Question

Calculate the change internal energy (δe) for a system that is giving off 45.0 kj of heat and is performing 855 j of work on the surroundings.

Answer 1

The change in internal energy of a system is given by (second law of thermodynamics)
$\Delta U = Q + W$
where Q is the heat absorbed by the system and W is the work done on the system.

In order to correctly evaluate the internal energy change, we must be careful with the signs of Q and W:
Q positive -> Q absorbed by the system
Q negative -> Q released by the system
W positive -> W done on the system by the surroundings
W negative -> W done by the system on the surroundings

In our problem, the heat released by the system is $Q=-45 kJ=-45000 J$ (with negative sign since it is released by the system), and the work done is $W=-855 J$ still with negative sign because it is performed by the system on the surrounding, so the change in internal energy is
$\Delta U = Q +W=-45000 J - 855 J=-45855 J$