A Carnot engine whose sink is at 300, K has a | vedclass

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(D) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_2$ is the sink temperature and $T_1$ is the source 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 $T_2$ is the sink temperature and $T_1$ is the source temperature.Given $\eta = 40\% = 0.4$ and $T_2 = 300\, K$.$0.4 = 1 - \frac{300}{T_1} \implies \frac{300}{T_1} = 0.6 \implies T_1 = \frac{300}{0.6} = 500\, K$.The efficiency is to be increased by $50\%$ of its original efficiency.New efficiency $\eta' = \eta + 0.5 	imes \eta = 0.4 + 0.5(0.4) = 0.4 + 0.2 = 0.6 = 60\%$.Let the new source temperature be $T_1'$.$0.6 = 1 - \frac{300}{T_1'} \implies \frac{300}{T_1'} = 0.4 \implies T_1' = \frac{300}{0.4} = 750\, K$.The increase in source temperature is $\Delta T = T_1' - T_1 = 750 - 500 = 250\, K$.

$A$ Carnot engine whose sink is at $300\, K$ has an efficiency of $40\%.$ By how much should the temperature of the source be increased so as to increase its efficiency by $50\%$ of its original efficiency? (in $K$)

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$A$ 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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