The efficiency of a Carnot engine is 25 %,whe | vedclass

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

Vedclass · Accepted Answer

$200$

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(C) For a Carnot engine,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature.Given $\eta_1 = 25\% = 0.25$ and $T_2 = 300 \ K$:$0.25 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.75 \implies T_1 = \frac{300}{0.75} = 400 \ K$.Now,we want the efficiency to become $\eta_2 = 50\% = 0.5$ by increasing the source temperature by $x$:$0.5 = 1 - \frac{300}{400 + x} \implies \frac{300}{400 + x} = 0.5 \implies 400 + x = \frac{300}{0.5} = 600 \ K$.$x = 600 - 400 = 200 \ K$.Thus,the required increase in the source temperature is $200 \ K$.

The efficiency of a Carnot engine is $25 \%$,when the temperature of the sink is $300 \ K$. The increase in the temperature of the source required for the efficiency to become $50 \%$ is (in $K$)

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