Guest Post by Willis Eschenbach
OK, a quick pop quiz. The average temperature of the planet is about 14°C (57°F). If the earth had no atmosphere, and if it were a blackbody at the same distance from the sun, how much cooler would it be than at present?
a) 33°C (59°F) cooler
b) 20°C (36°F) cooler
c) 8° C (15°F) cooler
The answer may come as a surprise. If the earth were a blackbody at its present distance from the sun, it would be only 8°C cooler than it is now. That is to say, the net gain from our entire complete system, including clouds, surface albedo, aerosols, evaporation losses, and all the rest, is only 8°C above blackbody no-atmosphere conditions.
Why is the temperature rise so small? Here’s a diagram of what is happening.
Figure 1. Global energy budget, adapted and expanded from Kiehl/Trenberth . Values are in Watts per square metre (W/m2). Note the top of atmosphere (TOA) emission of 147 W/m2. Tropopause is the altitude where temperature stops decreasing with altitude.
As you can see, the temperature doesn’t rise much because there are a variety of losses in the complete system. Some of the incoming solar radiation is absorbed by the atmosphere. Some is radiated into space through the “atmospheric window”. Some is lost through latent heat (evaporation/transpiration), and some is lost as sensible heat (conduction/convection). Finally, some of this loss is due to the surface albedo.
The surface reflects about 29 W/m2 back into space. This means that the surface albedo is about 0.15 (15% of the solar radiation hitting the ground is reflected by the surface back to space). So let’s take that into account. If the earth had no atmosphere and had an average albedo like the present earth of 0.15, it would be about 20°C cooler than it is at present.
This means that the warming due to the complete atmospheric system (greenhouse gases, clouds, aerosols, latent and sensible heat losses, and all the rest) is about 20°C over no-atmosphere earth albedo conditions.
Why is this important? Because it allows us to determine the overall net climate sensitivity of the entire system. Climate sensitivity is defined by the UN IPCC as “the climate system response to sustained radiative forcing.” It is measured as the change in temperature from a given change in TOA atmospheric forcing.
As is shown in the diagram above, the TOA radiation is about 150W/m2. This 150 W/m2 TOA radiation is responsible for the 20°C warming. So the net climate sensitivity is 20°C/150W-m2, or a temperature rise 0.13°C per W/m2. If we assume the UN IPCC canonical value of 3.7 W/m2 for a doubling of CO2, this would mean that a doubling of CO2 would lead to a temperature rise of about half a degree.
The UN IPCC Fourth Assessment Report gives a much higher value for climate sensitivity. They say it is from 2°C to 4.5°C for a CO2 doubling, or from four to nine times higher than what we see in the real climate system. Why is their number so much higher? Inter alia, the reasons are:
1. The climate models assume that there is a large positive feedback as the earth warms. This feedback has never been demonstrated, only assumed.
2. The climate models underestimate the increase in evaporation with temperature.
3. The climate models do not include the effect of thunderstorms, which act to cool the earth in a host of ways .
4. The climate models overestimate the effect of CO2. This is because they are tuned to a historical temperature record which contains a large UHI (urban heat island) component. Since the historical temperature rise is overestimated, the effect of CO2 is overestimated as well.
5. The sensitivity of the climate models depend on the assumed value of the aerosol forcing. This is not measured, but assumed. As in point 4 above, the assumed size depends on the historical record, which is contaminated by UHI. See Kiehl for a full discussion.
6. Wind increases with differential temperature. Increasing wind increases evaporation, ocean albedo, conductive/convective loss, ocean surface area, total evaporative area, and airborne dust and aerosols, all of which cool the system. But thunderstorm winds are not included in any of the models, and many models ignore one or more of the effects of wind.
