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

Must be negative.substance increases

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