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 - (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)) = (2 73)/( 473) = 0. 577`