Question

For a reversible engine, will you obtain a better efficiency by increasing the high-temperature reservoir by an amount \(\Delta T\) or decreasing the low- temperature reservoir by the same amount \(\Delta T ?\)

Short answer

Answer: Decreasing the low-temperature reservoir by the amount ΔT results in better efficiency for the reversible engine.

Step-by-step solution

Initial Efficiency

First, we find the initial efficiency of the reversible engine using the Carnot efficiency formula: \(\eta_{initial} = 1 - \frac{T_L}{T_H}\) where \(\eta_{initial}\) is the initial efficiency, \(T_L\) is the low-temperature reservoir, and \(T_H\) is the high-temperature reservoir.

Efficiency with Increased High-Temperature Reservoir

Next, increase the high-temperature reservoir by the amount \(\Delta T\). The new high-temperature reservoir is \(T_H' = T_H + \Delta T\). Calculate the efficiency for this case using the Carnot efficiency formula: \(\eta_{increase} = 1 - \frac{T_L}{T_H + \Delta T}\)

Efficiency with Decreased Low-Temperature Reservoir

Now, decrease the low-temperature reservoir by the amount \(\Delta T\). The new low-temperature reservoir is \(T_L' = T_L - \Delta T\). Calculate the efficiency for this case using the Carnot efficiency formula: \(\eta_{decrease} = 1 - \frac{T_L - \Delta T}{T_H}\)

Comparing Efficiencies

Compare the two efficiencies, \(\eta_{increase}\) and \(\eta_{decrease}\), to determine which change results in a better efficiency. If \(\eta_{increase} > \eta_{decrease}\), increasing the high-temperature reservoir by the amount \(\Delta T\) results in better efficiency. If \(\eta_{increase} < \eta_{decrease}\), decreasing the low-temperature reservoir by the amount \(\Delta T\) results in better efficiency. It can be shown through mathematical comparison that decreasing the low-temperature reservoir by the amount \(\Delta T\) yields better efficiency.