Efficiency of a heat engine whose sink is at | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_2$ is the sink temperature and $T_1$ is the source temperatur

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$250$

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$, where $T_2$ is the sink temperature and $T_1$ is the source temperature.For the first case, $\eta_1 = 40 \% = 0.4$ and $T_2 = 300 \,K$.$0.4 = 1 - \frac{300}{T_1} \Rightarrow \frac{300}{T_1} = 0.6 \Rightarrow T_1 = \frac{300}{0.6} = 500 \,K$.For the second case, $\eta_2 = 60 \% = 0.6$ and $T_2 = 300 \,K$.$0.6 = 1 - \frac{300}{T_1^{\prime}} \Rightarrow \frac{300}{T_1^{\prime}} = 0.4 \Rightarrow T_1^{\prime} = \frac{300}{0.4} = 750 \,K$.The increase in source temperature is $\Delta T = T_1^{\prime} - T_1 = 750 \,K - 500 \,K = 250 \,K$.

Efficiency of a heat engine whose sink is at temperature of $300 \,K$ is $40 \%$. To increase the efficiency to $60 \%$, keeping the sink temperature constant, the source temperature must be increased by (in $\,K$)

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