Chemestry heat and entropy

Question

If a system produces 111.0 kJ of heat and does 45.0 kJ of work on the surroundings, what is the ∆E for the surroundings, in kJ

Answer 1

∆E = ∆U = q + w

q = heat = -111.0 kJ (negative sign because the system is producing heat and losing it to the surroundings)

w = work = -45.0 kJ (negative sign because the system is doing work on the surroundings)

∆E = ∆U = (-111.0 kJ) + (-45 kJ) = 155.0 kJ