A reversible engine converts one-sixth of the heat input into work. When the temperature of the sink is reduced by `62^(@)C`, the efficiency of the engine is doubled. The temperature of the source and sink are

Options

A. `99^(@)C, 27^(@)C`
B. `80^(@)C, 37^(@)C`
C. `95^(@)C, 37^(@)C`
D. `90^(@)C, 37^(@)C`

Answer

A

Explanation

`eta_(1) = 1 - (T_(2))/(T_(1)) = (W)/(Q_(1)) = (1)/(6)` r `5T_(1) - 6T_(2) = 0` ….(i)
`eta_(2) = 1 - ((T_(2) - 62))/(T_(1)) = 2eta_(1) = (1)/(3)` or `2T_(1) - 3T_(2) = -186` …(ii)
Solving (i) and (ii), we get `T_(1) = 372 K = 99^(@)C`
`T_(2) = (5)/(6)T_(1) = (5)/(6) xx 372 K = 310 K = 37^(@)C`