Two Carnot engines $$A$$ and $$B$$ operate in series such that engine A absorbs heat at $$T_1$$ and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine $$A$$ and rejects heat to the sink at $$T_3$$. When workdone in both the cases is equal, to value of T is:

For every Carnot engine the efficiency is given by the well-known relation

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

Engine A works between the temperatures $$T_1$$ (source) and $$T$$ (sink). Let the heat absorbed from the hot reservoir be $$Q_1$$. Using the above formula we have for engine A

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

Therefore the work obtained from engine A is

$$W_A=\eta_A Q_1=\Bigl(1-\dfrac{T}{T_1}\Bigr)Q_1.$$

The heat rejected by engine A to the intermediate reservoir at temperature $$T$$ is found from the first law:

$$Q_2 = Q_1-W_A = Q_1 -\Bigl(1-\dfrac{T}{T_1}\Bigr)Q_1 = Q_1\dfrac{T}{T_1}.$$

According to the statement of the problem, engine B absorbs only half of this rejected heat, so

$$Q_B = \dfrac{Q_2}{2}= \dfrac{Q_1 T}{2T_1}.$$

Engine B now works between the temperatures $$T$$ (source) and $$T_3$$ (sink). Its Carnot efficiency is

$$\eta_B = 1-\dfrac{T_3}{T}.$$

Hence the work obtained from engine B is

$$W_B = \eta_B Q_B =\Bigl(1-\dfrac{T_3}{T}\Bigr)\dfrac{Q_1 T}{2T_1}.$$

The condition given is that both engines produce the same work, i.e.

$$W_A = W_B.$$

Substituting the expressions for $$W_A$$ and $$W_B$$ we get

$$\Bigl(1-\dfrac{T}{T_1}\Bigr)Q_1 =\Bigl(1-\dfrac{T_3}{T}\Bigr)\dfrac{Q_1 T}{2T_1}.$$

The factor $$Q_1$$ is present on both sides, so it cancels out. Multiplying the remaining equation by $$2T_1$$ we obtain

$$2T_1\Bigl(1-\dfrac{T}{T_1}\Bigr)=T\Bigl(1-\dfrac{T_3}{T}\Bigr).$$

Rewriting the terms inside the brackets gives

$$2T_1\Bigl(\dfrac{T_1-T}{T_1}\Bigr)=T\Bigl(\dfrac{T-T_3}{T}\Bigr).$$

Simplifying each side separately:

$$2(T_1-T)=T-T_3.$$

Expanding and collecting the temperature terms leads to

$$2T_1-2T = T - T_3.$$

Now shift all terms involving $$T$$ to one side and the constants to the other:

$$2T_1 + T_3 = 3T.$$

Finally, solving for the unknown intermediate temperature $$T$$ yields

$$T = \dfrac{2T_1+T_3}{3} = \dfrac{2}{3}T_1 + \dfrac{1}{3}T_3.$$

Hence, the correct answer is Option D.