Note that the climate sensitivity figure of half a degree per W/m2 is an average. It is not the equilibrium sensitivity. The equilibrium sensitivity has to be lower, since losses increase faster than TOA radiation. This is because both parasitic losses and albedo are temperature dependent, and rise faster than the increase in temperature:
a) Evaporation increases roughly exponentially with temperature, and linearly with wind speed.
b) Tropical cumulus clouds increase rapidly with increasing temperature, cutting down the incoming radiation.
c) Tropical thunderstorms also increase rapidly with increasing temperature, cooling the earth.
d) Sensible heat losses increase with the surface temperature.
e) Radiation losses increases proportional to the fourth power of temperature. This means that each additional degree of warming requires more and more input energy to achieve. To warm the earth from 13°C to 14°C requires 20% more energy than to warm it from minus 6°C (the current temperature less 20°C) to minus 5°C.
This means that as the temperature rises, each additional W/m2 added to the system will result in a smaller and smaller temperature increase. As a result, the equilibrium value of the climate sensitivity (as defined by the IPCC) is certain to be smaller, and likely to be much smaller, than the half a degree per CO2 doubling as calculated above.
Good stuff Willis!
Posted on June 28, 2009
http://wattsupwiththat.com/2009/06/27/new-paper-global-dimming-and-brightening-a-review/#comments
The sensitivity of global temperature to increased atmospheric CO2 is so small as to be inconsequential – much less than 1 degree C for a doubling of atmospheric CO2. CO2 feedbacks are negative, not positive. Climate model hindcasting fails unless false aerosol data is used to “cook” the model.
Brian W (22:14:31)
I am not interested in debating my diagram, or the reality of downwelling longwave radiation, on this thread. Those are subjects for some other thread, not here. Do a google search for “measured downwelling longwave radiation” (usually called DLR). There are hundreds of measurements of DLR made all the time. There are companies who make instruments specifically for measuring DLR. There are scientists who spend their lives studying the actual measurements of DLR. The idea that it doesn’t exist is a joke that is not worth debating. A good place to start fighting your lack of knowledge on this subject is here.
But please, this is not the place to discuss it. It is a good topic … but this is not the right thread for it. This thread is about climate sensitivity.
Thanks,
w.
Brian W:
B. Smith (21:16:39)
Yes.
I’m looking to find out the total effect of the atmosphere (GHGs, clouds, aerosols, etc.) on the temperature of the earth. So I’m taking the theoretical temperature of the earth without an atmosphere, but with the same albedo as the earth. I’m comparing that to the current conditions on the earth.
As regards your question about temperature variations, the high and low temperature variations of the moon and an atmosphere-less earth would be very different, because the earth cools at night for only 12 hours, while the moon cools during the lunar night for 14 days, and the same is true of the warming during an earth day and a lunar day.
So it is a good question, but that is not the point of this thread.
Matt (21:31:17)
Fixed, thanks.
w.
dp (21:52:12)
Yes and no. Surface albedo just sets the base temperature. Cloud albedo has a huge effect on the temperature, as it reflects some 70 W/m2 of the suns energy back into space.
But there are a host of processes that affect the temperature, among them thunderstorms, cosmic rays, wind strength, the list is long.
w.
dp….
“I think your science on this is not robust, or my understanding of what you mean by black body located in our celestial toroid is flawed.”
dp, Homer Simpson rules – we live on a celestial donut!
“Donuts. Is there anything they can’t do?”
Way, way, way off topic!
for JinOH
I will never forget this: I put birdseed on my deckrail. And of course the squirrels get into it. I was playing on the compute , came to a stopping point,and glanced out the sliding door to the deck. The squirrel, stealing birdseed, suddently started running back and forth then leapt off the deck. Out of nowhere came a Cooper’s hawk and snatched it out of the air. That is why Altamonte is such a crime.
MODERATORS! I know this is off topic. You have my complete permission to forward it to jinOH including my E-mail!!!!
(Something funny happened to my last post, I think because I used a less than sign..)
Brian W:
How about measurement of it?
