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

c. `eta = 1 - (T_(2))/(T_(1))`
`omega = (T_(2))/(T_(1) - T_(2)) = (T_(2) // T_(1))/(1 - (T_(2) // T_(1))) = ((1 - eta))/(eta) = (1)/(eta) - 1`