If a system has 325 kcal of work done to it, and releases 5.00 * 10^2 kJ of heat into its surroundings, what is the change in internal energy of the system?

The equation that relates the work, heat and internal energy is given below.

ΔU = Q - W

where ΔU is the change in internal energy, Q is the heat (positive when added to the system and negative when removed from the system), W is work ( - when done by the system and + when done on the system)

Substituting the known values,

ΔU = (5.00 x 10^2 kJ) x (1 kcal/4.18 kJ) - 325 kcal

ΔU = - 205.38 kcal