This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 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 ) < . Numerically, Q L = . 024 J . b) Notice that T L,f = 0 . 5 K now, not . 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 . We obtain the following equation a 4 b ( T 4 f T 4 L,i ) = ( T 1 f T 1 ) . 1 7B MT1 Sunday Review Session Solutions Patrick Varilly and Steven Byrnes Solving for T gives us the desired result. T = h T 1 f a 4 b ( T 4 f T 4 L,i ) i 1 Numerically, T = 0 . 25 K . This is a sensible answer: T < 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 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, S S = Z T S,f T S,i bT 2 S T S d T S = b 2 ( T 2 S,f T 2 S,i ) . Adding these two up and making the substitutions in (b), we obtain the final result: S universe = a 3 ( T 3 T 3 L,i ) b 2 ( T 2 T 2 S,i ) Numerically, S universe = +0 . 053 J K 1 . A positive entropy change of the universe is exactly whats expected when heat flows from a hot object to a cold one....
View Full
Document
 Spring '08
 Packard
 Heat

Click to edit the document details