The efficiency of a Carnot cycle is `1//6`. By lowering the temperature of sink by `65 K`, it increases to `1//3`. The initial and final temperature of the sink are

Options

A. `390 K and 325 K`
B. `450 K and 410 K`
C. `350 K and 275 K`
D. `400 K and 310 K`

Answer

A

Explanation

`eta = (T_1-T_2)/(T_1) = 1/6` ..(i)
`eta^(') = (T_(1)-(T_(2)-65))/(T_1) = 1/3` ..(ii)
From equation (i) and (ii)
`(eta')/(eta) =((T_(1)-T_(2)+65)/(T_1))((T_1)/(T_(1)-T_(2))) = ((1//3))/(1//6)) = 2`
or `(T_(1)-T_(2)+65)/(T_(1)-T_(2))=2`
or `T_(1) - T_(2) =65` ..(iii)
From equation (i), `(65)/(T_1) = 1/6 or T_(1) = 390K`
and from equation (iii) , `T_(2) = T_(1) -65 = 325K`.