A Carnot engine operates between a source and | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Given,initial efficiency $\eta_1 = 40\% = 0.4

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Given,initial efficiency $\eta_1 = 40\% = 0.4$ and $T_{	ext{sink}} = 27^{\circ}C = 300 K$.$0.4 = 1 - \frac{300}{T_{	ext{source},1}} \Rightarrow \frac{300}{T_{	ext{source},1}} = 0.6 \Rightarrow T_{	ext{source},1} = \frac{300}{0.6} = 500 K$.Now,for the second case,efficiency $\eta_2 = 50\% = 0.5$ with the same sink temperature.$0.5 = 1 - \frac{300}{T_{	ext{source},2}} \Rightarrow \frac{300}{T_{	ext{source},2}} = 0.5 \Rightarrow T_{	ext{source},2} = \frac{300}{0.5} = 600 K$.The increase in the temperature of the source is $\Delta T = T_{	ext{source},2} - T_{	ext{source},1} = 600 K - 500 K = 100 K$.

$A$ Carnot engine operates between a source and a sink. The efficiency of the engine is $40 \%$ and the temperature of the sink is $27^{\circ} C$. If the efficiency is to be increased to $50 \%$,then the temperature of the source must be increased by: (in $K$)

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