An ideal heat engine working between temperature TH and TL has efficiency η. If both the temperature are raised by 100 K each, the new efficiency of heat engine will be:
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a
equal to η
b
greater than η
c
less than η
d
greater or less than η depending upon the nature of the working substance
answer is C.
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Detailed Solution
η = TH-TLTHη' = (TH+100)-(tL+100)(TH+100) = TH-TLTH+100 ∴ η' < η