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

Idea: Use heat balance instead of writing the efficiency formula directly.

For a Carnot engine, heat exchanged is proportional to reservoir temperature: $\dfrac{Q_H}{Q_C} = \dfrac{T_H}{T_C}$. Efficiency can also be written as $\eta = 1 - \dfrac{Q_C}{Q_H} = 1 - \dfrac{T_C}{T_H}$, so equal efficiency means equal temperature ratio $\dfrac{T_C}{T_H}$ for both engines.

Let the common temperature ratio be $r$. For engine X the cold-to-hot ratio is $\dfrac{T}{1200}$, and for engine Y it is $\dfrac{300}{T}$. Setting them equal: \[\frac{T}{1200} = \frac{300}{T} = r.\]
Multiplying the two equal ratios chains nicely, but simplest is to note the ratios equal gives $T^2 = 1200 \times 300$. This is exactly the condition that $T$ is the geometric mean of $1200$ and $300$.
\[T = \sqrt{1200 \times 300} = \sqrt{360000} = 600\ \text{K}.\]
Check: $\eta_X = 1 - \tfrac{600}{1200} = 0.5$ and $\eta_Y = 1 - \tfrac{300}{600} = 0.5$. Both are $50\%$, so the answer is consistent. \[\boxed{T = 600\ \text{K}}\]