Two Carnot engines A and B are operated in su | vedclass

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(C) For a Carnot engine,efficiency $\eta = \frac{W}{Q_{in}} = \frac{T_{source} - T_{sink}}{T_{source}}$.Let $Q_1$ be the heat input to engine $A$. The

Vedclass · Accepted Answer

$550$

Vedclass · Answer

(C) For a Carnot engine,efficiency $\eta = \frac{W}{Q_{in}} = \frac{T_{source} - T_{sink}}{T_{source}}$.Let $Q_1$ be the heat input to engine $A$. The work done by engine $A$ is $W_1 = Q_1 \left( \frac{800 - T_2}{800} \right)$.The heat rejected by engine $A$ is $Q_2 = Q_1 - W_1 = Q_1 \left( 1 - \frac{800 - T_2}{800} \right) = Q_1 \frac{T_2}{800}$.This heat $Q_2$ is the input for engine $B$. The work done by engine $B$ is $W_2 = Q_2 \left( \frac{T_2 - 300}{T_2} \right) = \left( Q_1 \frac{T_2}{800} \right) \left( \frac{T_2 - 300}{T_2} \right) = Q_1 \left( \frac{T_2 - 300}{800} \right)$.Given that the work outputs are equal,$W_1 = W_2$:$Q_1 \left( \frac{800 - T_2}{800} \right) = Q_1 \left( \frac{T_2 - 300}{800} \right)$.$800 - T_2 = T_2 - 300$.$2T_2 = 1100$.$T_2 = 550\, K$.

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