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

Options

A. increases
B. decreases
C. remains constant
D. becomes 1

Answer

A

Explanation

`eta = 1 - (T_(2))/(T_(1)) = (T_(1) - T_(2))/(T_(1))`
where `T_(1)` and `T_(2)` are the temperatures of the source and sink respectively.
When `T_(1)` and `T_(2)` both are decreased by 100 K each, `(T_(1) - T_(2))` stays constant and `T_(1)` Decreases. `:. eta` increases.