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

Options

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

Answer

D

Explanation

Efficiency, `eta =(1- T_(2)/T_(1)) xx 100`
`eta. = 1-(T_(2) - 100)/(T_(1) - 100) xx 100 =(T_(1) -T_(2))/(T_(1) - 100) xx 100`
Comparing with `eta` we get, the efficiency increases.