We can measure downward longwave radiation (greater than 4um) at the earth’s surface. It’s not coming from the sun which radiates 99% of its radiation less than 4um.
Where is this radiation coming from?
You can see one example of some specific measurements at CO2 – An Insignificant Gas? Part Six – Visualization
Strangely enough the peak radiation of this downwards radiation occurs in those wavelengths that match the absorption spectra of CO2, CH4, water vapor, ozone.
And at the top of the atmosphere the outgoing (upward) longwave radiation has a “notch” in those same wavelengths.
Venus is the same size as earth and has a fairly uniform temperature of over 400C across the entire planet (day or night equator or pole) This is because of their atmosphere.
Sunlight on Mercury is four times stronger than Venus, yet temperatures average much cooler, because of the lack of an atmosphere.
My plane flight on Saturday descended 35,000 ft. and temperatures warmed by 150 degrees F as we descended – because of the blanket generated by the atmosphere.
Willis,
All the temperatures we have are in the air , and there is a large difference between ground and air. I do not know of any measured ground temperatures. Where do you take the 14C from?
SSTs might be OK, so maybe you should take the SST average temperature as the 0 atmosphere temperature. Earth is 75% ocean after all. the AMSU plots give a scale from -2 to 30, that would give 14, but I would think a true average would come higher.
The albedo of the moon is around 0.1 so it is not a good analogue for your purposes.
Sorry that is the http://weather.unisys.com/surface/sst.html
plots
The figure of 0.13 K per W/m^2 is interesting.
Spencer and Christy found that the LT temperature varies by 2.3K over the year due to the eccentricity of the orbit of the earth. There is little or no perceptible phase lag in this variation wrt the orbital pattern, suggesting this is a full figure and there is no “warming in the pipeline”.
The orbital eccentricity results in a variation of solar radiance of 90 W/m^2. Of course, this arrives over a pi*r disc and is spread over 4*pi*r surface of the sphere, resulting in an equivalent of 17 W/m^2 averaged over the surface of the earth.
This yields a climate sensitivity of 2.3/17 = 0.135 K per W/m^2, entirely independently of Willis’ calculation.
Coincidence?
Is this post meant to be parody, or is the humour unintentional?
re: my 23:59
Good grief, why don’t I check my posts before pressing submit 🙂
Firstly, the area of a circle is pi*r^2 and the surface of a sphere is 4*pi*r^2. Secondly, 90/4 is 22.5 not 17. I’m sure I got that number from somewhere, just can’t remember where…
This yields a climate sensitivity *lower* than Willis at 0.1 K per W/m^2. Not such a coincidence, but at least in the same ball park…
Thank you for clarifying, WE.
Willis :
enough (20:51:43) :
I follow you Willis. Good point and well done!
And as ‘enough’ pointed out about the 1/2 up and 1/2 down being rubbish, he is right. All of these energy balance charts are basically wrong, misleading, or incomplete but that doesn’t take one iota away from what you have just have shown us on the sensitivity. The one thing they all do agree on is output equal input.
You see, none of them take into account the dip of the horizon correction or just “the dip”. Any good sailor who knows celestial navigation is well aware of that factor and it applies whether radiation is inbound from stars or outbound from CO2 molecules. Half up and half down can only occur exactly at the sea’s surface and ALL radiation leaving randomly from atoms or molecules in the atmosphere must account for the dip factor.
If you even stand up and are 5 feet above sea level it is no longer half up and half down however tiny. At 20,000 feet it is sizeable. There is now much more going up to space and much less going down back to warm the surface and this factor increases with the height above sea level quickly at first and slower near satellite altitudes. At high tropic cloud top levels this imbalance can get rather large, shunting heat directly to space. If you want the equations, holler. I learned celestial navigation from a captain right out of college (had a dream of circumnavigating when young).
That reinforces your statements about the mention of wind speed on evaporation rates which feed the clouds which shunt the heat to space, but you can’t if “half is up and half is down” and that often used term IS rubbish.
