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billiards

Exam Question

Hi people! Back from my holidays in Barcelona and Leeds, had a fantastic time, hope you didn't miss me too much Very Happy

Was tidying my desk and stumbled across an exam paper I took last year, there's one question (that I chose not to do in the exam) that I think is quite interesting - so I will share it here and would love to see some answers/comments.

Quote:

A neswpaper recently claimed that the effect of heat released by fuel burning was being ignored in global warming predictions. Investigate this claim by determining the time taken for a small increase deltaT in temperature of the atmospheric column to radiate to space. By treating the atmosphere as a column of density p and thickness h, show that its temperature T satisfies

dT/dt=-sigma.T4/cpph

Use this equation to determine the time taken for the atmosphere to equilibrate. Comment on your answer.

Stefan's constant, sigma=5.7*10-8 J m-2 K-1 s-1

Specific heat of air, cp=1005 J kg-1 K-1

Density of air at 273 K = 1.3 kg m -3
scpg02

You were gone? Razz
sneez

Quote:
A neswpaper recently claimed that the effect of heat released by fuel burning was being ignored in global warming predictions. Investigate this claim by determining the time taken for a small increase deltaT in temperature of the atmospheric column to radiate to space. By treating the atmosphere as a column of density p and thickness h, show that its temperature T satisfies

dT/dt=-sigma.T4/cpph

Use this equation to determine the time taken for the atmosphere to equilibrate. Comment on your answer.

Stefan's constant, sigma=5.7*10-8 J m-2 K-1 s-1

Specific heat of air, cp=1005 J kg-1 K-1

Density of air at 273 K = 1.3 kg m -3





dQ/dt = emissivity*sigma*area*T^4 ------>radiative heating

dQ= m*cp*dT ----->definition of specific heat (m=mass)

V=m/rho ------>relationship between mass and density

let V= A*h -area time hight => m=A*h*rho
then:

dT/dt = [ emissivity*sigma*area*T^4 ] / [m* cp]

hence: dT/dt = [ emissivity*sigmaT^4 ] / [h*rho*cp] ------->the areas divide out

Integrating we would get time inverselly proportional to Temp. Thats quite interesting, so let me hear if I made a mistake here.
Andre

here is the empirical outcome:

http://www.knmi.nl/~laatdej/2006joc1292.pdf

Quote:
...We confirm the presence of a temperature change–industrialization correlation by analyzing the data with an additional statistical method and further confirm the absence of the above correlation in climate model simulations of enhanced GHG warming. Our findings thus provide an important test of climate model performance on regional scales.

These findings suggest that over the last two decades non-GHG anthropogenic processes have also contributed significantly to surface temperature changes. We identify one process that potentially could contribute to the observed temperature patterns, although there certainly may be other processes involve...
Derek

Hi All,
I just knew there was a reason why I did not become a scientist.....
To answer the exam questions means spouting and "proving" AGW rubbish.
What happened to the idea that the earth can loose 5 times more heat than we and it can generate / recieve.
It's like trying to prove a flea deciedes which direction the dog goes in.
In would have to be an enourmous, totally unrealistic flea wouldn't it.
A flea called "an IPCC approved climate model" no doubt...  Laughing  Laughing  Laughing
scpg02

Derek wrote:

A flea called "an IPCC approved climate model" no doubt...  Laughing  Laughing  Laughing


Sucking the blood of the tax payer.

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