A Carnot engine has an efficiency of 50% when | vedclass

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(B) The efficiency of a Carnot engine is given by the formula:$\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$Given efficiency $\eta = 50\% = 0.

Vedclass · Accepted Answer

$27$

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(B) The efficiency of a Carnot engine is given by the formula:$\eta = 1 - \frac{T_{	ext{sink}}}{T_{	ext{source}}}$Given efficiency $\eta = 50\% = 0.5$.The temperature of the source $T_{	ext{source}} = 327^{\circ}C = 327 + 273 = 600\,K$.Substituting the values into the formula:$0.5 = 1 - \frac{T_{	ext{sink}}}{600}$$\frac{T_{	ext{sink}}}{600} = 1 - 0.5 = 0.5$$T_{	ext{sink}} = 0.5 	imes 600 = 300\,K$.To convert the temperature back to Celsius:$T_{	ext{sink}}(^{\circ}C) = 300 - 273 = 27^{\circ}C$.

$A$ Carnot engine has an efficiency of $50\%$ when its source is at a temperature $327^{\circ}C$. The temperature of the sink is $.........^{\circ}C$.

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