A system receives 40.00 J of heat transfer while it does 10.00 J of work. Afterward, 25.00 J of heat transfer leaves the system while 4.00 J of work is done on the system. Using the sign convention in which heat into the system and work done by the system are both positive, what is the net change in internal energy of the system?

1. Net heat transfer Q is the heat entering minus the heat leaving: Q = 40.00 J - 25.00 J = 15.00 J. 2. Net work W is the work done by the system minus the work done on the system: W = 10.00 J - 4.00 J = 6.00 J. 3. Using the sign convention that heat into the system is positive and work done by the system is positive, the first law gives ΔU = Q - W. 4. Substituting the values found above: ΔU = 15.00 J - 6.00 J = 9.00 J. 5. Option A, 6.00 J, is only the net work W and stops short of applying the first law to it. 6. Option C, 15.00 J, is only the net heat transfer Q, mistaking it for the full change in internal energy. 7. Option D, 21.00 J, comes from adding Q and W instead of subtracting, which is the wrong sign for W in this convention. _Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.1 The First Law of Thermodynamics_