Answer
$5802KJ$
Work Step by Step
We can determine the total heat absorbed by the engine as follows:
As $Q_{Al}= m_{Al}C_{Al}\Delta T_{Al}$
We plug in the known values to obtain:
$Q_{Al}=(25\times 10^3) (0.900) (120-20)$
$\implies Q_{Al}=2250KJ$
Similarly $Q_{Fe}=m_{Fe}C_{Fe}\Delta T_{Fe}$
$\implies Q_{Fe} =(80\times 10^3) (0.444) (120-20)=3552KJ $
Now $Q=Q_{Al}+Q_{Fe} $
We plug in the known values to obtain:
$Q=2250+3552=5802KJ$