An ideal heat engine working between temperature `T_(H)` and `T_(L)` has efficiency `eta`. If both the temperature are raised by `100 K` each the new efficiency of heat engine will be

Options

A. equal to `eta`
B. greater than `eta`
C. less than `eta`
D. greater or less than `eta` depending upon the nature of the working substance

Answer

C

Explanation

`eta = (T_H-T_L)/(T_H)`
`eta^(') = ((T_H+100)-(T_L+100))/((T_H+100)) = (T_H-T_L)/(T_H+100) :. eta' lt eta` .