Answer:
28.8 %
Explanation:
The efficiency of an engine is given by:
[tex]\eta = \frac{W}{Q_{in}}[/tex]
where
W is the useful work done
[tex]Q_in[/tex] is the heat in input
For this machine, we have
W = 15.0 kJ is the work done
[tex]37.0 kJ[/tex] is the heat exchausted to the cold reservoir, so the total amount of heat in input is
[tex]Q_{in} = 15.0 kJ + 37.0 kJ=52.0 kJ[/tex]
And so the efficiency is
[tex]\eta=\frac{15.0 kJ}{52.0 kJ}=0.288[/tex]
which corresponds to 28.8 %.
