A cannot engine has efficiency `(1)/(6)`. If temperature of sink is decreased by `62^(@)C` then its efficiency becomes `(1)/(3)` then the temperature of source and sink:

Options

A. `33^(@)C, 67^(@)C`
B. `37^(@), 99^(@)C`
C. `67^(@)C, 33^(@)C`
D. `97K, 37 K`

Answer

B

Explanation

`(W)/(Q) = (1)/(6)`
`1 = (T_(L))/(T_(H)) = (1)/(6)`
`(T_(L))/(T_(H)) = n (5)/(6)`
If sink temp decrease by `62^(@)`C then
`1 - (T_(L) - 62)/(T_(H)) = (2)/(6) implies (T_(L) - 62)/(T_(H)) = (2)/(3)`
`2T_(H) = 3T_(L) - 186 implie 2T_(H) = 3 xx (5)/(6) T_(H) - 186`
`2T_(H) - (5)/(2) T_(H) = - 186 implies (5 - 4)/(2) T_(H) = 186`
`T_(H) = 186 xx 2 = 372 K = 99^(@) C`
`T_(L) = (5)/(6) xx 372 = 310 K = 37^(@)C`