A Carnot engine A working between temperature | vedclass

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Question

(B) For a Carnot engine, the efficiency $\eta$ is given by $\eta = 1 - \frac{T_{low}}{T_{high}}$.Also, the work done $W$ is given by $W = Q_{in} \cdot

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$500$

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(B) For a Carnot engine, the efficiency $\eta$ is given by $\eta = 1 - \frac{T_{low}}{T_{high}}$.Also, the work done $W$ is given by $W = Q_{in} \cdot \eta$, where $Q_{in}$ is the heat absorbed from the source.For engine $A$: $W_A = Q_A \left(1 - \frac{T}{600}ight)$.For engine $B$: $W_B = Q_B \left(1 - \frac{400}{T}ight)$.Since the engines are in series, the heat rejected by engine $A$ is the heat absorbed by engine $B$, so $Q_B = Q_A \left(\frac{T}{600}ight)$.Given $W_A = W_B$, we have $Q_A \left(1 - \frac{T}{600}ight) = Q_A \left(\frac{T}{600}ight) \left(1 - \frac{400}{T}ight)$.Simplifying the equation: $1 - \frac{T}{600} = \frac{T}{600} - \frac{400}{600}$.$1 + \frac{400}{600} = \frac{2T}{600}$.$1 + \frac{2}{3} = \frac{T}{300} \implies \frac{5}{3} = \frac{T}{300}$.$T = \frac{5 	imes 300}{3} = 500 \ K$.

$A$ Carnot engine $A$ working between temperatures $600 \ K$ and $T$ $(T < 600 \ K)$ and another Carnot engine $B$ working between temperatures $T$ $(T > 400 \ K)$ and $400 \ K$ are connected in series. If the work done by both the engines is the same, then $T =$ (in $K$)

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