The efficiency of a Carnot engine is 50% and | vedclass

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Initially,$\eta_1 = 50\% = 0.5$ and $T_{	ext

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$400$

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(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$.Initially,$\eta_1 = 50\% = 0.5$ and $T_{	ext{sink},1} = 500\,K$.$0.5 = 1 - \frac{500}{T_{	ext{source}}} \implies \frac{500}{T_{	ext{source}}} = 0.5 \implies T_{	ext{source}} = 1000\,K$.Now,the efficiency is raised to $\eta_2 = 60\% = 0.6$ while keeping $T_{	ext{source}}$ constant at $1000\,K$.$0.6 = 1 - \frac{T_{	ext{sink},2}}{1000}$.$\frac{T_{	ext{sink},2}}{1000} = 1 - 0.6 = 0.4$.$T_{	ext{sink},2} = 0.4 	imes 1000 = 400\,K$.

The efficiency of a Carnot engine is $50\%$ and the temperature of the sink is $500\,K$. If the temperature of the source is kept constant and its efficiency is raised to $60\%$,then the required temperature of the sink will be .... $K$.

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