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

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(A) The temperatures are given as $T_2 = 27 \, ^\circ C = 300 \, K$ and $T_1 = 927 \, ^\circ C = 1200 \, K$.Efficiency of a Carnot engine is given by

Vedclass · Accepted Answer

$16.8 	imes 10^{6} \, J$

Vedclass · Answer

(A) The temperatures are given as $T_2 = 27 \, ^\circ C = 300 \, K$ and $T_1 = 927 \, ^\circ C = 1200 \, K$.Efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_2}{T_1} = 1 - \frac{300}{1200} = 1 - 0.25 = 0.75 = \frac{3}{4}$.Also,efficiency $\eta = \frac{W}{Q_1}$,where $W$ is the work done and $Q_1$ is the heat absorbed from the source.Given $W = 12.6 	imes 10^{6} \, J$.Substituting the values: $\frac{3}{4} = \frac{12.6 	imes 10^{6}}{Q_1}$.Therefore,$Q_1 = \frac{4 	imes 12.6 	imes 10^{6}}{3} = 4 	imes 4.2 	imes 10^{6} = 16.8 	imes 10^{6} \, J$.

In a Carnot engine,the temperature of the sink is $27 \, ^\circ C$ and the temperature of the source is $927 \, ^\circ C$. If the work done by the engine to reject heat from the source to the sink is $12.6 \times 10^{6} \, J$,what is the heat absorbed from the source?

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