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Efficiency of a carnot engine is 50% when temperature of outlet is 500K. In order to increase efficiency upto 60% keeping temperature of in take the same. What is temperature of outlet in K _______.

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answer is 400.

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Detailed Solution

$η_{1} = \frac{50}{100} = 0.5$

$T_{2} = 500 K$

$η_{1} = 1 - \frac{T_{2}}{T_{1}}$

$0.5 = 1 - \frac{500}{T_{1}}$ .

$T_{1} = 1000 K$

$η_{1} = 1 - \frac{T_{2}^{1}}{T_{1}}$

$\frac{60}{100} = 1 - \frac{T_{2}^{1}}{1000}$

$T_{2}^{1} = 400 K$

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