A Carnot reversible engine converts `1//6` of heat input into work . When the temperature of the sink is reduced by 62 K, the efficiency of carnot's cycle becomes `1//3` . The temperature of the source and sink will be

Options

A. 372 K,310 K
B. 181 K,150 K
C. 472 K.410 K
D. None of these

Answer

A

Explanation

A heat engine is a device which converts heat energy into meachnical energy efficiency of heat engine is the fraction of total supplied to the engine whichis converted into work.
`eta=(W)/(Q)=1-(Q_(2))/(Q_(1))=1-(T_(2))/(T_(1))`
Where ,`T_(1)` is temperature of source and `T_(2)` is temperature of sink .
`"Given"," "eta_(1)=(1)/(6),eta_(2)=(1)/(3)`
`therefore" "(1)/(6)=(T_(1)-T_(2))/(T_(1))` ....(i)
`and " " (1)/(3)=(T_(1)-(T_(2)-62))/(T_(1))` ...(ii)
Solving Eqs .(i) and (ii) ,we get
`T_(1)=372KandT_(2)=310K`