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. `99^(@)C, 37^(@)C`
B. `80^(@)C, 37^(@)C`
C. `95^(@)C, 37^(@)C`
D. `90^(@)C, 37^(@)C`

Answer

A

Explanation

`eta_(1)=1-(T_L)/(T_H) = Q/(Q_1) = 1/6`
`or 5T_H - 6T_L =0` ..(1)
`eta_(2)=1-eta_(2) = 1-(T_L-62)/(T_H) = 2 eta_(2)=1/3`
or `2T_(H) - 3T_(L) = -186`..(2)
`:. T_(H)=372K = 99^@C`
`T_(L) = 5/6T_(H) = 5/6xx372K = 310K = 37^(@)C`.