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.
magicjava (07:06:00) :
[quote]
I think the skeptic community would be well served if we built out own energy balance model, rather than using one based on numbers James Hansen pulled out of his ass, which is all Trenberth’s model is.
Starting with data from the CERES satellite would be a good idea, IMHO. Building a model that changes with conditions rather than just giving a hard-coded answer would be a good idea too. And being honest about things we don’t know, like where extra energy is going, would also be a plus.
[/quote]
You mean something like this? http://www.palisad.com/co2/eb/eb.html
Sorry, that should be a correction factor of roughly
(albedo/emissivity) ^0.25 = (.85/,9)^.25 times the original temperature.
Another point, Using Trenbeth ballpark figures, if we get 240 watts from the sun, and the greenhouse effect makes that an effective 390 watts and a 33 C increase, isn’t the increase in temperature per delta watt increase in the ballpark of
33 C/(390-240) = about 0.22C per watt increase?
A thrid point using Trenbeth ballpark figures. The surface flux is about 490 rather than 390 watts, with 100 watts going into conduction and convection rather than sensible heat. Figuring that half of that is radiated to space, half back to earth, about half of 100 watts or 50 watts is removed from sensible heat by convection and conduction. 50 watts latent heat + 150 watt sensible heat implies that 25% of any past wattage increase has gone into latent heat. Shouldn’t that apply to the 3.8 watt increase for a doubling of CO2? Shouldn’t at least 0.95 watts go into latent heat, only 2.85 watts or less go into sensible heat?
A fourth point, since emissivity is less than 1, assuming it’s about 0.9, isn’t the actual greenhouse effect something like 100 latent+ 150*.9 = 135 watts sensible heat?
On a “small” climate sensitivity:
Estimates from Lindzen, Shaviv, Monckton, Schwarz, others, put the value BELOW what is known to be “natural” variability – partly stochastic, partly determinate.
My question is: if there is no way to observe a physical quantity, does it have meaning at all?
I can construct a thousand quantities that I can calculate in some way (e.g, change in Earth’s rotational angular momentum from buildings, transportation, etc) but the construct is entirely meaningless because there is no way to physically measure it
Willis,
there are some significant problems with the albedo numbers. If you go by K&T, it would seem the 0.3 total albedo is 0.22 for clouds and 0.08 for surface with about 0.62 cloud cover fraction. In essence, the 0.08 is the average albedo where there are no clouds present.
Your number of 0.15 would be fairly reasonable based upon the Moon and Mars. It is not reasonable for Earth as over 70% of the surface is covered by oceans and the oceans have less than 0.04 albedo for incoming solar radiation, at least for high angles of incidence (wrt horizon where the incoming flux is most significant).
Try these numbers:
surface (ocean – use 0.7 ocean fraction) 0.04
surface( land) 0.19 (includes snow cover, forests, sand etc)
average surface albedo 0.085 (corresponds to 29/342 on chart )
the land albedo is calculated from the 0.085 total surface and the 0.04 ocean.
clouds (62% cover, averaged value) 0.36
this corresponds to the assumption of 0.22 for cloud albedo and fractional contribution.
NOTE that these numbers do not add up.
The weighted average for 0.62 cloud cover leaves 0.38 surface contribution.
Using the 0.085 overall surface albedo and fraction visible, one gets
0.38*0.085 = 0.032 contribution to the overall albedo which conflicts with the chart.
Since the overall albedo listed is 0.307, the cloud contribution would actually be 0.307 – 0.032 = 0.275 at a 0.62 fractional cover.
That puts the cloud albedo at 0.275/0.62 = 0.44
Note that this leaves the clouds with what would seem a more realistic albedo value than the original 0.38 albedo value.
Your question of temperature drop really depends upon the conditions.
if the surface is stripped of ocean and ice and vegetation, then it’s going to have an albedo of around 0.17 or 0.15. That’s around 265K or -8 deg C or 22 C less (20 C less if 0.15). If the surface stayed the same, the drop would be 15 deg C. If the cloud albedo somehow stayed along with the surface albedo so that the total was 0.307 the temperature drop would be 34 deg. lower.
daniel says:
That Gerlich & Treuschchner saw the light of day in even a second-rate physics journal is extremely embarrassing, as its fundamental error regarding the 2nd Law of Thermodynamics is basic enough that one could lead a group of bright first-year physics students through the logic. (And, some of its other arguments are just plain bizarre.) A comment on G&T that I am a party to and that will essentially trash it completely has been accepted by the journal and should appear within the next few months.
