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In ideal Carnot engine, whose efficiency is 40%, receives heat at 500K. If its efficiency is 50%, then the intake temperature for the same exhaust temperature is

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Detailed Solution

$\begin{array}{l} η = 1 - \frac{T_{2}}{T_{1}} \Rightarrow η_{1} = \frac{40}{100} = 1 - \frac{T_{2}}{500} \Rightarrow T_{2} = 300 K \\ η_{2} = \frac{50}{100} = 1 - \frac{300}{T_{1}^{1}}; T_{1}^{1} = 600 K \end{array}$

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