Chem help please

Question

If a system has 5.00×102 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

∆E = U = q + w

q = heat = -5.00x10² kJ (negative sign because heat is released from system into surroundings)

w = work = 5.00x10² kcal (positive sign because work is done ON the system)

Now we have to get our units to match. To do so, we will convert 5.00x10² kcal to kJ as follows:

1 kcal = 4.184 kJ

5.00x10² kcal x 4.184 kJ/kcal = 2.09x10³ kJ

Finally, solve for ∆E/∆U

∆E/∆U = -500 kJ + 2090 kJ = +1592 kJ