- #1

castrodisastro

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## Homework Statement

One end of a metal rod is in contact with a thermal reservoir at

**695K**, and the other end is in contact with a thermal reservoir at

**113K**. The rod and reservoirs make up an isolated system.

**7190J**are conducted from one end of the rod to the other uniformly (no change in temperature along the rod).

a) What is the change in entropy of each reservoir?

b) What is the change in entropy of the rod?

c) What is the change in entropy of the system?

**T**

T

_{C}=113KT

_{H}=695K## Homework Equations

**Δs**

ΔS=[itex]\frac{Q}{T}[/itex]

_{total}=Δs_{1}+Δs_{2}ΔS=[itex]\frac{Q}{T}[/itex]

## The Attempt at a Solution

I was able to get answer part a) and part c) but I can't seem to understand why I can't figure out part b).

for

**ΔS**

_{H}=Q/T_{H}=(7160J)/(695K)=10.3453J/Kfor

**ΔS**

_{C}=Q/T_{C}=(7160J)/(113K)=63.63J/KIf the total entropy of the system is

**ΔS**

_{tot}=(-ΔS_{H})+ΔS_{C}Which is

**(-10.345J/K)+63.63J/K=53.28**

If I know that this value is the total entropy of the system(the online homework I'm doing counted this as correct), then the rod must have

**0**entropy right? Or else it would contribute somehow to the total entropy of the system and result in a different value than

**53.28 J/K**for total entropy. I have verified that

**53.28 J/K**is the correct answer for part c) so why is part b) not

**0 J/K**? Am I missing the concept of everything that takes place in a heat engine?? Thanks in advance.