Answer:
[tex]\eta = 0.411[/tex]
Explanation:
As we know that efficiency is defined as the ratio of output useful work and the input energy to the engine
So here we know that the
input energy = 441.3 kJ
energy rejected = 259.8 kJ
so we have
[tex]Q_1 - Q_2 = W[/tex]
[tex]W = 441.3 kJ - 259.8 kJ = 181.5 kJ[/tex]
now efficiency is defined as
[tex]\eta = \frac{W}{Q_1}[/tex]
[tex]\eta = \frac{181.5}{441.3}[/tex]
[tex]\eta = 0.411[/tex]