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Formula for heat transfer

  • Thread starter pkossak
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52
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The surface of the Sun is at approximately 5700 K and the temperature of the Earth's surface is about 290 K. What total entropy change occurs when 1000 J of heat energy is transferred from the sun to the earth?
a. 2.89 J/K b. 3.27 J/K c. 3.62 J/K d. 3.97 J/K

I don't know what formula to use. I tried heat transfer

deltaS = mcln(T/(T+deltaT))

but I don't know know what to use for mc?! Pretty confused. Thanks for any help!
 

Answers and Replies

52
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anyone? I would really appreciate it. I'm not looking for simply answers, I want to understand it. Exam coming up. Thanks a ton if you can help! Also, some please look at my other post - Ideal Gas.
 
9
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The forumula that you would want to use is Gibb's Free Energy Equation. deltaG = Heat of Formation - (TdeltaS) then with a little calculus you can reach the forumula S = e^-deltaG/RT and don't forget to change to the correct units.
 
Andrew Mason
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pkossak said:
anyone? I would really appreciate it. I'm not looking for simply answers, I want to understand it. Exam coming up. Thanks a ton if you can help! Also, some please look at my other post - Ideal Gas.
Entropy is the ratio of heat flow to temperature. If the flow all occurs at an arbitrarily small difference in temperature (ie reversibly as in the Carnot cycle) the entropy of the outflow is equal to the entropy of the inflow and there is no net change in entropy. When heat flows from a hot object to a cooler one (ie. non-reversibly), there is a change in entropy. So entropy is seen as a useful measure of the reversibility of the flow (or of the amount of work required to reverse the flow).

The entropy change (loss) of the sun is [itex]\Delta S_{sun} = -\Delta Q/T_{sun}[/itex]. The gain in entropy of the earth is [itex]\Delta S_{earth} = +\Delta Q/T_{earth}[/itex]

So, the total change in entropy is:

[tex]\Delta S = \Delta S_{earth} + \Delta S_{sun} = \Delta Q(\frac{1}{T_{earth}} - \frac{1}{T_{sun}})[/tex]

Work that out and you will have the answer.

AM
 

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