A Carnot engine takes 300 calories of heat fr | vedclass

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(B) Given: Heat taken from source,$Q_{1} = 300$ calories; Temperature of source,$T_{1} = 500 \,K$; Heat rejected to the sink,$Q_{2} = 150$ calories.We

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$250$

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(B) Given: Heat taken from source,$Q_{1} = 300$ calories; Temperature of source,$T_{1} = 500 \,K$; Heat rejected to the sink,$Q_{2} = 150$ calories.We need to find the temperature of the sink,$T_{2}$.For a Carnot engine,the efficiency $\eta$ is given by the ratio of heat exchange as $\eta = \frac{Q_{1} - Q_{2}}{Q_{1}}$ and also by the temperatures as $\eta = \frac{T_{1} - T_{2}}{T_{1}}$.Equating the two expressions for efficiency:$\frac{Q_{1} - Q_{2}}{Q_{1}} = \frac{T_{1} - T_{2}}{T_{1}}$Substituting the given values:$\frac{300 - 150}{300} = \frac{500 - T_{2}}{500}$$\frac{150}{300} = 1 - \frac{T_{2}}{500}$$\frac{1}{2} = 1 - \frac{T_{2}}{500}$$\frac{T_{2}}{500} = 1 - \frac{1}{2}$$\frac{T_{2}}{500} = \frac{1}{2}$$T_{2} = 500 	imes \frac{1}{2} = 250 \,K$.Thus,the temperature of the sink is $250 \,K$.

$A$ Carnot engine takes $300$ calories of heat from a source at $500 \,K$ and rejects $150$ calories of heat to the sink. The temperature of the sink is (in $\,K$)

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