A Carnot engine absorbs heat from a reservoir | vedclass

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(A) The thermal efficiency of a Carnot engine is given by the formula:$\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}$Given:Work done $W = 300 \,J$Heat ab

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$250$

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(A) The thermal efficiency of a Carnot engine is given by the formula:$\eta = \frac{W}{Q_1} = 1 - \frac{T_2}{T_1}$Given:Work done $W = 300 \,J$Heat absorbed $Q_1 = 400 \,J$Source temperature $T_1 = 1000 \,K$Sink temperature $T_2 = T$Substituting the values into the efficiency formula:$\frac{300}{400} = 1 - \frac{T}{1000}$$\frac{3}{4} = 1 - \frac{T}{1000}$Rearranging for $T$:$\frac{T}{1000} = 1 - \frac{3}{4} = \frac{1}{4}$$T = \frac{1000}{4} = 250 \,K$Therefore, the temperature of the sink is $250 \,K$.

$A$ Carnot engine absorbs heat from a reservoir maintained at temperature $1000 \,K$. The engine rejects heat to a reservoir whose temperature is $T$. If the magnitude of absorbed heat is $400 \,J$ and work performed is $300 \,J$, then the value of $T$ is (in $\,K$)

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