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. `37^(@)C`
B. `62^(@)C`
C. `99^(@)C`
D. `12^(@)C`

Answer

C

Explanation

Here, `eta_(1)= 1/6, eta_(2)= 2xx1/6`
From `eta_(1)= 1-(T_(1))/(T_(1))`
`1/6= 1-(T_(2))/(T_(1)), (T_(2))/(T_(1))= 5/6 or T_(2)= (5T_(1))/6`
Again, `eta_(2)= 1-((T_(2)-62))/(T_(1))`
`((T_(2)-62))/(T_(1))= 1-1/3= 2/3`
`3T_(2)-186= 2T_(1)`
`3xx5/6T_(1)-186= 2T_(1)`
`0.5T_(1)= 186`
`T_(1)=372K= 372-273= 99^(@)C`