1.

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)?

Answer»

`1.00`
`0.721`
`0.577`
`0.34`

SOLUTION :`eta_2=1 (-200+273)/(0+273)=200/273`
`eta_2 =1 - (-200+273)/(0+273) =200/273`
HENCE `eta_2//eta_1 = 0.577`


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