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Question

A Carnot's engine operates with an efficiency of 40% with its sink at 27 °C. The amount by which the temperature of the source be increased with an aim to increase the efficiency by 10% is K.

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Solution

Let T1 be the initial temperature of source, then
Using η=1T2T1
We have 40100=1(273+27KT1) or T1=500 K

For the efficiency to be 10 % more, i.e. 50 %, let T1 be the new temperature of the sink.
then, 50100=1(273+27 K)T1 or T1=600 K
The required increase in the temperature of the source is T1T=600 K500 K=100 K

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