In this popular answer, I invoked Newton's Law of Cooling/Heating:

$$\dot{q}=hA\Delta T\tag{1}$$ $$\dot{q}=\frac{\mathrm{d} Q}{\mathrm{d}t}\tag{2}$$ $$\dot{q}=\frac{\delta Q}{\mathrm{d}t}\tag{3}$$ $$\mathrm{d}U=\delta Q-p\mathrm{d}V\tag{4}$$ $$\dot{q}=-mc_p\frac{\mathrm{d}T}{\mathrm{d}t}\tag{5}$$ $$\delta Q= mc_p\mathrm{d} T\tag{6}$$ $$-mc_p\frac{\mathrm{d}T}{\mathrm{d}t}=hA\Delta T\tag{7}$$

$(1)$ leads to $(2)$, $(5)$ and $(7)$ which usually is an ODE with analytical solution.

At some point member '@EricDuminil' weighed in, in the comments, claiming I needed to use $(3)$ because of $(4)$ (First Law of Thermodynamics, no less!) Eric has since scrubbed his comments but did suggest an edit according to $(3)$, which I accepted.

My question is, was Eric right or is this $\delta Q$ lark just pedantry, at least in this specific context?

Edit: Following several useful answers, I've reverted back to $(3)$ in my own text.

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    $\begingroup$ @Jonas We're not talking about partial derivatives ($\partial$) but about inexact differentials ($\delta$). en.wikipedia.org/wiki/Inexact_differential $\endgroup$
    – Gert
    Feb 13, 2021 at 17:35
  • $\begingroup$ In (4) there is no time, because time is an extraneous parameter (non-thermodynamical), so $\delta Q (p, T, t)$ is the same as $ dQ (t)$. $\endgroup$
    – DanielC
    Feb 13, 2021 at 17:46
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/65724/2451 , physics.stackexchange.com/q/36150/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 13, 2021 at 20:06
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    $\begingroup$ As far as I'm concerned (for whatever that's worth), the use of $\delta Q$ notation is a lark in all cases; its introduction has done nothing but confuse endless numbers of students over the ages. It should have been sufficient to say that Q and W are both functions of path, while the thermodynamic functions such as U are equilibrium physical properties of the material (i.e., functions of state). $\endgroup$ Feb 13, 2021 at 20:50
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    $\begingroup$ @ChetMiller It should have been sufficient to say that Q and W are both functions of path, To be fair, that's what Eric wrote but then insisted I use the $\delta Q$ notation. $\endgroup$
    – Gert
    Feb 13, 2021 at 20:53

4 Answers 4


I suppose what you consider pedantry is ultimately a matter of personal opinion. I would guess that I tend towards the more pedantic end of PhysSE users, so I'll give my point of view.

To me, the notation $\mathrm dQ$ means "the differential of some function $Q$"; that is, there exists some function $Q$, and $\mathrm dQ$ is a tiny change in its value. When we write the first law as $$dU = \delta Q - \delta W\qquad (\star)$$ we change the notation because $Q$ and $W$ are not functions. If they were, it would imply that it makes sense to talk about the heat or work present in a system, which it of course does not. Instead, $(\star)$ reads

An infinitesimal change in the internal energy function during some process is equal the heat added to the system during the process minus the work done by the system during the process.

$\delta Q$ is not to be interpreted as "the $\delta$ of some function $Q$"; rather, $\delta Q$ is a primitive symbol in its own right which denotes an infinitesimal bit of heat added to the system.

Now with that being said, one could define a function $q(t)$ which gives e.g. the total heat added to the system since time $t=0$. $\mathrm dq = \dot q \mathrm dt$ is perfectly well-defined in this case. Furthermore, when the process in question is "the system is supplied with heat over a time interval $\mathrm dt$," we have that $\delta Q = \dot q \mathrm dt$.

Though it is tempting to write $\delta Q/dt = \dot q$ and then say "ah, well then $Q=q$, let's just use the same symbol for both" or some such thing, I would regard that as an abuse of notation. Leaving it as $\delta Q = \dot q \mathrm dt$ makes it (more) clear that the tiny bit of heat added to the system ($\delta Q$) is given by the differential of your "cumulative heat function" $q$.

