Question

Two Carnot engines, X and Y, are operating in series. The engine X receives heat at \(1200\ \text{K}\) and rejects to a reservoir at a temperature \(T\). The second engine, Y, receives the heat rejected by X and, in turn, rejects to a heat reservoir at \(300\ \text{K}\). What is the temperature \(T\) (in Kelvin) for the situation when the efficiency of the engines is the same?

• A. \(600\ \text{K}\)
• B. \(750\ \text{K}\)
• C. \(0\)
• D. \(450\ \text{K}\)

Hint:

Equal Carnot efficiency means equal cold-to-hot temperature ratio, so \(T/1200 = 300/T\). The middle temperature is the geometric mean.

Answer 1

The Correct Option is A

Step 1: The efficiency of a Carnot engine working between a hot reservoir at temperature \(T_H\) and a cold reservoir at temperature \(T_C\) is \[\eta = 1 - \frac{T_C}{T_H}.\]
Step 2: For engine X, the hot side is \(1200\ \text{K}\) and the cold side is \(T\): \[\eta_X = 1 - \frac{T}{1200}.\] For engine Y, the hot side is \(T\) and the cold side is \(300\ \text{K}\): \[\eta_Y = 1 - \frac{300}{T}.\]
Step 3: The two efficiencies are equal, so \[1 - \frac{T}{1200} = 1 - \frac{300}{T}.\] Cancel the 1 on each side: \[\frac{T}{1200} = \frac{300}{T}.\]
Step 4: Cross multiply: \[T^2 = 1200 \times 300 = 360000.\] So \(T\) is the geometric mean of the two reservoir temperatures: \[T = \sqrt{360000} = 600\ \text{K}.\]
Step 5: The physical temperature must be positive, so we keep the positive root. \[\boxed{T = 600\ \text{K}}\]