Answer: 57.1 %
Explanation:
The efficiency of the engine is given by:
[tex]\eta=\frac{Q_{used}}{Q_{in}}[/tex]
where the numerator corresponds to the heat used by the engine, while the denominator corresponds to the heat in input to the engine.
In this problem, the heat in input is
[tex]Q_{in}=7000 J[/tex]
while the heat used is
[tex]Q_{used}=7000 J-3000 J=4000 J[/tex]
So, the efficiency is
[tex]\eta=\frac{4000 J}{7000 J}=0.571=57.1\%[/tex]
