The efficiency of a Carnot cycle is 1/6. If on reducing the temperature of the sink by 650C, the efficiency becomes 1/3, the initial temperatures between which the cycle is working are

Given  η1=16,η2=13If the temperatures of the source and the sink between which the cycle is working are T1 and T2, then the efficiency in the first case will beη1=1−T2T1=16In the second case η2=1−T2−65T1=13Solving T1 = 390 K and T2 = 325 K.