When the absolute temperature of the source of a Carnot heat engine is increased by 25%, its efficiency increases by 80%. The new efficiency of the engine is

Options

A. `12%`
B. `24%`
C. `48%`
D. `36%`

Answer

D

Explanation

`eta` is engine initial efficiency
new efficiency `eta +(80)/(100)eta`
`= eta +0.8 eta =1.8 eta`
`eta =(T_1 - T_2)/(T_1)`
`eta =(100 - T_2)/(100)`
`eta =(125-T_2)/(125)`
From equation (1) and (2) , we get ,
`(eta)/(eta)=(100-T_2)/(100)xx(125)/(125-T_2)`.
`=(1)/(1-8)=(100-T_2)/(100)xx(125)/(125-T_2)`.
`(5)/(9)=(100-T_2)/(4)xx(5)/(125-T_2)`
`=900-9T_2 =500 -4T_2`
`=900-500=5T_2`
`=400= 5T_2`
`T_2 =(400)/(5)=80`
`eta =1 -(T_2)/(T_1)`
`=1-(80)/(125)`
`=(125-80)/(125)=(45)/(125)`
`eta^(11)%=(45)/(125)xx100=36%`.