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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 100K. Then, the efficiency of the engine

A

increases

B

decreases

C

remains constant

D

becomes 1

Text Solution

Verified by Experts

The correct Answer is:
A
`eta=1-(T_(2))/(T_(1))=(T_(1)-T_(2))/(T_(1))`
When `T_(1)` and `T_(2)` both are decreased by 100 each `(T_(1)-T_(2))` stay constant.
`T_(1)` decreases
`:.eta` increase.
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