In a Carnot engine,the temperature of the res | vedclass

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

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$16.8 	imes 10^6 \ J$

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(A) The efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1},$ where $T_1$ is the temperature of the reservoir and $T_2$ is the temperature of the sink in Kelvin.Given $T_1 = 927\ ^oC = 927 + 273 = 1200 \ K$ and $T_2 = 27\ ^oC = 27 + 273 = 300 \ K.$Substituting these values,$\eta = 1 - \frac{300}{1200} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75.$The efficiency is also defined as $\eta = \frac{W}{Q_1},$ where $W$ is the work done and $Q_1$ is the heat absorbed from the reservoir.Given $W = 12.6 	imes 10^6 \ J,$ we have $0.75 = \frac{12.6 	imes 10^6}{Q_1}.$Therefore,$Q_1 = \frac{12.6 	imes 10^6}{0.75} = 16.8 	imes 10^6 \ J.$

In a Carnot engine,the temperature of the reservoir is $927\ ^oC$ and that of the sink is $27\ ^oC.$ If the work done by the engine is $12.6 \times 10^6 \ J,$ the quantity of heat absorbed by the engine from the reservoir is:

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