Consider a Carnot cycle operating between T_1 | vedclass

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Substituting the given values: $\eta = 1 - \frac{300}{500} = 1 -

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$2500$

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(B) The efficiency $\eta$ of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Substituting the given values: $\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$ or $\frac{2}{5}$.The efficiency is also defined as the ratio of work done $W$ to the heat absorbed $Q$ from the source: $\eta = \frac{W}{Q}$.Given $W = 1\,kJ = 1000\,J$,we can find $Q$ as $Q = \frac{W}{\eta}$.$Q = \frac{1000}{2/5} = 1000 	imes \frac{5}{2} = 2500\,J$.Thus,the heat transferred to the engine is $2500\,J$.

Consider a Carnot cycle operating between $T_1 = 500\,K$ and $T_2 = 300\,K$ producing $1\,kJ$ of mechanical work per cycle. Find the heat transferred to the engine by the reservoir at $T_1$ in Joules.

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