A Carnot engine operates between two reservoi | vedclass

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(C) For a Carnot engine,the ratio of heat exchanged is equal to the ratio of temperatures of the reservoirs:$\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$Here,$T

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$600$

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(C) For a Carnot engine,the ratio of heat exchanged is equal to the ratio of temperatures of the reservoirs:$\frac{Q_1}{Q_2} = \frac{T_1}{T_2}$Here,$T_1 = 900 \; K$ (source temperature),$T_2 = 300 \; K$ (sink temperature),and $W = 1200 \; J$.From the first law of thermodynamics,the work done is $W = Q_1 - Q_2$,so $Q_1 = Q_2 + W$.Let $Q_2 = Q$. Then $Q_1 = Q + 1200$.Substituting these values into the efficiency equation:$\frac{Q + 1200}{Q} = \frac{900}{300}$$\frac{Q + 1200}{Q} = 3$$Q + 1200 = 3Q$$2Q = 1200$$Q = 600 \; J$.Thus,the heat energy delivered to the low-temperature reservoir is $600 \; J$.

$A$ Carnot engine operates between two reservoirs of temperatures $900 \; K$ and $300 \; K$. The engine performs $1200 \; J$ of work per cycle. The heat energy (in $J$) delivered by the engine to the low-temperature reservoir in a cycle is:

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