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Which one true in First law of thermodynamics:

  1. $Q = \Delta U \pm W = \Delta U \pm p\Delta V$? (where $\Delta U$ is change of internal energy, $W$ work made by system and $Q=cm\Delta T $ heat made by system.)

  2. or $\Delta U= \Delta Q + \Delta W $? (where $\Delta U$ is change of internal energy, $W$ work made by system and $Q=cm\Delta T $ heat made by system.)

First one from one the finnish physics texbook another is from here.

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1 Answer 1

up vote 2 down vote accepted

It is pretty much a matter of convention regarding who is doing work on whom. For me the most conceptually clear picture is the wikipedia version, $$\Delta U=\Delta Q+\Delta W,$$ i.e., that the change in the internal energy of the system equals the heat delivered to it plus the work performed on the system. (Note, however, the difference from what you quote!) If one takes $\Delta W$ to be the work performed by the system then its sign will change, but this does not of course change the physical content of the law. If in doubt, put it in words! Once you're clear on what each symbol means the signs will follow automatically.

A couple of caveats, though: note that $Q$ as such is a misleading term. One can only assign heat quantities to processes, which is emphasized by the notation $\Delta Q$.

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I feel that $\Delta Q$ is not a particularly obvious notation to emphasise that point eighter. It suggests that a particular something, "a heat", is changed and so rather enforces the wrong idea. Rather than $\Delta Q$, I think $\Delta_H U$ would be a better notation for example. –  NikolajK Oct 11 '12 at 15:05
One should never write $\Delta W$ or $\Delta Q$. Work and heat are not state variables that can be measured at the beginning and end of a process. They are path-dependent processes that can only be assigned to the processes. $\Delta Q$ implies that you are taking some sort of $Q_{f}-Q_{i}$, which is nonsense. –  Jerry Schirmer Oct 11 '12 at 15:39
@Jerry but that's not what $\Delta$ means in this case. It refers to the amount of heat transferred in or out of the system. I don't think the notation is so bad as long as we are clear about what it means. –  David Z Oct 11 '12 at 16:24
@DavidZaslavsky: that is exactly what $Q$ and $W$ refer to. And it's confusing when you have $\Delta$'s that mean EXACTLY that set to equality with them. $\Delta S \equiv S_{f}-S_{i}$, $\Delta V \equiv V_{f}-V_{i}$, and $\Delta U \equiv U_{f}-U_{i}$. If you use $\Delta W$ to mean what you say, then $\Delta W = -P\Delta V$ is a baffling 'equation' –  Jerry Schirmer Oct 11 '12 at 16:34
@Jerry fair enough, and I can definitely see the benefit in pushing no-Δ equations in an educational context if that's what it takes to make students understand what's really going on. –  David Z Oct 11 '12 at 19:10

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