What is the source temperature of the carnot engine reuired to get 70 % efficiency ?

Options

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

Answer

D

Explanation

Given , efficiency , `eta=70%`
Sink temperature , `T_(2)=27" " 273=300K`
Source temperature ,`T_(1)` =?
As we kown that efficiency is given by
`eta=1-(T_(2))/(T_(1))rArr70%=1-(300)/(T_(1))`
`(70)/(100)=1-(300)/(T_(1))rArr0.7=1-(300)/(T_(1))`
`(300)/(T_(1))=1-0.7rArrT_(1)=(3000)/(3)`
`T_(1)=1000K=1000-273=727^(@)C`