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GR9677 #74
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Thermodynamics}Entropy

Recall the definition of entropy to be . The heat is defined here as , and thus .

One is given two bodies of the same mass. One mass is at and the other is at before they're placed next to each other. When they're put next to each other, one has the net heat transferred being 0, thus .

The entropy is thus , as in choice (B).  Alternate Solutions
 asdfman2009-11-05 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.Reply to this comment 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.
 physicsphysics2011-10-11 08:24:13 Sorry. I confused two cases. Just reversible case is zero. irreversible case is S>=0. 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. 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.
 kroner2009-10-11 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. 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.
 hungrychemist2007-10-07 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.      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$