Thank you, Mister Troll Shore, for your erudite analysis of Gerlich and Tscheuschner’s paper, whether you believe it or not, the analysis is correct, and if you think Phys B is second rate, don’t look at it, read some first rate stuff like “Scientific American” instead, the favored bedtime reading material of the first-rate watermelon
Joel Shore (17:26:23) : edit
Dang, here I am agreeing with Joel again … will wonders never cease? G&T is bad, bad, bad.
w.
Alan DMcIntyre says:
Well, the issue of latent heat is really one of distribution of the temperature change with height in the atmosphere since what evaporates at the surface has to condense higher in the atmosphere, and this in turn can affect the radiative balance since most of the emission back out into space escapes from the upper troposphere. But, now the punchline: What you are describing is, I believe, just another way of talking about the (negative) lapse rate feedback, i.e., the notion that because of the condensation of water vapor, the upper troposphere tends to warm faster than the surface. This is a feedback that is already included in all the climate models…and, in fact, because much of the physics governing this feedback also governs the (positive) water vapor feedback, models that have a higher magnitude for the water vapor feedback tend to have a higher magnitude for the lapse rate feedback…and hence the sum of these two feedbacks tends to vary less from model-to-model than each of them individually.
Willis Eschenbach (17:36:52) :
G&T is bad, bad, bad.
I’ll agree with that.
Willis says:
Watch out…It could be addicting! 😉
yan Stephenson (07:28:41) :
Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius.
—————–
Geothermal is supposed to play its biggest role in warming the bottom of the oceans, as the crust under the ocean floor is much thinner than under most land. The amounts of geothermal energy and its variations and distributions are largely unknown. Considering how close it is from us, it’s amazing how little we really know about what is going on at the earth’s core. Some interesting new theories are emerging:
http://tinyurl.com/y8pguwr
Beginning in 1969, astronomers discovered that three of the giant planets, Jupiter, Saturn, and Neptune, each radiate about twice as much energy as they receive from the Sun. Those planets each contain a powerful energy source which was inexplicable until J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source for those planets
http://tinyurl.com/y8pguwr
Leif Svalgaard (13:23:50) :
A questions perhaps you could answer please?
Albedo refers to reflectivity.
The normally referred albedo is to the visible light reflectance. But surely the reflectance in the LW IR will be different. Is this known?
Emissivity of a surface is not he same as albedo. Having used an IR camera a few times I know that the IR temperature of an object changes with emissivity. and corrections can be incorporated in the IR to temp conversion (this is why Mr Watts IR photos in his surface station project are not very valid – he need to correct for surface emissivity.) When using IR cameras one usually coats the objects with a thin layer of mat paint to equalise (approx) emissivity.
Surely albedo controls the absorbed solar radiation, but emissivity controls the radiated LW energy.
If so then one needs to know the emissivity of the earth in the 4u range not its albedo.
/harry
“Surface emissivity” must be considered; both may not have equal emissivity figures … which will affect radiation.
An Emissivity Primer
Emissivity Coefficients of some common Materials
Examples, emissivity:
Aluminum Foil – 0.04
Beryllium —- 0.18
Granite —– 0.45
Sand —- 0.76
Wrought Iron – 0.94
Plaster —– 0.98
.
.
Francisco (17:52:28) :
J. Marvin Herndon, pictured at left, demonstrated in 1992 the feasibility of natural, nuclear fission reactors as the energy source
Recent measurements of geo-neutrinos rule out the hypothesis that most of the planet’s internal heat is generated by a uranium-fuelled nuclear geo-reactor in the Earth’s core.:
http://www.physorg.com/news187946006.html
yan Stephenson (07:28:41) :
Interestingly I don’t see any part of this diagram to account for geothermal heat from underground. Even under the UK (not known for its volcanic activity) the ground at 3000meters depth is about 90Celsius.
———————-
See also letter by Herndon in Current Science, titled “Variables unaccounted for…”
http://tinyurl.com/yfx4dax
Nothing impedes clear physical reasoning than a wrong conceptual model. A blackbody Earth without an atmosphere might be a good starting point for teaching students, but no realistic analysis of actual temperatures can be made without fully mastering the implications of enthalpy. That atmospheric pressure is a vital variable seems to have been forgotten.
