An ideal heat engine based on the Carnot cycl | vedclass

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Question

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature

Vedclass · Accepted Answer

$1.2 	imes 10^{4} \, cal$

Vedclass · Answer

(C) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the source temperature and $T_2$ is the sink temperature in Kelvin.Given: $T_1 = 227 + 273 = 500 \, K$ and $T_2 = 127 + 273 = 400 \, K$.Heat absorbed from the source $Q_1 = 6 	imes 10^{4} \, cal$.The efficiency is also defined as $\eta = \frac{W}{Q_1}$,where $W$ is the net work done.Therefore,$W = Q_1 \left( 1 - \frac{T_2}{T_1} ight)$.Substituting the values: $W = 6 	imes 10^{4} \left( 1 - \frac{400}{500} ight)$.$W = 6 	imes 10^{4} \left( 1 - 0.8 ight) = 6 	imes 10^{4} 	imes 0.2$.$W = 1.2 	imes 10^{4} \, cal$.

An ideal heat engine based on the Carnot cycle operates between temperatures of $227 \, ^\circ C$ and $127 \, ^\circ C$. If it absorbs $6 \times 10^{4} \, cal$ of heat from the high-temperature reservoir,the net work done by the engine is:

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