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GR9677 #74
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Alternate Solutions |
asdfman
2009-11-05 00:11:38 | Used POE.
Expect there to be a change in entropy - E is out.
Know the typical integral to find S will involve a  - A is out.
Expect the entropy to increase for the system - D is out.
Based on the numbers, expect there to be a ratio as an argument of the - C is tentatively out.
I'd come back to it if there was time and run the math to see if B is actually correct. If not this logic narrows it down to 2 choices. |  |
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Comments |
q
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2011-10-11 08:21:33 | I think this problem is a little bit tricky. To obtain the equation form of integral, the system should be changed reversibly. If this system is irreversible, the entropy change of this system is - zero.
physicsphysics
2011-10-11 08:24:13 |
Sorry. I confused two cases. Just reversible case is zero. irreversible case is S>=0.
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| | mrTrig
2010-11-05 13:50:32 | yosun, you can reduce the logarithms much faster by simply knowing that addition of two results in multiplication of arguments. } +\ln{3}) = mc\ln{(\frac{9}{5})}) | | asdfman
2009-11-05 00:11:38 | Used POE.
Expect there to be a change in entropy - E is out.
Know the typical integral to find S will involve a - A is out.
Expect the entropy to increase for the system - D is out.
Based on the numbers, expect there to be a ratio as an argument of the - C is tentatively out.
I'd come back to it if there was time and run the math to see if B is actually correct. If not this logic narrows it down to 2 choices. | | dstahlke
2009-10-09 10:53:51 | But isn't it true that  with equality only in the case where the process is reversible? This process doesn't seem reversible to me.
kroner
2009-10-11 20:16:00 |
In the context of the whole system it's not a reversible process. From that perspective dQ = 0 so you are correct that dS > dQ/T = 0. Clearly the change in entropy is positive.
But considering the two objects separately, they're each undergoing a reversible process (being uniformly heated or cooled). The change in entropy for each can be found by setting dS = dQ/T where dQ is the heat flowing into that object. Then you sum the changes contributed by each object.
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|  | vinograd19
2007-02-02 10:27:17 | I think there is a mistake in solution. One integral is positive, the other is negative. But in solution they are both positive.
hungrychemist
2007-10-07 21:34:32 |
Solution is correct. The sign of integral is determined solely from the limits of integration. Notice ln(3/5) itself is a minus number as expected.
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