An engine has an efficiency of `1/6`. When the temperature of the sink is reduced by `62^(@)C`, its efficiency becomes `1/3`, then the temperature of the source is

Options

A. `62^(@)C`
B. `99^(@)C`
C. `37^(@)C`
D. `124^(@)C`

Answer

B

Explanation

Efficiency of an engine `eta=1-(T_(2))/(T_(1))`
`T_(1)=` temperature of source
`T_(2)=` temperature of sink
`:.1/6=1-(T_(2))/(T_(1))`
`implies(T_(2))/(T_(1))=5/6`……..i
When temperature of sink is decreased by `62^(@)C` (or 62K) efficiency become double, as temperature of source remains unchanged.
`:.2xx1/6=1,((T_(2)-62))/(T_(1))`
`implies1/3=1-((T_(2)-62))/(T_(1))`.........ii
Solving equation i and ii
We have `T_(1)=372K=99^(@)C`