Maximum work done by a Carnot engine

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Taking the engine to be a Carnot engine,

$\eta = \frac { T_h – T_c } {T_h} = \frac { W} {Q_h}$ ...(1)

$Q_h = C(T_h – T_f) ...(2) \\ Q_c = C(T_f – T_c) ...(3) \\ W = Q_h – Q_c = C(T_h + T_c – 2 T_f)$ ...(4)Solving the above equations,

$T_f = \frac { 2T_c T_h }{T_h +T_c}$ ...(5)This solution does not lead me to any of the given option.

Is this correct?

 

[TSny]

Does $\eta$ remain constant during the process?

 

[Chestermiller]

Algebraically, in terms of $T_h$, $T_f$, and the heat capacity C, what is the entropy change of the hot system?

 

[Pushoam]

Does $\eta$ remain constant during the process?

As I understand: $\eta$ is not defined during a process. $\eta$ is defined for a process. For a Carnot engine working between temperatures $T_h$ and $T_f$, $\eta$ is defined as $\frac W{Q_h}$. This is what I have used.

I am not getting the intention behind this question. Please give some more hint.

 

[Pushoam]

The entropy change of the hotter system is equal and opposite to that of the colder as it is a reversible process.

$dS = C \frac { dT } T$ ...(1)

$\Delta S_h = - C \ln \frac { T_h}{T_f} = - \Delta S_c = - C \ln \frac { T_f}{T_c}$ ...(2)

$T_f = \sqrt{ T_h T_c}$ ...(3)

$W = Q_h – Q_c = C(T_h + T_c – 2 T_f)$ ...(4)

$W = C(\sqrt{ T_h } - \sqrt{ T_c})^2$ ...(5}

So, the answer is option (d).Why is using the efficiency equation not a correct step?

The engine is working between the two temperatures through a reversible cycle. So, it is a Carnot engine.

For calculating maximum work done, I have to take maximum efficiency. This is what I did in the OP.

 

[Chestermiller]

The entropy change of the hotter system is equal and opposite to that of the colder as it is a reversible process.

$dS = C \frac { dT } T$ ...(1)

$\Delta S_h = - C \ln \frac { T_h}{T_f} = - \Delta S_c = - C \ln \frac { T_f}{T_c}$ ...(2)

$T_f = \sqrt{ T_h T_c}$ ...(3)

$W = Q_h – Q_c = C(T_h + T_c – 2 T_f)$ ...(4)

$W = C(\sqrt{ T_h } - \sqrt{ T_c})^2$ ...(5}

So, the answer is option (d).Why is using the efficiency equation not a correct step?

The engine is working between the two temperatures through a reversible cycle. So, it is a Carnot engine.

For calculating maximum work done, I have to take maximum efficiency. This is what I did in the OP.

The temperatures of the two reservoirs are changing during the process. So you can't use the results for a single Carnot cycle with constant reservoir temperatures. You can't get from the initial state of this system to the final state reversibly with a single Carnot cycle. You need to use multiple Carnot cycles, each with slightly different pair of reservoir temperatures. And, for each of these multiple Carnot cycles, the efficiency is different.

 

[Pushoam]

The temperatures of the two reservoirs are changing during the process. So you can't use the results for a single Carnot cycle with constant reservoir temperatures. You can't get from the initial state of this system to the final state reversibly with a single Carnot cycle. You need to use multiple Carnot cycles, each with slightly different pair of reservoir temperatures. And, for each of these multiple Carnot cycles, the efficiency is different.

I didn't realize that the temperatures of the reservoirs are changing and the efficiency eqn is valid for constant reservoir temperatures even after solving the question.
Thanks for it.