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% ?

Options

A. 225 K
B. `128^(@)C`
C. 580 K
D. 145 K

Answer

D

Explanation

`eta = 1 - (T_(2))/(T_(1))`
Initially, `(50)/(100) = 1 - (273 + 17)/(T_(1))` or `(290)/(T_(1)) = (1)/(2)` or `T_(1) = 580 K`
Finally, `(60)/(100) = 1 - (273 + 17)/(T._(1))` or `(290)/(T._(1)) = (2)/(5)` or `T._(1) = 725 K`
`:.` Change in source temperature `= (725 - 580) K = 145 K`