saturday-review-session-solutions-1

# saturday-review-session-solutions-1 - 7B MT1 Sunday Review...

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7B MT1 Sunday Review Session Solutions Patrick Varilly and Steven Byrnes 1 Liphardt, Spring 05, Prob 2 a) Notice that c L and c S are plain heat capacities, not speciﬁc (per kg) heat capacities (look at their units). Deﬁne the following quantities. Q L := heat gained by liquid helium (negative if heat is lost by the helium); T L := instantaneous temperature of liquid helium; T L,i := initial temperature of the liquid helium; T L,f := ﬁnal temperature of the liquid helium; S L := entropy of the liquid helium . Also, deﬁne analogous quantities for the solid, with an S instead of an L subscript. Now, since no work is being done, we have d Q L = c L d T L = aT 3 L d T L . Integrate this equation from the initial to the ﬁnal temperatures to get Q L = Z T L,f T L,i aT 3 L d T L = a 4 ( T 4 L,f - T 4 L,i ) < 0 . Numerically, Q L = - 0 . 024 J . b) Notice that T L,f = 0 . 5 K now, not 0 . 7 K. The calculation here is analogous to (b): Q S = Z T S,f T S,i bT - 2 S d T S = - b ( T - 1 S,f - T - 1 S,i ) . Since the system is isolated, there is no net heat ﬂow: Q L + Q S = 0 . Substituting our previous results in this last equation yields a 4 ( T 4 L,f - T 4 L,i ) - b ( T - 1 S,f - T - 1 S,i ) = 0 . When the liquid helium and the solid reach equilibrium, their temperatures will be equal. We thus set T L,f = T S,f = T f . Also, set T S,i = T 0 . We obtain the following equation a 4 b ( T 4 f - T 4 L,i ) = ( T - 1 f - T - 1 0 ) . 1

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7B MT1 Sunday Review Session Solutions Patrick Varilly and Steven Byrnes Solving for T 0 gives us the desired result. T 0 = h T - 1 f - a 4 b ( T 4 f - T 4 L,i ) i - 1 Numerically, T 0 = 0 . 25 K . This is a sensible answer: T 0 < T f < T L,i . c) The ﬂow of heat from the “hot” liquid helium to the “cold” solid is irreversible. However, we can exactly calculate the entropy change for the process with the following argument. The entropy of the liquid helium is a state function. Thus, the change in its entropy is equal for any process that cools the liquid helium, in particular, a reversible one. The change in the entropy of the solid can be calculated in the same way. The total entropy change is just the sum of the entropy changes of the liquid helium and the solid. So, the inﬁnitesimal entropy change of the liquid helium is given by the reversible process formula d S L = d Q L T L . Substituting d Q L = c L d T L as above and integrating, we get Δ S L = Z T L,f T L,i aT 3 L T L d T L = a 3 ( T 3 L,f - T 3 L,i ) . Similarly, for the solid,
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## This note was uploaded on 06/30/2010 for the course PHYSICS ?? taught by Professor Zettle during the Spring '08 term at University of California, Berkeley.

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saturday-review-session-solutions-1 - 7B MT1 Sunday Review...

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