## 9.101 (b) [ENDORSED]

$\Delta S = \frac{q_{rev}}{T}$

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Andy Liao 1B
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### 9.101 (b)

Problem:
9.101 A heat pump heats a house in winter by extracting heat from the cold outdoors and releasing it into the warm interior. For the transfer of a given amount of heat, (a) how do the entropies of the interior and exterior of the house change (increase or decrease)? (b) Which change is greater? Assume that the temperatures inside and outside the house do not change. Explain your answers.

I understand part (a) because the answer for this part relates to the idea that heating increases thermal disorder. However, I can't seem to grasp the concept behind part (b). The solution is ΔSinterior < ΔSexterior. Can someone please explain how the solutions manual arrived at this answer?

Chem_Mod
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### Re: 9.101 (b)  [ENDORSED]

Remember in class I gave the example of heat going from a piece of hot copper, T(hot), to the surroundings, T(cold).

Since ΔS = q/T and T(hot) > T(cold) then ΔS(copper) < ΔS(surroundings).

Same for ΔS(interior) < ΔS(exterior) because T(interior) > T(exterior).

This is a great topic to discuss in Workshops, Discussion Sections, Step-Up Sessions, Office Hours, Review Sessions, Student Study Groups, etc.

Vincent Kim 2I
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### Re: 9.101 (b)

Just further developing on the previous reply, since delta S is q/T, the greater T is, the smaller delta S is. It's simply an inverse relationship

Sally Nason - 1K
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### Re: 9.101 (b)

I had trouble with this one as well. You would think that since the outdoors is so much larger that any temperature change wouldn't affect the entropy as much but the question really just wants you to look at the equation.

April P 1C
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### Re: 9.101 (b)

Also from chegg, it states that entropy change depends on space for disorder, so since there is more space in the exterior there will be more disorder

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