A Carnot engine will sink temperature at 17^(@)C has 50% efficiency. By now much should its source temperature by changed to increase its efficiency to 60% ?

A Carnot engine will sink temperature at 17^(@)C has 50% efficiency. B

1.	A Carnot engine will sink temperature at 17^(@)C has 50% efficiency. By now much should its source temperature by changed to increase its efficiency to 60% ?
Answer» <html><body><p>225 K<br/>`128^(@)C`<br/>580 K<br/>145 K</p>Solution :`eta = 1 - (T_(2))/(T_(1))` <br/> <a href="https://interviewquestions.tuteehub.com/tag/initially-516121" style="font-weight:bold;" target="_blank" title="Click to know more about INITIALLY">INITIALLY</a>, `(<a href="https://interviewquestions.tuteehub.com/tag/50-322056" style="font-weight:bold;" target="_blank" title="Click to know more about 50">50</a>)/(<a href="https://interviewquestions.tuteehub.com/tag/100-263808" style="font-weight:bold;" target="_blank" title="Click to know more about 100">100</a>) = 1 - (273 + 17)/(T_(1))` or `(290)/(T_(1)) = (1)/(2)` or `T_(1) = 580 K` <br/> Finally, `(<a href="https://interviewquestions.tuteehub.com/tag/60-328817" style="font-weight:bold;" target="_blank" title="Click to know more about 60">60</a>)/(100) = 1 - (273 + 17)/(T._(1))` or `(290)/(T._(1)) = (2)/(5)` or `T._(1) = 725 K` <br/> `:.` Change in source <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> `= (725 - 580) K = 145 K`</body></html>