Entropy Changes

  • #1

Homework Statement


A solid with constant heat capacity is warmed from 20 C to 100 C in 3 different ways:
a) by placing it in contact with a large reservoir at 100 C
b) by placing it first in cantact with a large reservoir at 50 C until it reaches that temperature and then in contact with a reservoir at 100 C
c) by operating a reversible heat engine between it and the reservoir at 100 C

For each of these cases, what are the entropy changes of the solid, the reservoirs and the total entropy change?


Homework Equations


[tex]\int_{T_i}^{T_f} \frac{C} {T} dT [/tex]


The Attempt at a Solution



Because entropy is a state function, it is path independent and thus shouldn't the 'total' entropy change for all 3 be zero? Also, to actually calculate the entropy change in the gas we can simply integrate over the temperature difference. And the entropy change in the reservoirs should be the negative of that. Corrections please! :!!)
 

Answers and Replies

  • #2
Dick
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The entropy is indeed a state function. So you are correct that entropy in the solid doesn't depend on the path. But entropy is not conserved. So you can't conclude that anything is the negative of anything.
 
Last edited:
  • #3
Andrew Mason
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Because entropy is a state function, it is path independent and thus shouldn't the 'total' entropy change for all 3 be zero? Also, to actually calculate the entropy change in the gas we can simply integrate over the temperature difference. And the entropy change in the reservoirs should be the negative of that. Corrections please! :!!)
Why? dS = dQ/T where dS is the change in entropy of the object, dQ is the heat flow into or out of the object, and T is the temperature of the object. If I have an object at 0 degrees C (273K) in contact with a reservoir at 100 degrees C (373K), the change in entropy of the object is dQ/273 and the change in entropy of the reservoir is -dQ/373.

AM
 
  • #4
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Entropy is a state function, so it doesn't depend on the path. But what do depend on the path is the number of items that will suffer a change in entropy. First case is the system and the hot reservoir, second case has a second reservoir that will be involved and in the third case you will have some work produced.
There is another and important thing to take into account: changes on S are usually calculated through the heat transfer, but heat transfer is dependent on path.
 
  • #5
In response to Andrew Mason:

Sorry I'm having a little trouble understanding this concept of entropy change. When we have the entropy change of changing temperture, we just throw in the initial temperature of the object to the equation dS = dQ/T? Don't need need to integrate? If we just solve that dS question, how do we bring the heat capacity into the equation? For dQ, do we sub in dE (from first law) and C = dE/dT? Thanks!
 
  • #6
Okay. I found the expression for the entropy change of the solid in the first path:

dS = C ln ( Tf/Ti )

and i see that the entropy change in the reservoir must be dS = dQ / 373. But is that as 'simple' as I can get the answers? Or can I actually dQ?

Edit: I found a way get dQ: C = dQ/dT thus dQ = C( Tf - Ti )
 
Last edited:
  • #7
Thanks guys, I got all the answers!

For a) I found:

Change in S total = C (0.026)

For b) : dS = C (0.013)

For c) : dS = 0
 

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