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

Options

A. `270^(@)C`
B. `1000^(@)C`
C. `727^(@)C`
D. `90^(@)C`

Answer

C

Explanation

Efficiency of the Carnot engine,
`eta=1-(T_(2))/(T_(1))`
where `T_(1)` and `T_(2)` are the temperatures of source and sink respectively.
Here, `eta=70%= (70)/(100)=0.7`
`T_(1)=27^(@)C=(27+273)K=300K`
`:.0.7=1-(300)/(T_(1))`
`(300)/(T_(1))=1-0.7=0.3`
`T_(1)=(300)/(0.3)K=1000K=(1000-273)""^(@)C`
`=727^(@)C`