A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?

A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By h

1.	A carnot.engine whose sink is at 300 K has an efficiency of 40 %. By how much should the temperature of source be increased so as to increase its efficiency by 50% of original efficiency ?
Answer» <html><body><p><br/></p>Solution :<a href="https://interviewquestions.tuteehub.com/tag/efficiency-20674" style="font-weight:bold;" target="_blank" title="Click to know more about EFFICIENCY">EFFICIENCY</a> of a carnot engine, <br/> `eta=(T_1-T_2)/T_1` , where `T_1`=<a href="https://interviewquestions.tuteehub.com/tag/temperatureof-2293939" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATUREOF">TEMPERATUREOF</a> the <a href="https://interviewquestions.tuteehub.com/tag/source-1219297" style="font-weight:bold;" target="_blank" title="Click to know more about SOURCE">SOURCE</a> <br/> `T_2`=temperature of the sink <br/> `<a href="https://interviewquestions.tuteehub.com/tag/therefore-706901" style="font-weight:bold;" target="_blank" title="Click to know more about THEREFORE">THEREFORE</a> 0.40=(T_1-300)/T_1` <br/> `therefore T_1-300=0.40T_1` <br/> `therefore 0.6T_1=300 rArr T_1=300/0.6=3000/6`=500 K <br/> Now, 50% efficiency increases , <br/> `therefore 0.60=(T_1-300)/T_1 rArr T_1-300=0.6T_1` <br/> `therefore 0.4T_1=300 rArr T_1 =300/0.4=(300xx10)/4` =750 <br/> Increase in temperature of source =750-500=250 K</body></html>