Ok here’s one problems I have with AGW.?

According to the IPCC the simplified expression radiative forcing for CO2 is

http://www.grida.no/publications/other/ipcc_tar/?src=/climate/ipcc_tar/wg1/222.htm#tab62

(yahoo won't let me do the symbols correctly so I link to the IPCC report)

If I did my math correctly that means we should have warmed up by about 5c.


Since we haven't warmed up by anything close to that, did I do my math incorrectly, did the IPCC revise the math, is something else happening to prevent that forcing, or is the IPCC just wrong?

2009-09-02T16:09:12Z

to James E

Did they correct this in the fourth assessment?

2009-09-02T16:10:45Z

Its at 6.3.5 Simplified Expressions Table 6.2: scroll down a little and you'll see it.

2009-09-02T16:37:31Z

OK here you go

C = 380

C0 = 280

Which give us

5.35*ln(380/280) = 1.63 Wm-2

Now 1 Watt is one Joule.second right?

365.25*24*60*60 = 31,557,600 number of seconds in a year (Note I use 365.25 to allow for leap year)

1.63 * 31,557,600 = 51,438,888

It takes 1,005 Joules to warm 1 kg air 1c

So 51,438,888/1005 = 51182.973

One-meter squared column of the atmosphere has a mass of 10,000kg

So 51182.973 / 10000 = 5.1182


Right?

2009-09-02T17:30:39Z

to gcnp58

But wouldn't that just mean there would be a lot more evaporation and with water vapor being a green house gas cause even more warming, so shouldn't we be seen even more warming then the 0.78c that's been reported and would we be seeing a lot more rain, expectantly in area close to the ocean?

2009-09-02T17:48:19Z

By the way I'm working out how much the ocean temperature should have risen assuming the numbers are close, but it's late so I'll work on it later.

2009-09-02T18:14:03Z

Sorry it seems you already did, thanks.

2009-09-02T18:21:01Z

To bucket22

No I didn’t get them from junkscience.com Steven Milloy, it was brought up in another board and they didn’t work it out they linked to the IPPC page and stated we should have warmed up by 5c. If they got it from junkscience.com they didn’t say, I worked out the math. That’s why I took the simplest formula :)

But if it’s from junkscience.com then they should at least get the credit for pointing this out, even if they are wrong. I’ll search for it and if I find it I’ll post a link.

Thanks.

2009-09-02T18:22:39Z

Looks like you already found the link, thanks.

2009-09-02T18:34:19Z

Just so you know here's the links I did use

I got the one watt = one joules from http://en.wikipedia.org/wiki/Watt

http://www.engineeringtoolbox.com/specific-heat-capacity-d_391.html

is where I got 1005 joules to raise the temp of 1 kg of air.

And the mass of a column of air I got from
http://onlinephys.com/pressure4.html

The last two links I got from online friends when I was working on this, had I known someone else had done the work it would have save me a lot of time.

2009-09-02T19:09:15Z

To Keith P

I used 380 because it made the math simpler, I figured it would get me close, had I been way off from what that person had said I would have gone back and used 386. Like I said I’m trying to work out how much the ocean should have warmed, I know that it takes about 4,200 joules to warm one kg of water 1c and that the sun can only warm the top 100 meters. So now I’m working out how many kg or water in a column of water 1 meter sq by 100 meters. That is assuming the ocean will be absorbing the heat at the same rate as the formula says. That might be a faulty assumption, but I’ll work it out anyway.

Keith P2009-09-02T18:48:36Z

Favorite Answer

A for effort!

You're reading the science, and doing the math correctly. In particular, your computation of CO2 radiative forcing is correct. (I could quibble a bit and point out that CO2 is currently 386, and also that equation 3 is actually a bit more accurate than equation 1 ... but the differences are small.)

The problem you have is that the extra 1.6 W/m² isn't going directly into the air. I't's going mostly into the sea. And that creates a lot of complications, because then you have to figure how deep a surface layer of ocean is directly affected, and how much of that heat goes from the surface layer into the air, and how much goes into the deeper ocean.

