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A heat engine has an efficiency eta. Tem...

A heat engine has an efficiency `eta`. Temperatures of source and sink are each decreased by 100 K. The efficiency of the engine

A

increases

B

decreases

C

remains

D

becomes 1

Text Solution

Verified by Experts

The correct Answer is:
A

`eta =1=1-(T_(1)-T_(2))/(T_(1))`
When `T_(1)` and `T_(2)` both are decreased by 100 K each, `(T_(1)-T_(2))` stays constant . `T_(1)` decreases
`eta` increases.
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Knowledge Check

  • A heat engine has an efficiency eta . Temperature of source and sink are each decreased by 100 K. Then, the efficiency of the engine

    A
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    B
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    A
    `T_(1)+5,T_(2)+5`
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