Discussion
Water at 300 K can be brought to 320 K using one of the following processes.
Process 1: Water is brought in equilibrium with a reservoir at 320 K directly.
Process 2: Water is first brought in equilibrium with a reservoir at 310 K and then with the reservoir at 320 K.
Process 3: Water is first brought in equilibrium with a reservoir at 350 K and then with the reservoir at 320 K.
The corresponding changes in the entropy of the universe for these processes are \(\Delta S_{1}\), \(\Delta S_{2}\) and \(\Delta S_{3}\), respectively. Then
Answer: B
The order is \(\Delta S_{3} \gt \Delta S_{1} \gt \Delta S_{2}\).
When water of heat capacity \(C\) goes from \(T_{1}\) to \(T_{2}\) in contact with a reservoir at \(T_{r}\), the entropy of the universe changes by \(C\ln(T_{2}/T_{1}) - C(T_{2} - T_{1})/T_{r}\). In units of \(C\):
- Process 1: \(\ln(320/300) - 20/320 = 0.0645 - 0.0625 = 0.0020\).
- Process 2: \(\ln\frac{310}{300} - \frac{10}{310} \approx 0.0005\) for the first stage plus \(\ln\frac{320}{310} - \frac{10}{320} \approx 0.0005\) for the second, about 0.0010 in all.
- Process 3: heating to 350 K gives \(0.1542 - 0.1429 = 0.0113\); cooling back to 320 K gives \(-0.0896 + 0.0938 = 0.0041\); total about 0.0154.
Smaller temperature steps make the heat transfer closer to reversible, so process 2 generates the least entropy. Overshooting to 350 K and coming back adds two irreversible transfers, so process 3 generates the most.
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