Answer
(a) $4.24MJ/mi$
(b) decrease
Work Step by Step
(a) We can find the required heat as follows:
$\Delta U=-Q-W$
This can be rearranged a:
$Q=-\Delta U-W$
We plug in the known values to obtain:
$Q=-(\frac{-1.19\times 10^8}{25})-5.20\times 10^5$
$Q=4.24MJ/mi$
(b) Due to an increase in miles per gallon, the efficiency is improved and thus there is a decrease in heat released to the atmosphere.
