Question

A Carnot engine, whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K,

Answer 1

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Answer 2

Question ⤵

️

A Carnot engine whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60% Then, the intake temperature for the same exhaust(sink) temperature must be:

Answer⤵

$n(effiency) = 1 - \frac{TL}{Th}$

$TL \: = \: Low \: \: temperature(sink)$

$\frac{400}{100} = 1 - \frac{TL}{500}$

$=> \frac{TL}{500} = \frac{60}{100}$

$= > \: TL = 300$

$Now,$

$\frac{60}{100} = 1 - \frac{300}{TH}$

$\frac{300}{TH} = 1 - \frac{60}{100}$

$\frac{300}{TH} = \frac{40}{100}$

$TH = \frac{300 \times 100}{40} \\ = 750k$