If you do 1,000 J of work on a system and remove 500 J of heat from it, what is the change in its internal energy

Question

ybarrap · Answer

Doing 1000 J of work means that the system is doing negative work. Removing 500 J of heat from the system means that the system is losing heat:

U=Q-W=-500-(-1000)=-500+1000=500J

The internal energy (U) increases by 500 J, since the quantity U is positive.

anonymous · Answer

thank you I had a feeling it might be 500 J just wasn't sure :)

ybarrap · Answer

One important correction (which doesn't change our answer).

If heat is added to the system q > 0. Similarly, if work is done to the system, w > 0. Both of which increase internal energy. So $U=Q+W$ NOT $U=Q-W$. However, because I indicated that work done by the system was negative, we still got the right answer, but I think its much easier to think of work $\color{red}{added}$ and heat $\color{red}{added}$ $\color{red}{to}$ the system as being positive:

$$
U=Q\color{red}{+}W\
=-500+1000=500~J
$$

Now  (and I hope that this does not to make things too confusing) If we know we have a reversible system, then $w=-\int PdV$. Which makes internal energy equal to
$$
U=Q-\int_{v_1}^{v_2} PdV
$$
(which is where the negative sign comes in)

Where if work were done to the system then $v_2 < v_1$, which would increase internal energy, because then the integral part of U would be positive (i.e. $-\int_{v_1}^{v_2} PdV > 0$)