Two carnot engines $$A$$ and $$B$$ are operated in series. The first one, $$A$$, receives heat at $$T_1 (= 600K)$$ and rejects to a reservoir at temperature $$T_2$$. The second engine $$B$$ receives heat rejected by the first engine and, in turn, rejects to a heat reservoir at $$T_3 (= 400K)$$. Calculate the temperature $$T_2$$ if the work outputs of the two engines are equal:

For a reversible Carnot engine, the efficiency is given by the well-known relation

$$\eta = 1 - \dfrac{T_C}{T_H},$$

where $$T_H$$ is the absolute temperature of the hot reservoir and $$T_C$$ is that of the cold reservoir. Another equally useful Carnot relation is the proportionality of heats to temperatures,

$$\dfrac{Q_C}{Q_H} = \dfrac{T_C}{T_H},$$

because the entropy change $$\dfrac{Q}{T}$$ is the same for both reservoirs in a reversible cycle.

Let the heat absorbed by engine $$A$$ from the source at $$T_1 = 600 \text{ K}$$ be $$Q_1$$. This engine rejects heat $$Q_2$$ to the intermediate reservoir at the unknown temperature $$T_2$$. Applying the heat-temperature ratio to engine $$A$$ we write

$$\dfrac{Q_2}{Q_1} = \dfrac{T_2}{T_1} \quad\Longrightarrow\quad Q_2 = Q_1 \dfrac{T_2}{T_1}.$$

The work output of engine $$A$$ is the difference between the heat absorbed and the heat rejected:

$$W_A = Q_1 - Q_2 = Q_1 - Q_1 \dfrac{T_2}{T_1} = Q_1\left(1 - \dfrac{T_2}{T_1}\right) = Q_1\dfrac{T_1 - T_2}{T_1}.$$

The heat $$Q_2$$ becomes the input to engine $$B$$, whose hot reservoir is at $$T_2$$ and whose cold reservoir is at $$T_3 = 400 \text{ K}$$. For engine $$B$$ the Carnot heat ratio gives

$$\dfrac{Q_3}{Q_2} = \dfrac{T_3}{T_2} \quad\Longrightarrow\quad Q_3 = Q_2 \dfrac{T_3}{T_2}.$$

The work output of engine $$B$$ is therefore

$$W_B = Q_2 - Q_3 = Q_2 - Q_2 \dfrac{T_3}{T_2} = Q_2\left(1 - \dfrac{T_3}{T_2}\right) = Q_2\dfrac{T_2 - T_3}{T_2}.$$

Substituting $$Q_2 = Q_1\dfrac{T_2}{T_1}$$ into this expression yields

$$W_B = \left(Q_1\dfrac{T_2}{T_1}\right) \dfrac{T_2 - T_3}{T_2} = Q_1\dfrac{T_2 - T_3}{T_1}.$$

The problem states that the two engines deliver equal work, so we set $$W_A = W_B$$:

$$Q_1\dfrac{T_1 - T_2}{T_1} = Q_1\dfrac{T_2 - T_3}{T_1}.$$

The common factor $$\dfrac{Q_1}{T_1}$$ cancels out, leaving a simple linear equation in $$T_2$$:

$$T_1 - T_2 = T_2 - T_3.$$

Rearranging gives

$$2T_2 = T_1 + T_3 \quad\Longrightarrow\quad T_2 = \dfrac{T_1 + T_3}{2}.$$

Substituting the numerical values $$T_1 = 600 \text{ K}$$ and $$T_3 = 400 \text{ K}$$, we find

$$T_2 = \dfrac{600 + 400}{2} = \dfrac{1000}{2} = 500 \text{ K}.$$

Hence, the correct answer is Option A.