 210C, in a period of about 292 hours. If Luna had a 24 hour day, and an average temperature of 255K, then it would be about 251K shortly after dawn, and about 259K in the late afternoon. This gives a range of about 8C.
We can check that by calculating cooling rates for Luna. To cool from 259K to 251K would take about 13.5 hours. Since Luna would, in a 24 hour day, take just under 10 hours to go from minimum to maximum, these numbers are definitely in the ballpark. Note that this gives a daily range for Luna that is LESS than the daily range for Earth.
beng – check out variation of pressure with horizontal displacement vs. with variation with vertical displacement. Check out temperature and pressure readings at any weather station as we move from night to day and back again.
I think you have misunderstood what happens wrt to lapse rate in the atmosphere but I do not have the time to argue it through with you. Check your definition of “adiabatic”. Search for “dry adiabatic lapse rate”, “hydrostatic equilibrium” and “hypsometric equation”. Apologies if you are on top of this and I have misunderstood your pov.
Stephen Wilde says:
December 27, 2011 at 5:56 pm :
Thanks Stephen – references appreciated.
@John, Being
Neither of you has any cause to be sorry. You are both a lot further along than the typical climate scientist, who builds his AGW chimera in some parallel radiative universe were convection does not exist.
” if I were an AGC and/or CAGC advocate, I would try to suppress critiques such as the one given above.”
if you think that article is nasty- you ain’t been around.
be a big boy. suppress your butthurt instead. nobody needs more censors. also- give up subjunctives- you are not a CAGW so stop with the fairy tales altogether.
on the other hand, encourage the fools to indulge their mistakes and fantasies – it helps them move further down the road to extinction.
Let’s wave the magic wand and take out all greenhouse gases, but leave the surface as it is.
THe surface albedo is about 0.125 (Trenberth, 2008). So you’re actually looking at an equilibrium temperature of ~270 K. Assuming your thermally conducting sphere, the whole thing would freeze. A planet entirely covered in sea ice has a much higher albedo. Enceladus is an ice ‘planet’ – and it has a Bond albedo close to 1.
So the temperature would fall much, much further below freezing. You’d end up calculating that the greenhouse effect is many tens of degrees more useful than we think. Using a simplistic approach, I could argue that without it, we’d be hundreds of degrees worse off.
Another way of looking at it is to keep the gases, but magically remove their greenhouse effect. You cool Earth, ice spreads, it cools, clouds eventually disappear (too cold), ice spreads, ice spreads, albedo is close to 1, temperature falls far, far below the 255 K given.
So you’re right, this quantificatin of the greenhouse effect is only a rough one, in reality the presence of the greenhouse effect keeps us much warmer than the typical ‘back of the envelope’ calculation you gave, which is common in introductory planetary physics textbooks.
The question is, would the (surface air-) temperature change, if earth had a much denser atmosphere, of equal composition?
Can’t we test it approximately by putting two bottles in the sun; one with compressed air and one with standard pressure?
Stephen Wilde says:
December 27, 2011 at 5:56 pm
Stephen – for info: I couldn’t get your link to the JoC article to work from your comment or your paper.
This one worked for me:
http://journals.ametsoc.org/doi/pdf/10.1175/1520-0442%282001%29014%3C3117%3ATTITPR%3E2.0.CO%3B2
“The Tropopause in the Polar Regions”
GUNTHER ZANGL
Meteorologisches Institut der Universitat Munchen, Munich, Germany
KLAUS P. HOINKA
Institut fur Physik der Atmosphare, DLR Oberpfaffenhofen, Wesseling, Germany
Journal of Climate 15 JULY 2001
Matter says:
December 28, 2011 at 9:32 am
“Let’s wave the magic wand and take out all greenhouse gases, but leave the surface as it is.”
I think your reasoning has several problems.
You can’t take water wapor out from the atmosphere and leave the sea in place. Even if you could, day time heat from the sun would keep the sea from freezing anyway. Without any clouds acting as shades it would be obvious. And 95% percent of the atmosphere would still be there (nitrogen and oxygen) and do their job via convection and conduction giving night time insulation.
Reed Coray says:
December 27, 2011 at 8:38 pm
“It would be informative for the AGW and CAGW advocates who have commented on this thread to read that critique and assess its relevance to what I wrote”
Reed, you don’t really want to ask this do you?
I believe the author of the article described you as “an ill-informed but enthusiastic amateur”. Do you object to this description?
Now there are many commentators here to which the same could be applied, and most of those support you. Not because they understand what you wrote, but because as Judith Curry once said, they decide if the article is for us or against us. That is the limit of their reasoning.
Bomber_the_Cat. You may be right–asking AGW/CAGW proponents to read what others on their side of the issue say is probably counterproductive. However, I believe not all of the pro-AGWCAGW commenters on Anthony’s blog respond based solely on the criterion that “the article is for us or against us.” As far as the “ill-informed but enthusiastic amateur” comment goes, I agree with the “enthusiastic amateur”, but not necessarily with the “uniformed” part. I’ve been following much of the discussion for four years. There is some truth in what “gnomish” (December 28, 2011 at 9:04 am) said. I should probably have kept my mouth shut. However, I thought at the time that some commenters, in particular Joel Shore, might be interested the blog.
Thomas L. said:
“To cool from 259K to 251K would take about 13.5 hours. Since Luna would, in a 24 hour day, take just under 10 hours to go from minimum to maximum, these numbers are definitely in the ballpark. ”
_____
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.
The University of Chicago ‘MODTRAN radiation code’ utility can be used to calculate the *raw* effect of changing the amounts of greenhouse gases in the atmosphere. For my own purposes, I usually use the default conditions and tropical locality with no clouds or rain. This program, using atmospheric component, absorption data, normally calculates radiant energy flow at various levels in the atmosphere as a function of surface temperature entered as an offset from 299.70 deg K.
For greenhouse effect calculations, I usually search for the temperature offset entry required to obtain a nominal energy flow, seen looking down from 70 km up of 292.993 W/m²—a number the program likes to produce. This standard value was based on my estimate of the incoming average daily solar radiation.
Starting with default conditions, except for a CO2 concentration of 396 PPM, I find that a surface temperature of 301.18 deg K is required to produce the nominal radiant energy flow (offset: 1.48).
If I set the CO2 concentration to zero, the nominal flow surface temperature drops to 293.58 deg K (offset: -6.14), an apparent 7.62-degree decrease.
Next, if I change the nominal CH4 from 1.7 ppm to zero, the nominal flow surface temperature drops to 293.17 deg K (offset: -6.53), an apparent 0.39-degree decrease.
Then, if I change the tropospheric ozone from 28 to zero and the stratospheric ozone scale from one to zero, the surface temperature for nominal flow drops to 291.75 deg K (offset: -7.955), an apparent 1.42-degree net decrease.
Finally, if I change the water vapor scale from one to zero, the required surface temperature drops to 275.86 deg K (offset: -23.839) for a final 15.89-degree decrease. Thus a 25.32-degree total calculated temperature decrease is obtained when all program optional greenhouse gases are removed.
Note that 275.86 deg K is *not* a calculation of the “Earth’s characteristic-emission temperature,” a hypothetical construct.