1.

What is the source temperature of the Carnot engine required to get 70% efficiency ? Given, sink temperature = 27^(@)C.

Answer»

`1000^(@)C`
`90^(@)C`
`270^(@)C`
`727^(@)C`

SOLUTION :Given,
Efficiency `eta=70%`
Sink temperature `T_(1)=27^(@)C+273=300 K`
Source temperature `T_(2)=?`
As we KNOW that efficiency is given by
`eta=1-(T_(2))/(T_(1))`
`70% =1-(300)/(T_(1))`
`(70)/(100)=1-(300)/(T_(1))`
`(300)/(T_(1))=1-0.7`
`T_(1)=(300)/(0.3)`
=1000 K
or `T_(1)=1000-273`
`=727^(@)C`.
So, correct choice is (d).


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