This is only my opinion, wink.
I am missing something. Everywhere I have read the calculations result in a blackbody temp of 255K.
No atmosphere , no blanket, the temp would be 255K. About 30K less than it is. 1C = 1K.
anna v:
You have a point when you say:
The air temperature is taken as the closest proxy to the land surface temperature. During the day it is a reasonably close value because the land heats up, heats the air directly over it and this air then expands and rises. The result is convective mixing which means that the air temperature six feet off the ground is a reasonable estimate.
At night it’s a different story. The temperature can vary significantly in the first few feet because the bottom layer is now colder and there is no convective mixing. “No wind” conditions give very different results from wind conditions. For this reason and UHI issues we should just give up on trying to measure land temperatures
But given what we do have our best estimate of annual average global surface temperature is around 15’C.
Hi Willis, great post for opening up discussion. The ‘atmospheric window’ is only 13W/M^2 according to Trenberth. You commented that thunderstorms would increase with increasing temperature. Presumably that would widen the window quite a bit. Would it also stick quite a lot more water vapour into the higher part of the troposphere? What effect would that have?
You should not compare to a black body of an homogeneus temperature. Because of the emisivity dependence on t^4, two black bodies with the same average temperature but different distribution of it along their surface radiate different ammounts of energy. For a given average temperature, the more temperature differences you have along the surface, the more you radiate to space.
To continue with the moon example, if the average of the moon’s daily temperature is 107C and the average night temperature is -153C, you can be sure that the emisivity total is quite higher than that of a black body with a uniform average temperature of -23C, i.e. the same average temperature. That’s because the emisivity you gain with the +130C of the illuminated side is much bigger than the emisivity you lose with the -130C of the dark side. Actually, it would be equivalent to a black body with a uniform temperature of 320K (47C). Quite hot huh? But this is considering a moon with one half at some uniform 107C and the other half at some uniform -153C. In reality, there are also differences of temperature along both surfaces leading to those averages. And remember: any inhomogeneity in the temperature leads to more emisivity than that of a black body of the same average temperature. So the equivalence of the emisivity would be to a black body of even more than 47C average temperature.
Calculations:
Daily temperature = 107C = 380K
Night temperature = -153C = 120K
The black body with equivalent emisivity would have a temperature equal to the 4th root of ((380^4+120^4)/2) = 320K = 47C
(The day side, with 3.16 times the temperature of the night side, radiates more than 3.16^4= 100.5 times the energy radiated by the night side.
anna v (23:48:31)
HadCRUT absolute temperatures (averages 1961-1990). 3 Mb zipped ascii file.
Regarding the moon, I’m not the one who brought it up.
Willis Eschenbach (22:57:51)
Well, lets see, you used the diagram to show us that “this is what’s happening”.
I and many others do not agree. If the diagram is wrong then using w/m2 and blackbody physics will not give you the right answer. If the numbers are wrong then how do you calculate anything? Yes indeed a very impressive link but since incoming sunlight is comprised of 10% ultraviolet, 44.8% visible (shortwave) and 45.2% INFRARED how is it these wonderful scientists can accurately separate upwelling from downwelling without proper hard experiments in the atmosphere? Sensitivity based on what, fancy algorithms with built in beliefs. Explaining part or all of the diagram shows me that you know what you are talking about, and you avoided it so I’m dubious.
As to the sensitivity. CO2 with a specific heat less than N2, O2 and even aluminum, poor absorption compared to air, fast emission and most importantly a concentration of .038% by volume contributes NO sensible (usable) heat to our atmosphere. A doubling of CO2 will not give even .1 degree increase. I wish people would stop fixating on CO2. So the models do overestimate sensitivity by a huge amount. The whole AGW/CO2 thing is a stinking scientific fraud. Period.
Sou (00:09:06)
Not a parody. What humour?
It was not a Cooper’s hawk. it was an Eschwnbach hawl!