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Efficiency of a cannot engine is $50 %$ when temperature of outlet is 500K. In order to increase efficiency up to $60 %$ keeping temperature of intake the same what is temperature of outlet

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400 K

800 K

200 K

600 K

answer is C.

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Detailed Solution

$η = 1 - \frac{T_{2}}{T_{1}}$ $= 0.5 = 1 - \frac{500}{T_{1}} \Rightarrow T_{1} = 1000 K$

now $η = 0.6 \Rightarrow 0.6 = 1 - \frac{T_{2}^{'}}{1000}$

on solving we get $T_{2}^{'} = 400 K$

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