We can solve the problem by using the first law of thermodynamics:
[tex] \Delta U = Q-W [/tex]
where
[tex] \Delta U [/tex] is the change in internal energy of the system
[tex] Q [/tex] is the heat absorbed by the system
[tex] W [/tex] is the work done by the system on the surrounding
In this problem, the work done by the system is
[tex] W=-225 kcal=-941.4 kJ [/tex]
with a negative sign because the work is done by the surrounding on the system, while the heat absorbed is
[tex] Q=-5 \cdot 10^2 kJ=-500 kJ [/tex]
with a negative sign as well because it is released by the system.
Therefore, by using the initial equation, we find
[tex] \Delta U=Q-W=-500 kJ+941.4 kJ=441.4 kJ [/tex]
