If a system has 425 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

∆U = w + q

∆U = change in internal energy = ?

w = work = +425 kcal (positive sign because work is done on the system)

q = heat = -500 kJ (negative sign because heat is released from system to surroundings)

NOTE: not sure you meant to have different units (kcal and kJ), but I'll assume you did for the sake of solving the problem. If they are supposed to be the same, e.g. kJ, then just don't do the conversion.

+425 kcal x 4.184 kJ/kcal = +1778 kJ of work = w

∆U = +1778 + (-500 kJ)

∆U = +1278 kJ

Answer 2

ΔU_sys = E_in - E_out = 275 kcal - ( 425kJ)(0.239 kcal/kJ) = 173.425 kcal

 

 

Since the energy balance of the system and surroundings is constant, ΔU_sys + ΔU_sur = 0.

 

 

So, ΔU_sur = -ΔU_sys = ( 173.425kcal)(1 kJ/0.239 kcal) = - 725kJ