An ideal heat engine has an efficiency `eta`. The cofficient of performance of the engine when driven backward will be

Options

A. `1-((1)/(eta))`
B. `eta-((1)/(eta))`
C. `((1)/(eta))-1`
D. `((1)/(1-eta))`

Answer

C

Explanation

`eta = 1-(T_2)/(T_1)`
and `omega = (T_2)/(t_1-T_2) = (T_2//T_1)/(1-(T_2//T_1)) = (1-eta)/(eta) = 1/(eta)-1`.