The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

The efficiency of Carnot engine is 50% and temperature of sink is 500

1.	The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be
Answer» <html><body><p>100 K<br/>600 K<br/>400 K<br/>500 K</p>Solution :The efficiency of a Carnot <a href="https://interviewquestions.tuteehub.com/tag/engine-25861" style="font-weight:bold;" target="_blank" title="Click to know more about ENGINE">ENGINE</a> is `eta = <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a> - (T_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>))/(T_(1))` where `T_(1)` is the temperature of the source and `T_(2)` is the temperature of the <a href="https://interviewquestions.tuteehub.com/tag/sink-1210432" style="font-weight:bold;" target="_blank" title="Click to know more about SINK">SINK</a>. <br/> Here, `T_(2) = 500 K, eta = 50%, T_(1) = ?` <br/> `:. 0.5 = 1 - (500)/(T_(1))` or `T_(1) = 1000 K` <br/> Let `T._(2)` be the new sink temperature. Then <br/> `eta. = 1- (T._(2))/(T_(1)) :. 0.6 = 1 - (T._(2))/(1000)` <br/> or `T._(2) = 400 K`</body></html>