The efficiency of Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

Options

A. 100 K
B. 600 K
C. 400 K
D. 500 K

Answer

C

Explanation

The efficiency of a Carnot engine is `eta = 1 - (T_(2))/(T_(1))` where `T_(1)` is the temperature of the source and `T_(2)` is the temperature of the sink.
Here, `T_(2) = 500 K, eta = 50%, T_(1) = ?`
`:. 0.5 = 1 - (500)/(T_(1))` or `T_(1) = 1000 K`
Let `T._(2)` be the new sink temperature. Then
`eta. = 1- (T._(2))/(T_(1)) :. 0.6 = 1 - (T._(2))/(1000)`
or `T._(2) = 400 K`