Question

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at \(1400 \mathrm{K}\) and rejects the waste heat to another reservoir at temperature \(T\) The second engine receives this energy rejected by the first one, converts some of it to work, and rejects the rest to a reservoir at \(300 \mathrm{K}\). If the thermal efficiencies of both engines are the same, determine the temperature \(T .\)

Short answer

Answer: The temperature of the reservoir between the two Carnot heat engines is approximately \(37.4 \mathrm{K}\).

Step-by-step solution

Write the efficiency equations for both engines.

The efficiency of the first engine is: \(\eta_1 = 1 - \frac{T}{1400}\) The efficiency of the second engine is: \(\eta_2 = 1 - \frac{300}{T}\)

Set both equations equal to each other.

We are given that the efficiencies of both engines are the same, so we set \(\eta_1 = \eta_2\): \(1 - \frac{T}{1400} = 1 - \frac{300}{T}\)

Solve for the temperature T.

To solve for T, first subtract 1 from both sides: \(-\frac{T}{1400} = -\frac{300}{T}\) Next, multiply both sides by -1 to cancel the negative signs: \(\frac{T}{1400} = \frac{300}{T}\) Now, we'll multiply both sides by 1400T to get rid of the denominators: \(T^2 = 300 \cdot 1400\) Divide both sides by 300: \(T^2 = 1400\) Finally, take the square root of both sides: \(T = \sqrt{1400} = 37.4 \mathrm{K}\) The temperature \(T\) of the reservoir between the two Carnot heat engines is approximately \(37.4 \mathrm{K}\).