Question

5, A Carnot heat engine has an efficiency of 40%. If the high temperature is raised 10%, what is the new efficiency, keeping the same low temperature?

Answer 1

given that

efficiency $\eta$1 = 0.4

initial high temperature = T1

from the relation

efficiency $\eta$ = 1-T2/T1

here 0.4 = 1-T2/T1

T2/T1 = 1-0.4 = 0.6

T2 = 0.6T1

now high temperature is increase by 10%

that is T1' = 1.01T1

new efficiency $\eta$2 = 1-T2/T1'

$\eta$2 = 1-0.6T1/1.01T1 = 1-0.6/1.01

$\eta$2 = 0.406

that is 40.6%