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(C) The efficiency of a Carnot engine is given by $\eta = (1 - \frac{T_L}{T_H}) 	imes 100$,where $T_L$ is the sink temperature and $T_H$ is the sourc

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$750$

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(C) The efficiency of a Carnot engine is given by $\eta = (1 - \frac{T_L}{T_H}) 	imes 100$,where $T_L$ is the sink temperature and $T_H$ is the source temperature.For the first case: $40 = (1 - \frac{T_L}{500}) 	imes 100$.$0.4 = 1 - \frac{T_L}{500} \implies \frac{T_L}{500} = 0.6 \implies T_L = 300 \ K$.For the second case,the sink temperature $T_L$ remains constant at $300 \ K$ and efficiency $\eta$ becomes $60\%$:$60 = (1 - \frac{300}{T_H'}) 	imes 100$.$0.6 = 1 - \frac{300}{T_H'} \implies \frac{300}{T_H'} = 0.4$.$T_H' = \frac{300}{0.4} = 750 \ K$.

$A$ Carnot engine having an efficiency of $40\%$ takes heat from a source maintained at a temperature of $500 \ K$. If the efficiency is to be increased to $60\%$ while keeping the sink temperature constant,what should be the new source temperature in $K$?

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