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

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Question

(A) Given:Temperature of the reservoir,$T_{1} = 527^{\circ} C = 527 + 273 = 800 \; K$.Temperature of the sink,$T_{2} = 200 \; K$.Work done,$W = 12000

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(A) Given:Temperature of the reservoir,$T_{1} = 527^{\circ} C = 527 + 273 = 800 \; K$.Temperature of the sink,$T_{2} = 200 \; K$.Work done,$W = 12000 \; kJ = 12 	imes 10^{6} \; J$.Efficiency of a Carnot engine is given by $\eta = 1 - \frac{T_{2}}{T_{1}}$.Also,efficiency is defined as $\eta = \frac{W}{Q_{1}}$,where $Q_{1}$ is the heat absorbed from the reservoir.Equating the two expressions for efficiency:$1 - \frac{200}{800} = \frac{12 	imes 10^{6}}{Q_{1}}$$1 - \frac{1}{4} = \frac{12 	imes 10^{6}}{Q_{1}}$$\frac{3}{4} = \frac{12 	imes 10^{6}}{Q_{1}}$$Q_{1} = \frac{12 	imes 10^{6} 	imes 4}{3}$$Q_{1} = 16 	imes 10^{6} \; J$.Thus,the value of $x$ is $16$.

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