What should be the temperature of the source to increases the efficiency to `30%`

Options

A. `298.4^(@)C`
B. `298.4K`
C. `400K`
D. `600K`

Answer

A

Explanation

`eta'= 30%, T_(2)= 400K, T_(1)=?`
As `eta'=1 -(T_(2))/(T_(1))`
`(T_(2))/(T_(1))=1-eta'= 1-(30)/(100)= (70)/(100)`
`T_(1)=(10)/7T_(2)=(10)/7xx400= 571.4K`
`T_(1)= 571.4-273 = 298.4^(@)C`.