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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equal to η

greater than η

less than η

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

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