Entropy of a System: Answers Explained

[shally]

Homework Statement

A piece of metal of heat capacity 500 J/K assumed to be independent of temperature is at 500 K. The metal piece is cooled to 300K in two steps: it is first plunged into a liquid bath at 400 K. After cooling it is plunged into a colder liquid bath at 300 K. The metal piece is then heated to 500 K in two steps: it is plunged into a liquid bath at 400 K first and then into a liquid bath at 500 K. During the cooling-heating process, the metal piece and the liquid baths gain or lose entropy. The total change in entropy of the system (the metal piece and the liquid baths) is
1. 1500 ln (5/3)J/K
2. Zero
3. –200 J/K
4. +200J/K

Homework Equations

Delta S = Delta Q / T

The Attempt at a Solution

Don't get one of the options provided.

Please help.

 

[Mapes]

Hi shally, welcome to PF. I suspect there's an error in the question or the answers. Did you get 66.7 J/K?

 

[shally]

Thank you for your effort. My ans. was -66.7 J/K.

I used the formula Delta S = Delta Q/T

Delta S = 500 (-100/400 -100/300 + 100/400 +100/500) = -200/3 J/K = -66.7J/K

 

[Mapes]

So the entropy of the entire system decreases? This would be quite unusual.

 

[D_Tr]

Hi Shally, I think you got the signs wrong and Mapes is correct. Because in the two first terms, the heat goes TO the baths and in the two last terms the heat goes FROM the baths to the metal, so the first 2 terms should be + and the two last should be -. I am also getting 66.7 J/K, since the total entropy change of the metal is zero and we only have the heat flows to/from the baths...