Good evening. I've happened to be sitting down today and just couldn't wrap my head around this question which seems rather simple at first. From reading about the first law and sign conventions as it deals with work done by a system and it's surroundings, when the system does work it has a positive sign convention and when the surroundings does work it has a negative sign convention.
The first Law of thermodynamics for a closed system reads $Q-W=U_2 - U_1$.
So let's assume the system undergoes some random process of which $Q=0$, $K.E.=0$ and $P.E.=0$ (for the system itself), and W is positive (indicating that the system is doing work on its surroundings). The First Law Equation then becomes $-W=U_2 - U_1$ (whereas, by convention, $U_2 - U_1$ would be taken as an increase in the internal energy of the system). We know that this is normally associated with an increase in temperature because in most cases, temperature increases are linked to increases in molecular interactions, which are internal to the system.
My question here (based on every accepted convention and equation) is why is there a decrease in the internal energy of a system if it is doing work on the surroundings? To provide a little more clarity here, suppose you were pushing a heavy load up a ramp. You(the system) would be doing work on the surrounding object(the heavy load). But at the same time, the temperature within your body is continuing to increase for as long as you exert a force for the total length of application. So therefore, rather than having a decrease in your bodily internal energy, that internal energy should be increasing, correct? In other words the work done by the system is generally increasing the internal energy of the system, rather than the opposite. Can anyone explain to me where the flaw is in my understanding of this, because I feel like I'm missing something.