In a Carnot engine when `T_(2) = 0^(@)C` and `T_(1) = 200^(@)C` its efficiency is `eta_(1)` and when `T_(1) = 0^(@)C` and `T_(2) = -200^(@)C`. Its efficiency is `eta_(2)`, then what is `eta_(1)//eta_(2)`?

Options

A. `0.577`
B. `0.733`
C. `0.638`
D. Cannot be calculated

Answer

A

Explanation

`eta = 1-(T_2)/(T_1)=(T_1-T_2)/(T_1) implies eta_(1) = ((473-273)/473) = 200/473`
and `eta_(2) = (273-73)/(273) = 200/273`
So required ratio = `(eta_1)/(eta_2) = (273)/(473) = 0.577`.