The temperatures of the source and the sink of a Carnot engine are 500 K and 300 K respectively. If the temperature of the source diminished to 450 K, what will be the percentage change in efficiency?

The temperatures of the source and the sink of a Carnot engine are 500

1.	The temperatures of the source and the sink of a Carnot engine are 500 K and 300 K respectively. If the temperature of the source diminished to 450 K, what will be the percentage change in efficiency?
Answer» <html><body><p></p>Solution :In the first <a href="https://interviewquestions.tuteehub.com/tag/case-910082" style="font-weight:bold;" target="_blank" title="Click to know more about CASE">CASE</a>, <a href="https://interviewquestions.tuteehub.com/tag/efficiency-20674" style="font-weight:bold;" target="_blank" title="Click to know more about EFFICIENCY">EFFICIENCY</a> <br/> `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>) = 1 - (T_2)/(T_1) = 1 - (300)/(<a href="https://interviewquestions.tuteehub.com/tag/500-323288" style="font-weight:bold;" target="_blank" title="Click to know more about 500">500</a>) = 2/5` <br/> In the second case, efficiency <br/> `eta_(2) = 1-(T_2)/(T_1) = 1 - (300)/(450) = 1/3` <br/> `:.` Fractional change in efficiency <br/> `=(eta_(2) - eta_(1))/(eta_1) = (eta_2)/(eta_1)-1 = (1//3)/(2//5)-1 = 5/6-1 = -1/6` <br/> `:.` <a href="https://interviewquestions.tuteehub.com/tag/percentage-13406" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENTAGE">PERCENTAGE</a> change = `-1/6 xx 100 = -16.67%` <br/> So, efficiency decreases by 16.67%.</body></html>