Question

Two Carnot heat engines are operating in series such that the heat sink of the first engine serves as the heat source of the second one. If the source temperature of the first engine is \(1300 \mathrm{K}\) and the sink temperature of the \(\sec\) ond engine is \(300 \mathrm{K}\) and the thermal efficiencies of both engines are the same, the temperature of the intermediate reservoir is \((a) 625 \mathrm{K}\) (b) \(800 \mathrm{K}\) \((c) 860 \mathrm{K}\) \((d) 453 \mathrm{K}\) \((e) 758 \mathrm{K}\)

Short answer

Answer: (a) 625K

Step-by-step solution

Write down given information

The given information is as follows: Source temperature of the first engine \(T_{S1} = 1300K\), Sink temperature of the second engine \(T_{S2} = 300K\), Efficiency of both engines is equal. Let's denote the intermediate reservoir temperature as \(T_m\).

Determine the efficiency formula for Carnot engines

The efficiency of a Carnot engine, \(η\), can be determined from the temperatures of the source and sink as follows: \(η = 1 - \frac{T_{sink}}{T_{source}}\).

Write down the efficiencies for both engines

The efficiencies are equal for both engines. Using the efficiency formulas, we can write down the efficiencies: \(η_1 = 1 - \frac{T_m}{T_{S1}}\) and \(η_2 = 1 - \frac{T_{S2}}{T_m}\). As the efficiencies are equal, we can set them equal and solve for \(T_m\): \(η_1 = η_2 \Rightarrow 1 - \frac{T_m}{T_{S1}} = 1 - \frac{T_{S2}}{T_m}\).

Solve for the intermediate reservoir temperature \(T_m\)

From the equation above, we can solve for \(T_m\): \(\frac{T_m}{T_{S1}} + \frac{T_{S2}}{T_m} = 1 \Rightarrow T_m^2 = T_{S1}\cdot T_{S2}\) Now, we can find the value of \(T_m\) by substituting \(T_{S1} = 1300K\) and \(T_{S2} = 300K\) into the equation: \(T_m = \sqrt{1300 \times 300} = \sqrt{390000} \approx 625K\) Therefore, the temperature of the intermediate reservoir is: (a) \(625K\).