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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

Correct answer

682

Step-by-step solution

The efficiency of heat engine is given by η = 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, η 1 = 1 6 , η 2 = 1 3 ∴ 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 = 372 K and T 2 = 310 K

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