Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, T$$_1$$ and T$$_2$$. The temperature of the hot reservoir of the first engine is T$$_1$$ and the temperature of the cold reservoir of the second engine is T$$_2$$. T is temperature of the sink of first engine which is also the source for the second engine. How is T related to T$$_1$$ and T$$_2$$, if both the engines perform equal amount of work?

First we recall the Carnot‐engine efficiency formula, valid for all temperatures measured on the absolute (kelvin) scale:

$$\text{Efficiency} = 1 - \dfrac{T_\text{cold}}{T_\text{hot}}.$$

The first engine works between the hot reservoir at temperature $$T_1$$ and the intermediate reservoir at temperature $$T$$. Its efficiency is therefore

$$\eta_1 = 1 - \dfrac{T}{T_1}.$$

Let us denote by $$Q_1$$ the heat absorbed from the hot reservoir $$T_1$$. The work produced by this first engine is

$$W_1 = \eta_1 Q_1 = \left(1 - \dfrac{T}{T_1}\right)Q_1.$$

The heat rejected by the first engine becomes the input heat for the second engine. Using the definition $$Q_\text{rejected} = Q_\text{absorbed} - W$$ we write

$$Q_{\text{rejected},1} = Q_1 - W_1.$$

Substituting the value of $$W_1$$ we get

$$Q_{\text{rejected},1} = Q_1 - \left(1 - \dfrac{T}{T_1}\right)Q_1 = Q_1 \left[1 - 1 + \dfrac{T}{T_1}\right] = Q_1 \dfrac{T}{T_1}.$$

This rejected heat is the heat absorbed by the second engine, so we set

$$Q_2 = Q_{\text{rejected},1} = Q_1 \dfrac{T}{T_1}.$$

The second engine operates between the source at $$T$$ and the sink at $$T_2$$. Its efficiency is

$$\eta_2 = 1 - \dfrac{T_2}{T}.$$

Hence the work done by the second engine is

$$W_2 = \eta_2 Q_2 = \left(1 - \dfrac{T_2}{T}\right) \left(Q_1 \dfrac{T}{T_1}\right).$$

We are told that both engines perform equal amounts of work, so we impose the condition

$$W_1 = W_2.$$

Substituting the expressions for $$W_1$$ and $$W_2$$, we have

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

Dividing both sides by $$Q_1$$ (which is non-zero) gives

$$1 - \dfrac{T}{T_1} = \left(1 - \dfrac{T_2}{T}\right)\dfrac{T}{T_1}.$$

Now we multiply every term by $$T_1$$ to clear the denominators:

$$T_1 - T = \left(1 - \dfrac{T_2}{T}\right)T.$$

Expanding the right-hand side:

$$T_1 - T = T - \dfrac{T_2 T}{T}.$$

Simplifying the last term $$\dfrac{T_2 T}{T} = T_2$$, we obtain

$$T_1 - T = T - T_2.$$

Bringing like terms together:

$$T_1 + T_2 = 2T.$$

Finally we solve for the intermediate temperature $$T$$:

$$T = \dfrac{T_1 + T_2}{2}.$$

Thus the required temperature is simply the arithmetic mean of $$T_1$$ and $$T_2$$.

Hence, the correct answer is Option B.