A Carnot engine of efficiency 40% , takes heat from a source maintained at a temperature of 500 K . If is desired to have an engine of efficiency 60% . Then , the source temperature for the same sink temperature must be

Options

A. 650 K
B. 750 K
C. 550 K
D. 850 K

Answer

B

Explanation

%Efficiency of carnot engine,
`eta=(1 - (T_(2))/(T_(2)))xx 100`
where, `T_(2)` = sink temperature
and `T_(1)` = source temperature
For 40% efficiencyof carnot engine , `T_(1) = 500 K`
`rArr (40)/(100) = (1 -(T_(2))/(500))rArr 0 . 4 = 1 - (T_(2))/(500)`
`rArr (T_(2))/(500) = 0 . 6 rArr T_(2) = 300 K`
For 60% efficiency, the source temperature of Carnot engine for the same sink temperature,
`rArr (60)/(100) = 1- (T_(2))/(T_(1)) rArr (T_(2))/(T_(1)) = 1 - (6)/(10)`
`rArr (T_(2))/(T_(1)) = (4)/(10) rArr T_(1) =(10)/(4) T_(2)`
or `T_(1) = (10)/(4) xx 300 = 750 K`