Change in entropy of a heat engine

[castrodisastro]

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_C=113K
T_H=695K

Homework Equations

Δs_total=Δs₁+Δs₂
ΔS=\frac{Q}{T}

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/K

for ΔS_C=Q/T_C=(7160J)/(113K)=63.63J/K

If 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.

 

[Andrew Mason]

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_C=113K
T_H=695K

Homework Equations

Δs_total=Δs₁+Δs₂
ΔS=\frac{Q}{T}

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/K

That should be -10.3453 J/K

for ΔS_C=Q/T_C=(7160J)/(113K)=63.63J/K

If the total entropy of the system is

ΔS_tot=(-ΔS_H)+ΔS_C

ΔS_tot = ΔS_H+ΔS_C

Since ΔS_H is negative, the answer is correct.

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.

Your math is correct. But this may be a trick question.

Entropy is a state function. But a state is defined as an equilibrium state. Is the rod in an equilibrium state?

AM

 

[castrodisastro]

Entropy is a state function. But a state is defined as an equilibrium state. Is the rod in an equilibrium state?

AM

Well the rod has uniform temperature. Since it isn't changing I want to say that it is in an equilibrium state otherwise it would be heading towards equilibrium and the rod's temperature will head towards the lower temperature.

 

[Andrew Mason]

Well the rod has uniform temperature. Since it isn't changing I want to say that it is in an equilibrium state otherwise it would be heading towards equilibrium and the rod's temperature will head towards the lower temperature.

I interpret the statement "no change in temperature along the rod" to mean that the temperature gradient along the rod length did not change with time. You appear to interpret it as the temperature of the rod is the same at all points along the rod i.e. there is no temperature gradient along the rod length. But if that was the case, there would be no heat conducted through the rod - it would have to be a perfect insulator.

AM

 

[castrodisastro]

I interpret the statement "no change in temperature along the rod" to mean that the temperature gradient along the rod length did not change with time. You appear to interpret it as the temperature of the rod is the same at all points along the rod i.e. there is no temperature gradient along the rod length. But if that was the case, there would be no heat conducted through the rod - it would have to be a perfect insulator.

AM

I see what you are saying. So if 7190 J are conducted through the rod then I should calculate the change in entropy for the rod as Q/T but to know the temperature don't I need to know the length of the rod, the material of the rod, and the thermal conductivity of said material?

 

[Chestermiller]

Maybe the issue is something simple like significant figures? 0.00 J/K? Just a crazy thought.

Chet

 

[Andrew Mason]

I see what you are saying. So if 7190 J are conducted through the rod then I should calculate the change in entropy for the rod as Q/T but to know the temperature don't I need to know the length of the rod, the material of the rod, and the thermal conductivity of said material?

I am thinking that the correct answer may be: the change in entropy of the rod is undefined because the rod is not in an equilibrium state i.e. it is a trick question.

[Addendum: The change in entropy of the rod would be defined as ∫|dQ|_rev-in/T_in - ∫|dQ|_rev-out/T_out. But Tin and Tout refer to the temperature of the whole rod assuming an equilibrium state, not the temperature of the rod at the point of contact. The rod does not have an equilibrium temperature. So we really cannot determine the change in entropy of the rod from this process. ]

AM

 

[castrodisastro]

Thanks! Getting the answer correct was just a matter of significant digits. It was 0.00 J/K. I now understand why. Thanks again.