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Question

A Carnot engine operates at an efficiency of 40% with the sink at 27C. Find the amount by which the temperature of the source must be increased in order to increase the efficiency by 10%.

A
600 K
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B
100 K
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C
500 K
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D
300 K
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Solution

The correct option is B 100 K
Let T1 be the initial temperature of the source hence efficiency of heat engine is given by,
η=40%=40100
η=1T2T1 ...(i)
40100=1(273+27)T1
T1=500 K
For the efficiency to be increased by 10% i.e. new efficiency is η=50%
let T1 be the new temperature of the source
again from eq.(i),
50100=1(273+27)T1
or T1=600 K
The required increase in the temperature of the source,
ΔT=T1T1=600 K500 K
ΔT=100 K

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