Q 11-11-113JEE MainJEE Main 2021 (27 Jul, Shift 2)Medium
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 work done in both the cases is equal, the value of $T$ is:
Answer: (D) $\dfrac23T_1 + \dfrac13T_3$
Let engine $A$ reject heat $Q$ at $T$. Then it absorbs $Q\dfrac{T_1}{T}$, so $W_A = Q\left(\dfrac{T_1}{T} - 1\right)$.
Engine $B$ absorbs $\dfrac Q2$ at $T$: $W_B = \dfrac Q2\left(1 - \dfrac{T_3}{T}\right)$.
$$\frac{T_1}{T} - 1 = \frac12 - \frac{T_3}{2T} \Rightarrow 2T_1 + T_3 = 3T \Rightarrow T = \frac23T_1 + \frac13T_3$$
Solution by Sreeraj P, M.Sc Physics