An ideal heat engine working between temperatures TH and TL has an efficiency η . If both the temperatures are raised by 100 K each, the new efficiency of the heat engine will be :
A
equal to η
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B
greater than η
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C
less than η
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D
greater or less than η depending upon the nature of the working substance.
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Solution
The correct option is C less than η The efficiency of a heat engine is given by, η=TH−TLTH Increasing both the temperatures by 100, η′=TH−TLTH+100 Thus, the efficiency will effectively decrease.