Hi Ken,
Thanks very much for that. I'm particularly interested in your
parenthetic comment at the end:
"(I think that 75 was the ratio of current atmospheric CO2 radiative
forcing to direct heating from fossil fuel burning, but I would need to
go back to check.)"
I had been wondering about the fact that solar panels increase albedo.
But if they are saving on fossil fuels, i.e. CO2 in atmosphere, then
that outweighs the albedo effect, by around 75 times (taking the figure
that you quote).
Looking at it another way, if a solar panel is replaced by a sun
reflector of about 75 times the area, this would offset as much CO2 as
the solar panel would have saved through replacing fossil fuel power.
Is that right? Can you check the "75" figure?
If it is right, we have a good argument for covering vast areas of
desert with something to make them more reflective. How much would we
have to increase Earth's albedo to offset current CO2 or net forcing
(assuming both around 1.6 Watts per square metre)?
Of course, putting a reflective covering on deserts is not going to
cool the Arctic quickly enough to save the Arctic sea ice and Greenland
ice sheet. But would a covering over the ice to stop pools of water
forming in summer be helpful, and by how much?
Note that saving the Arctic sea ice has to be our number one priority
in all this. Sea ice disappearance could be a point of no return. So,
if we fail to save it, then there may not be anything else that can be
done to stop us (our civilisation) sliding down a slippery slope into
oblivion. But I've yet to convince some of you about that!
Cheers,
John
---
Ken Caldeira wrote:
1 digit calculations just for orders of magnitude:
If we assume a doubling of CO2 is 4 W / m2 and the earth is 5 x 10^14
m2, a doubling of CO2 traps about 2 x 10^15 W.
If we assume 2 GtC / ppm, and think it takes say 300 ppm to double CO2,
that is 600 GtC, 600 x 10^12 kgC = 6 * 10^14 GC, so each kgC in the
atmosphere traps around 3 W.
Oil is about 4.5 x 10^7 J / kg. If we pretend oil is CH2, then we can
assume that most of this mass is carbon, but a lot of the energy comes
for the hydrogen. So by this reckoning it would take ( 4.5 x 10^7 J /
kg ) / (3 W / kgC) = 1.5 * 10^7 s or less than half a year for the
greenhouse gas to heat up as much as the thermal heating from the oil.
Of course, this CO2 is accumulating in the atmosphere.
If you think the airborne fraction on the margin, is around 0.5 over
the first thousand years, giving you about the radiative heating each
year equivalent to the chemical heating from burning. Then you get a
few hundred thousand years with several fold less heating, with a
cumulative radiative heating on the order of 100,000 times the direct
chemical heating. (I am not going to quibble about small integer
multipliers one way or the other.)
Of course, all of this heat will not go into melting ice.
(I think that 75 was the ratio of current atmospheric CO2 radiative
forcing to direct heating from fossil fuel burning, but I would need to
go back to check.)
___________________________________________________
Ken Caldeira
Carnegie Institution Dept of Global Ecology
260 Panama Street, Stanford, CA 94305 USA
[email protected];
[email protected]
http://dge.stanford.edu/DGE/CIWDGE/labs/caldeiralab
+1 650 704 7212; fax: +1 650 462 5968
On Thu, Nov 19, 2009 at 3:08 PM, Ron Larson <[email protected]>
wrote:
Dave
(cc Ken and list):
Thanks to Dave.
1. Since I doubt very much that the computation shown included
anything on CO2 effects, I hope Ken can weigh in on this, per the
discussion last week re: http://climateprogress.org/wp-content/uploads/2009/11/Warming-burning-091018.pdf
2. The answer might be 100,000 times larger - but that might exhaust
the supply of glaciers.
3. Would Exxon today say that one day's worth of melting was
calculated properly. That we are only talking of an insignificant
addition of only about 75/365 (only about another 20%, assuming we
don't worry about whether today's energy consumption is impacting any
glacier tomorrow.) (Ken had a factor of 75 for 1 year).
4. I haven't had any luck logging on to to leave a comment at the
Grist site, so hope someone will. One chap has shown a multiplicative
factor of 65 - which looks like he has calculated for a year.
Ron
Hawkins, Dave wrote:
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