# Change in entropy of a heat engine

• castrodisastro
In summary, the conversation discusses the change in entropy of a metal rod in contact with two thermal reservoirs at different temperatures. The total change in entropy of the system is calculated, but there is confusion over the change in entropy of the rod itself due to the lack of an equilibrium state. It is suggested that the correct answer may be undefined or 0 J/K.
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?

TC=113K
TH=695K

## Homework Equations

Δstotal=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 ΔSH=Q/TH=(7160J)/(695K)=10.3453J/K

for ΔSC=Q/TC=(7160J)/(113K)=63.63J/K

If the total entropy of the system is

ΔStot=(-ΔSH)+ΔSC

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.

castrodisastro said:

## 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?

TC=113K
TH=695K

## Homework Equations

Δstotal=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 ΔSH=Q/TH=(7160J)/(695K)=10.3453J/K
That should be -10.3453 J/K

for ΔSC=Q/TC=(7160J)/(113K)=63.63J/K

If the total entropy of the system is

ΔStot=(-ΔSH)+ΔSC
ΔStot = ΔSH+ΔSC

Since ΔSH 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

Last edited:
1 person
Andrew Mason said:
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.

castrodisastro said:
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

Andrew Mason said:
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?

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

Chet

1 person
castrodisastro said:
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/Tin - ∫|dQ|rev-out/Tout. 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

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

## 1. What is entropy and how does it relate to heat engines?

Entropy is a thermodynamic quantity that measures the disorder or randomness of a system. In heat engines, entropy is related to the efficiency of the engine, as an increase in entropy results in a decrease in efficiency.

## 2. How does a heat engine produce change in entropy?

A heat engine produces change in entropy by converting heat energy into mechanical work. This conversion leads to a change in the distribution of energy within the system, resulting in an increase in entropy.

## 3. What factors affect the change in entropy of a heat engine?

The change in entropy of a heat engine is affected by the temperature difference between the hot and cold reservoirs, the amount of heat energy input, and the efficiency of the engine.

## 4. Can the change in entropy of a heat engine be reversed?

No, according to the second law of thermodynamics, the change in entropy of a heat engine cannot be fully reversed. Some energy will always be lost to the surroundings in the form of heat, resulting in an increase in entropy.

## 5. How can the change in entropy of a heat engine be calculated?

The change in entropy of a heat engine can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the heat energy input, and T is the temperature of the hot reservoir.

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