Energy vs. entropy plot for a Carnot cycle

[quasar_4]

Homework Statement

I'm trying to understand how one would plot this.

Homework Equations

dS = dQ/T
dU = dQ-dW (W is work done by the system, Q is heat absorbed by system, U is energy)
For a closed cycle, dU = 0.
Carnot Cycle:
1- isothermal expansion at Th
2- adiabatic expansion to Tc (lower temp)
3- isothermal compression at Tc
4- adiabatic compression to Th

The Attempt at a Solution

So in the perfectly idealized case, no new entropy is created during the Carnot cycle. For steps 1, 3, the change in entropy would be given by dS = (1/T) dQ, which could theoretically be integrated if you knew the amount of work being done by the system at each step. And overall for the whole system dU = 0 means dQ = dW for the entire system.

For steps 2, 4, the fact that they're adiabatic means no heat is transferred and there is thus no change in entropy in these steps.

So I can see that for steps 2, 4, the entropy remains a constant. Because dU = 0, the area under the U-S curve must be zero. But how do I deal (qualitatively) with steps 1,3 without knowing anything else about Q or W in these two steps? I know that for a closed cycle, the area enclosed should equal the work done by the system. Do I somehow use this?

 

[mark726]

To plot this, you can use a graph with temperature on the x-axis and entropy on the y-axis. For steps 1 and 3, you can plot a horizontal line at the temperature Th, since there is no change in entropy during these steps. For steps 2 and 4, you can plot a vertical line at the temperature Tc, since there is no change in temperature during these steps. This will give you a rectangle on the graph, representing the Carnot cycle.

To qualitatively deal with steps 1 and 3, you can consider the fact that for an isothermal process, the change in entropy is directly proportional to the heat absorbed by the system. So, if you know the heat absorbed during these steps, you can plot a line with a positive slope connecting the horizontal lines for steps 1 and 3. The exact value of the slope will depend on the amount of heat absorbed.

You can also use the fact that for a closed cycle, the area under the U-S curve must be zero. This means that the area of the rectangle on the graph must also be zero, which can help you determine the relative magnitudes of the heat absorbed during steps 1 and 3.

Overall, plotting the Carnot cycle on a temperature-entropy graph can help you visualize the different processes and their effects on the system's energy and entropy. Keep in mind that this is a simplified representation and in reality, there may be other factors at play that can affect the plot.