An ideal heat engine operates in a Carnot cyc | vedclass

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(C) For a Carnot cycle,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Given temperatures in Kelvin: $T_1 = 127 + 273 =

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$1.25 	imes 10^4 	ext{ cal}$

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(C) For a Carnot cycle,the efficiency $\eta$ is given by $\eta = 1 - \frac{T_2}{T_1} = \frac{W}{Q_1}$.Given temperatures in Kelvin: $T_1 = 127 + 273 = 400 	ext{ K}$ and $T_2 = 27 + 273 = 300 	ext{ K}$.Given heat absorbed $Q_1 = 5 	imes 10^4 	ext{ cal}$.Substituting the values into the efficiency formula:$\eta = 1 - \frac{300}{400} = 1 - 0.75 = 0.25$.Since $\eta = \frac{W}{Q_1}$,we have $W = \eta 	imes Q_1$.$W = 0.25 	imes 5 	imes 10^4 	ext{ cal} = 1.25 	imes 10^4 	ext{ cal}$.Thus,the amount of heat converted to work is $1.25 	imes 10^4 	ext{ cal}$.

An ideal heat engine operates in a Carnot cycle between $127^{\circ} C$ and $27^{\circ} C$. It absorbs $5 \times 10^4 \text{ cal}$ of heat at the higher temperature. The amount of heat converted to work is:

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