- #1

pkossak

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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!

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- Thread starter pkossak
- Start date

- #1

pkossak

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- 0

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!

- #2

pkossak

- 52

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- #3

dlaszlo88

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- #4

Andrew Mason

Science Advisor

Homework Helper

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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).pkossak said:

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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