The efficiency of a Carnot cycle is `1//6`. If on reducing the temperature of the sink by `65^@C`, the efficiency becomes `1//3`, find the source and sink temperatures between which the cycle is working.

Options

A. 390K,325K
B. 780K,325K
C. 390K,162K
D. 300K,100K

Answer

A

Explanation

Given `eta_(1)=1/6,eta_(2)=1/3`
If the temperatures of the soure and the sink between which the cycle is working are `T_(1)` and `T_(2)`, then th efficiency in the first case will be
`eta_(1)=1-(T_(2))/(T_(1))=1/6`
in the second case `eta_(2)=1-(T_(2)-65)/(T_(1))=1/3`
Solving `T_(1)=390K` and `T_(2)=325K`