Further, the fraction that does end up in the air doesn't go all the way up to the top of the atmosphere. The whole atmosphere isn't warming, just the surface and the troposphere, while the stratosphere is cooling (because more greenhouse gases means less IR is getting up that high).

And yet another complication is the effects of various feedbacks in the climate system, which amplify the direct forcing. These feedbacks are both positive and negative, although when taken together the net is positive.

All that is pretty complex, which is why climate scientists use complex models of the climate system. If you want a simpler way to do things, you can just apply a factor known as "climate sensitivity" to your radiative forcing result. But because of all these issues, however, the value of climate sensitivity isn't well known either. The IPCC TAR gave it as .5 K/W/m² ± 50%. Which means we should expect 1.6 x 0.5 = 0.8° C increase, about what we have actually seen.

David2009-09-02T17:28:34Z

I'm sure someone else will be able to give a better answer, but the first obvious problem I see is that you multiplied it by the number of seconds in a year. Why?

The radiative forcing is a measurement of the flux imbalance. You can do a quick and dirty estimate of the expected temperature difference with the Stefan Boltzmann law of blackbody radiation.
http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Planetary_Equilibrium_Temperature

j* = (5.67 E -8)(T^4)

Starting with an average Earth temperature of 288 K, we get a radiative flux of 390.079 w/m^2--a bit higher than the actual average flux of 342, but it'll do.

390.079 + 1.63 = 391.709 W/m^2. Working backwards:

391.709 = (5.67 E -8)(T^4), solve for T to get 288.3 K.

So by the extremely simplified calculation you can expect a +0.3 C rise from the 1.63 w/m^2 forcing. Of course this is complicated by feedbacks, the earth's albedo, the lag time for RF to equate with temp, band saturation of CO2, etc.

But at least with this most simple of calculation you can see that the estimates are not horribly wrong.

I cant believe this website2009-09-02T17:39:41Z

First I would like to say I am no expert on this subject or could even do the math your link contains. Your calculation of 5c may be correct. I recently seen an article about how the pollution in the air is actually counter-acting the effects of global warming, thus slowing the rate of temperature rise by blocking sunlight. Other things contribute to this effect such as when Mt. St. Helen's erupted and spewed ash into the atmosphere. Just not on the grand scale of when the dinosaurs became extinct. Like I said,I couldn't do the math, but wondered if the equation takes this into consideration. Just thought this might be why 5c is high.

bucket222009-09-02T17:42:02Z

EDIT

I was thinking that these calculations were Richard's original thought. Looks like it's a junkscience.com Steven Milloy thing. At the very least Richard should credit the tobacco lobbyist for the material.

http://www.junkscience.com/Greenhouse/forcing.html

-----------------

Your implied estimate of climate sensitivity is absolutely enormous! Thank goodness the situation isn't that dire.

Your radiative forcing to global mean temperature change analysis is quite simplistic. See the Stefan-Boltzmann law for starters.

http://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law#Temperature_of_the_Earth

From here we can calculate about 1 C of forcing from a doubling of CO2. Then, of course, there's feedbacks, which get us in the 2-4.5 C range. Of course, you then have to consider equilbrium - the time lag between the forcing and the temperature change. Lastly, we have to consider other forcings, such as the cooling effect of anthropogenic sulfates.

http://www.realclimate.org/index.php/archives/2006/11/cuckoo-science/

gcnp582009-09-02T17:20:48Z

Your calculations are kind of right, but you need to factor in the thermal mass of the ocean and time lag for the heat to mix into it. The forcing isn't uniform with latitude, so neither will the heating.

The disconnect you're observing is because the planet hasn't yet equilibrated to the rise in the forcing. Once that happens, the expected rise in temperature of the tropics, for instance, will be about 5 K, just as you calculated. The equilibration will take a while, on order of a few hundred years.

http://www.ncdc.noaa.gov/paleo/globalwarming/temperature-change.html

There is nothing fundamentally wrong with the science. However, a lot of people want there to be because they can't accept it's correct.

Show more answers (5)