A Carnot's engine whose sink is at temperature of 300K has an efficiency of `40%` By how much should the temperature of the source be increased so as to increase the efficiency to 60%?

Options

A. 750K
B. 200 K
C. 300K
D. 275K

Answer

A

Explanation

Efficiency, `eta=1-(T_(2))/(T_(1))`
`eta=1-(T_(2))/(T_(1))=(40)/(100)=1-(300)/(T_(1))`
=0.6
`T_(1)=500K`
Final: `(60)/(100)=1-(300)/(T_(1)')=0.4`
`(T_(1)')/(T_(1))=(3)/(2)` ltBrgt `T_(1)'=(3)/(2)xx500=750Kk`