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In a Carnot engine when T(2) = 0^(@)C an...

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)`?

A

`1.00`

B

`0.721`

C

`0.577`

D

`0.34`

Text Solution

Verified by Experts

The correct Answer is:
C
`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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