A Carnot engine whose heat sink is at 27,^{ci | vedclass

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

Vedclass · Accepted Answer

Increase by $200\,^{\circ} C$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_L}{T_H}$,where $T_L$ is the sink temperature and $T_H$ is the source temperature.Given $T_L = 27\,^{\circ} C = 300\, K$ and $\eta = 25\% = 0.25$.Substituting the values: $0.25 = 1 - \frac{300}{T_H} \implies \frac{300}{T_H} = 0.75 \implies T_H = \frac{300}{0.75} = 400\, K$.If the efficiency is increased by $100\%$ of its original value,the new efficiency $\eta' = \eta + 100\% 	ext{ of } \eta = 0.25 + 0.25 = 0.50$ or $50\%$.Let the new source temperature be $T_H'$.$0.50 = 1 - \frac{300}{T_H'} \implies \frac{300}{T_H'} = 0.50 \implies T_H' = \frac{300}{0.50} = 600\, K$.The change in temperature is $\Delta T = T_H' - T_H = 600\, K - 400\, K = 200\, K$.Since a change in temperature of $1\, K$ is equivalent to a change of $1\,^{\circ} C$,the temperature should be increased by $200\,^{\circ} C$.

$A$ Carnot engine whose heat sink is at $27\,^{\circ} C$ has an efficiency of $25 \%$. By how many degrees should the temperature of the source be changed to increase the efficiency by $100 \%$ of the original efficiency?

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