Thermal imaging and emissivity here
http://en.wikipedia.org/wiki/Thermal_imaging
“Since there is no such thing as a perfect black body, the infrared radiation of normal objects will appear to be less than the contact temperature. The rate (percentage) of emission of infrared radiation will thus be a fraction of the true contact temperature. This fraction is called emissivity.
Some objects have different emissivities in long wave as compared to mid wave emissions. ”
/harry
Update, pls:
Hmmm … which space-based sensor (and what wavelength) can allow measurement, can allow this to be actively seen during convective T-storm activity?
I’m going to posit the theory that a largish amount of thermal energy is _not_ radiated directly from the troposphere (from the gases constituting the air e.g. CO2 which are *capable* of EM radiation) into space, but rather becomes part of Hadley Cell circulation and the bulk of that sensible heat is released back into space is at latitudes greater than ~45 degrees due to *surface cooling* …
Also note that thermal budgets show a ‘deficit’ of incoming solar energy at those latitudes (for the average temps seen), so it is circulation from the lower latitudes that keeps those temperatures elevated relative to what might otherwise be experienced over ~50 latitude.
.
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Willis tells us “The average temperature of the planet is about 14°C.” Of course that’s disregarding the massive cold of the ocean which is more like 4°C – averaged out. What is it that makes the oceans so cold? Two miles down on the continents the earth is quite warm, but two miles down in the oceans it’s quite cold. It suggests to me there is a refrigeration process of some sort extracting heat from the depths. Also, do I remember my grade school science correctly when I say the total mass of the atmosphere is represented by the mass of just 33 feet of water – the theoretical lifting limit of a suction pump?
cba (17:25:23)
I love posting here, people always force me to refine my figures.
I went and got the ERBE figures for clear-sky albedo here (Table 1). They give the clear sky albedo as 0.123.
Using this figure instead of the 0.15 I used earlier gives a warming from a no-atmosphere situation of ~ 18°C rather than the ~ 20°C I used above. This changes the climate sensitivity 0.43°C per doubling, rather than the 0.5°C per doubling I quoted in the lead post.
Which, of course, makes absolutely no difference to my argument, which is that the average climate sensitivity is much, much lower than the canonical IPCC value, and that the equilibrium sensitivity is lower yet.
That’s why I’m not particularly concerned about the final digits in my estimates for TOA radiation. My calculations shown graphically above put it at 147 W/m2. Kiehl/Trenberth put it at 169 W/m2. ERBE satellite data puts it at around 170 W/m2.
But none of these change my point. Use whichever TOA and albedo figures you want, you won’t get anything near the UN IPCC canonical sensitivity of 3°C per doubling. Instead, you’ll get something well below their lowest estimate of 1.5°C per doubling, and the IPCC says sensitivity is very unlikely to be less than 1.5°C.
Willis Eschenbach (19:16:37) :
a warming from a no-atmosphere situation of ~ 18°C rather than the ~ 20°C I used above.
I like your number now.
Willis and Leif think Gerlich & Tscheuschner’s analysis is bad.
Sorry folks, it’s good. Two legitimate criticisms leveled at it: it does not account for convection heat transfer in the atmosphere (considers conduction only), and it does not consider clouds.
But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer), and all radiation in the atmosphere is eventually lost via degradation to 20+ micrometer wavelengths, for which everything in the lower atmosphere is transparent
– or it increases the internal energy of the atmosphere, the contribution (via increasing the average k.e. of molecules) is negligible.
Ladies and Gentlemen: Either Gerlich and Tschneuschner are correct, or the atmosphere is a perpetual motion machine.
If it is, then copy the principle with a scale model and make yourself wealthy. Power your invention up a little bit with cold fusion.
Brian G Valentine says:
That’s a “head-I-win, tails-you-lose” style of argumentation. Actually, either Gerlich and Tschneuschner are correct and the greenhouse effect describes a perpetual motion machine or they are wrong and it doesn’t. It is very easy to show that the second of these possibilities is in fact the case…And, I think Brian that you are intelligent enough to comprehend this.
Willis Eschenbach says:
But, I’d like to see you address the more fundamental objections that I raised in my post of (12:34:08), namely that I don’t see how your calculation includes feedback effects. I think you are just calculating the sensitivity before feedbacks and somehow getting the wrong number (as I don’t understand where the 150 W/m^2 value you use comes from).
Brian G Valentine (19:25:34) :
But “convection” heat transfer is nothing more than conduction heat transfer (through a thermal boundary layer)
It is a lot more. Conduction is mediated by micro-scale vibration without any bulk transfer of matter, while convection is macro-scale movements of matter.