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 temperatures of the source and sink are

Options

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

Answer

A

Explanation

Initially the efficient of the engine was `1/6` which increases to `1/3` when the sink temperature reduces by `62^(@)` C
`eta =1/6 =1 -T_(2)/T_(1)`, when `T_(2)` = sink temperature
`T_(1)` = source temperature `rArr T_(2)= 5/6 T_(1)`
Secondly,
`1/3=1 -(T_(2)-62)/T_(1) =1-T_(2)/T_(1) +62/T_(1) =1-5/6 + 62/T_(1)`
or `T_(1) = 62 xx 6 = 372 K = 372 - 273 = 99^(@)` C
& `T_(2) = 5/6 xx 372 = 310 K = 310 - 273 = 37^(@)` C