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

Recall the definition of entropy to be $dS = dQ/T$. The heat is defined here as $dQ = m c dT$, and thus $S = \int mcdT/T$.

One is given two bodies of the same mass. One mass is at $T_1=500$ and the other is at $T_2=100$ 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 $Q_1 = -Q_2 \Rightarrow T_f = (T_1+T_2)/2=300$.

The entropy is thus $S = \int^{T_f}_{T_1} mcdT/T + \int^{T_f}_{T_2} mcdT/T = mc \left ( \ln(3/5) + \ln(3) \right) = 2mc \ln 3 - mc \ln 5 = mc(\ln 9 - \ln 5) = mc \ln(9/5)$, 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 $ln$ - 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 $ln$ - 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
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physicsphysics
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 -$Delta q/T + \Delta q/T =$ 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. $mc(\ln{(\frac{3}{5})} +\ln{3}) = mc\ln{(\frac{9}{5})}$
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 $ln$ - 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 $ln$ - 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 $dS \ge dQ/T$ 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.

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