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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 sum of temperature (in kelvin) of the source and sink will be
PhysicsThermodynamicsJEE Main
Solution:
2117 Upvotes Verified Answer
The correct answer is: 682
The efficiency of heat engine is given by

   η=WQ=1-Q2Q1=1-T2T1

where T1 is temperature of source and T2 is temperature of sink.

Given, η1=16,   η2=13

              16=T1-T2T1 ...(i)

and 13=T1-(T2-62)T1 ...(ii)

Solving Eqs. (i) and (ii), we get

          T1=372 K

and T2=310 K

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