If a system has 475 kcal of work done to it, and releases 5.00×102 kJ of heat into its surroundings, what is the change in internal energy (Δ𝐸 or Δ𝑈) of the system?

Answer 1

Work (w) done ON the system has a positive value = + 475 cal

Heat (q) released to the surroundings has a negative value = -5.00x10² kJ = -500 kJ

Change in internal energy ∆E or ∆U = q + w

∆E = -500 kJ + 475 kcal

Now we need to get the units to be the same

1 cal = 4.184 J

475 cal x 4.184 J/cal = 1987 J = 1.987 kJ

∆E = -500 kJ + 1.987 kJ

∆E = -498 kJ