GR9677 #74



Alternate Solutions 
asdfman 20091105 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.  

Comments 
q 20191009 14:08:02  Sevilla: 1224, 39 pointsFollow Damon Salvadore sur TwitterFOR PLUS NOUVELLES ET MISES à JOUR SPORTS, cliquez sur ce lien pour visiter LATIN POST.\r\nq http://www.eurotrends.it/en6.asp   physicsphysics 20111011 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 20111011 08:24:13 
Sorry. I confused two cases. Just reversible case is zero. irreversible case is S>=0.

  mrTrig 20101105 13:50:32  yosun, you can reduce the logarithms much faster by simply knowing that addition of two results in multiplication of arguments.   asdfman 20091105 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 20091009 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 20091011 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.

  vinograd19 20070202 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 20071007 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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