Calculation of entropy change

  • #1

Homework Statement


A solid metallic cube of heat capacity S is at temperature 300K. It is brought in contact with a reservoir at 600K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the universe after reaching the thermal equillibrium is

A. 0.69S
B. 0.54S
C. 0.27S
D. 0.19S

[Answer : 0.19S]

Homework Equations


(heat supplied)=SΔT

Q = SΔT

Change in entropy = (change in heat Q)/T

ΔE = Q/T

[I have taken entropy as E rather than usual S since S is already taken for heat capacity]

The Attempt at a Solution


[/B]
Should I take reservoir as an infinite pool of temperature? Then, I get

Q = SΔT
Q=S*(600-300) since, at thermal equillibrium the temperatures are same
Q=300S
ΔE=(300/600)=0.5 which is not the answer

If I take reservoir which loses temperature, I am unable to continue with the problem since I do not know the heat capacity of it.

Is there a different approach to this problem?
 

Answers and Replies

  • #2
Quantum Defect
Homework Helper
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Homework Statement


A solid metallic cube of heat capacity S is at temperature 300K. It is brought in contact with a reservoir at 600K. If the heat transfer takes place only between the reservoir and the cube, the entropy change of the universe after reaching the thermal equillibrium is

A. 0.69S
B. 0.54S
C. 0.27S
D. 0.19S

[Answer : 0.19S]

Homework Equations


(heat supplied)=SΔT

Q = SΔT

Change in entropy = (change in heat Q)/T

ΔE = Q/T

[I have taken entropy as E rather than usual S since S is already taken for heat capacity]

The Attempt at a Solution


[/B]
Should I take reservoir as an infinite pool of temperature? Then, I get

Q = SΔT
Q=S*(600-300) since, at thermal equillibrium the temperatures are same
Q=300S
ΔE=(300/600)=0.5 which is not the answer

If I take reservoir which loses temperature, I am unable to continue with the problem since I do not know the heat capacity of it.

Is there a different approach to this problem?
Bits are right, but a lot of what you have is incorrect.

BTW this is a HORRIBLE problem -- it seems like the person who wrote the problem is trying to confuse you with the heat capacity = "S' garbage.

dS = dqrev/T

For the block, to calculate Delta S, you will need to integrate the above over the temperature range.

For the bath, assuming that the bath has a huge heat capacity, and does not change temperature (T is constant), then Delta S bath = Qbath/T

To find Qbath, you need to find Qblock (Qbath = - Qblock -- energy is conserved). Qblock can be found from "S" Delta T, as you have noted.

Delta S Universe = Delta S block + Delta S bath
 
  • #3
Thanks a lot. I got it.

This is actually from a question paper that I was practicing.
 

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