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 heat engine has an efficiency eta. Temperature of source and sink ar

1.	A heat engine has an efficiency eta. Temperature of source and sink are each decreased by 100 K. Then, the efficiency of the engine
Answer» INCREASES decreases remains constant becomes 1 Solution :`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.