Consider a Carnot engine working between T_1 | vedclass

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(A) The efficiency 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 =

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$2.5$

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(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$.Substituting the given values: $\eta = 1 - \frac{300}{500} = 1 - 0.6 = 0.4$.The efficiency is also defined as the ratio of work done $(W)$ to the heat absorbed from the source $(Q_1)$: $\eta = \frac{W}{Q_1}$.Given $W = 1 \ kJ$,we have $0.4 = \frac{1}{Q_1}$.Therefore,$Q_1 = \frac{1}{0.4} = 2.5 \ kJ$.The heat absorbed by the engine from the source is $2.5 \ kJ$.

Consider a Carnot engine working between $T_1 = 500 \ K$ and $T_2 = 300 \ K$ producing $1 \ kJ$ of mechanical work per cycle. Calculate the heat absorbed by the engine from the source. (in $kJ$)

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