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. Can not be calculated

Answer

A

Explanation

`eta_1=1-(273)(473)` and `eta_2=1-(73)/(273)`
`eta_1=(200)/(273)` and `eta_2=(200)/(273)`
`:. (eta_1)/(eta_2)=(273)/(473)=0.577`