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.


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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    $\begingroup$ 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. $\endgroup$ – Nikolaj-K Oct 11 '12 at 15:05
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    $\begingroup$ 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. $\endgroup$ – Jerry Schirmer Oct 11 '12 at 15:39
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    $\begingroup$ @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' $\endgroup$ – Jerry Schirmer Oct 11 '12 at 16:34
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    $\begingroup$ @DavidZaslavsky: I'm probably just cranky after having to explain this out of student's heads one too many times. I still feel that this notation would be almost the only place in all of science where $\Delta$(quantity) does not mean "change in (quantity)". $\endgroup$ – Jerry Schirmer Oct 11 '12 at 19:05
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    $\begingroup$ @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. $\endgroup$ – David Z Oct 11 '12 at 19:10

What's true is that it's law of conservation of energy. It particularly states: Q= Del(U) + W,

U: internal energy. That is, the total energy given to a system does two things, first it makes the system do the desired useful work and second it changes the internal energy of the system. The work can be positive or negative according to W=q(dv). Example: If the gas in piston expands and when the gas in the piston is under compression then the work done would be positive and negative respectively. :)


The previous answer, in my opinion is half-baked. The use of the delta term is quite misleading, as pointed out in the comments. Now, according to almost every leading textbook and my Professor, who is also a leading author, the I Law of Thermodynamics can be stated as:

Q = ΔE + W

where Q is the heat transfer across the system boundary, W is the work transfer across the system boundary and ΔE is the change in energy of the system. Here, the energy of the system 'E' includes the Macroscopic Energy mode(Kinetic and Potential Energies of the bulk) and the Microscopic Energy Mode(Internal energy that included the translational kinetic energy, rotational kinetic energy, vibrational potential energy, chemical energy etc. of each molecule). The sign of Q and W depend on the sign convention used. Hence, you don't need to include the +- signs in the equation as it will only tend to confuse you!

The above statement gives a whole picture of the I Law of Thermodynamics for a 'Closed System' undergoing a 'Change of State'.

  • $\begingroup$ Like seriously??? I would seriously like an explanation for that downvote!!! Its the most complete answer you can find for that question. $\endgroup$ – don_Gunner94 Apr 18 '15 at 10:35
  • $\begingroup$ Your points are all correct, but you seem to miss the spirit of the question. The sign of $Q$ and $W$ depend on the sign convention used, indeed, and the question is how to implement the different sign conventions, which you have not addressed. (Not my downvote btw.) $\endgroup$ – Emilio Pisanty Apr 18 '15 at 11:16
  • $\begingroup$ If you would go through my answer, you would find that I have mentioned about the sign convention affecting the + and - signs! I have also mentioned that these signs need not be used, as they only tend to confuse! $\endgroup$ – don_Gunner94 Apr 18 '15 at 11:25
  • $\begingroup$ On the contrary, your answer does contain a + sign, which only holds true if one does remember that W is the work performed by, and not on, the system. Your answer does little to alleviate this confusion, which is the heart of the OP's question. $\endgroup$ – Emilio Pisanty Apr 19 '15 at 2:16
  • $\begingroup$ There's a minus - sign right next to the + sign! Which everybody seems to have rather conveniently ignored $\endgroup$ – don_Gunner94 Apr 19 '15 at 2:24

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