A Carnot engine has an efficiency of 40%. The | vedclass

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

Vedclass · Accepted Answer

$250$

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(D) The efficiency of a Carnot engine 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 $\eta = 0.40$ and $T_2 = 300 \ K$,we have $0.40 = 1 - \frac{300}{T_1}$.$\frac{300}{T_1} = 0.60 \implies T_1 = \frac{300}{0.60} = 500 \ K$.Now,the efficiency is increased by $50\%$ of its original value. The new efficiency $\eta' = \eta + 0.50 	imes \eta = 0.40 + 0.20 = 0.60$.Keeping $T_2 = 300 \ K$ constant,the new source temperature $T_1'$ is given by $\eta' = 1 - \frac{T_2}{T_1'}$.$0.60 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.40$.$T_1' = \frac{300}{0.40} = 750 \ K$.The increase in source temperature is $\Delta T = T_1' - T_1 = 750 \ K - 500 \ K = 250 \ K$.

$A$ Carnot engine has an efficiency of $40\%$. The temperature of its sink is $300 \ K$. To increase the efficiency by $50\%$ of its original efficiency while keeping the sink temperature constant,by how many $K$ must the source temperature be increased?

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