A reversible Carnot heat engine converts `(1)/(4)` th of its input heat into work. When the temperature of the sink is reduced by 50 K its efficiency becomes `33 (1)/(3)%`. The initial temperatures of the source and the sink respectively are

Options

A. 600 K, 550 K
B. 600 K , 450 K
C. 300 K , 150 K
D. 450 K, 350 K

Answer

B

Explanation

Given work W =` (Q)/(A)`
where Q = input heat.
Efficiency `eta = (W)/(Q) =(Q//A)/(Q) implies eta = (1)/(4)`
Also `eta =1 -(T_(2))/(T_(1))`
where `T_(2)` = temperature of sink and `T_(1)`= temperature of source.
`1-(T_(2))/(T_(1))=(1)/(4) cdots (i)`
` implies` When temperature of sink is reduced by 50 K efficiency becomes `(100)/(3) %`
`eta= (100)/(3) xx (1)/(100) = (1)/(3)`
So, `1-(T_(2)-50)/(T_(1))= (1)/(3)`
`1 -(T_(2))/(T_(1)) + (50)/(T_(1))= (1)/(3) cdots (ii)`
Subtracting Eq (ii) from Eq. (i) we get
`(50)/(T_(1))= (1)/(3)-(1)/(4)= (1)/(12)`
`T_(1)= 600 K`
`T_(2)= (3)/(4) xx600 = 450 K`