A Carnot engine takes 3 times 10^6 , text{cal | vedclass

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

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$8.4 	imes 10^6 \, 	ext{J}$

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(B) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1}$,where $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink in Kelvin.Given: $T_1 = 627^{\circ}C = 627 + 273 = 900 \, 	ext{K}$,$T_2 = 27^{\circ}C = 27 + 273 = 300 \, 	ext{K}$,and $Q_1 = 3 	imes 10^6 \, 	ext{cal}$.Converting heat to Joules: $Q_1 = 3 	imes 10^6 	imes 4.2 \, 	ext{J} = 12.6 	imes 10^6 \, 	ext{J}$.The efficiency is $\eta = 1 - \frac{300}{900} = 1 - \frac{1}{3} = \frac{2}{3}$.Since $\eta = \frac{W}{Q_1}$,the work done is $W = \eta 	imes Q_1 = \frac{2}{3} 	imes 12.6 	imes 10^6 \, 	ext{J} = 2 	imes 4.2 	imes 10^6 \, 	ext{J} = 8.4 	imes 10^6 \, 	ext{J}$.

$A$ Carnot engine takes $3 \times 10^6 \, \text{cal}$ of heat from a reservoir at $627^{\circ}C$ and gives it to a sink at $27^{\circ}C$. The work done by the engine is:

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