Stepping down off of my soapbox, I would express things as follows. If $q(t)$ is the total heat added to the system by time $t$, then $\dot q(t)$ is the rate at which heat is being added at time $t$. Assuming that the system has temperature $T(t)$ and its surroundings have temperature $T_0$ (assumed constant for simplicity), we would have

$$\dot q(t) = P_{in}(t)-hA\big(T(t) - T_0\big)$$

where $P_{in}(t)$ is the power being added at time $t$ (in your linked answer, from the microwave). By definition of the specific heat capacity, the addition of some heat $\delta Q$ causes a corresponding increase in temperature given by

$$\delta Q = mc \mathrm dT$$

Since $\delta Q = \dot q\mathrm dt$, we obtain

$$mcT'(t) = \dot q = P_{in} - hA(T-T_0)$$

which is the ODE to which you refer.

  • $\begingroup$ Actually, using the notation $\delta Q$ one should write $\dot q = \frac{{\mathrm d}\delta Q}{{\mathrm d}t}$. $\endgroup$ Feb 13, 2021 at 18:21
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    $\begingroup$ @GiorgioP I disagree. $\delta Q$ is not a function of time - it's just the symbol for the tiny bit of heat being added during whatever process is under consideration. When that process is "the system is being heated for time $\mathrm dt$", then $\delta Q = \dot q \mathrm dt$. $\endgroup$
    – J. Murray
    Feb 13, 2021 at 18:26
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    $\begingroup$ @GiorgioP Having read your answer, I regard the symbols $\delta Q$ and $\delta W$ as meaning the same thing as your symbols $q$ and $w$ from your last line. $\endgroup$
    – J. Murray
    Feb 13, 2021 at 18:30
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    $\begingroup$ I added my last comment before reading your reply. I think that we agree beyond notational differences. $\endgroup$ Feb 13, 2021 at 18:36
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    $\begingroup$ @EricDuminil Thank you for your kind words :) $\endgroup$
    – J. Murray
    Feb 13, 2021 at 23:28

The suggestion of using $(3)$ is not pedantry, but it is plainly wrong. And my statement remains true whatever is the status of work and heat, whether exact differentials or not.

The reason is the following. Whatever is the attitude about the expression of the first principle, the differentials appearing there correspond to a linear approximation of the variation of the corresponding function as a function of state variables. This is a different functional dependence as the time dependence. More explicitly, while we cannot write in general. $$ {\mathrm d} U = {\mathrm d} Q +{\mathrm d} W, $$ we can write $$ \frac{{\mathrm d} U}{{\mathrm d} t} = \frac{{\mathrm d} Q}{{\mathrm d} t} + \frac{{\mathrm d} W}{{\mathrm d} t}. $$ In the first case, a state variable dependence is implied. In the second, it is just required the time dependence of quantities like $Q(t)$ and $W(t)$ which may or may not be exact differentials as a function of the state variables. Said in another way, it is impossible to speak meaningfully about differences of heat or work in general. Still, it is always possible to take differences of time-varying functions between two different times.

Notice that all I wrote above cannot be considered a matter of opinion, but it is sound math. The only side of this problem that could be considered opinion is the way of writing equation $(3)$. I add that, following people who knew quite well thermodynamics, like Max Planck, I prefer to write the first principle as $$ {\mathrm d} U = q + w $$ eliminating the need to introduce an ill-defined entity like inexact differentials and make it easier to understand time variations.

  • $\begingroup$ Thanks for your answer. Couldn't we avoid $\delta Q$ by writing $\frac{\mathrm{d}Q}{\mathrm{d}t}\mathrm{d}t$? I mean, here $Q(t)$ is well-defined. And what do your $q$ and $w$ stand for? $\endgroup$
    – Gert
    Feb 13, 2021 at 19:00
  • $\begingroup$ @Gert, that is a possibility and it is the way Kondepudi&Prigogine interpret what they write as ${\mathrm d} Q$. $q$ and $w$ in the final part of my answer stand for the quantities you indicate as $\delta Q$ and $-pdV$. $\endgroup$ Feb 13, 2021 at 20:30
  • $\begingroup$ OK, thank you, Giorgio. $\endgroup$
    – Gert
    Feb 13, 2021 at 20:36
  • $\begingroup$ I agree that insisting on writing $\frac{\delta Q}{dt}$ in this specific case is wrong, but I don't think that writing $\frac{\delta Q}{dt}$ is wrong in itself. $\endgroup$ Feb 13, 2021 at 23:07
  • $\begingroup$ @EricDuminil, I said it is wrong on the basis of the existing use of the notation. Do you have examples of that notation in the scientific literature? If you are introducing a new notation, that is perfectly possible but you should state clearly that it is a new proposal and I would comment that it would be quite a misleading notation. $\endgroup$ Feb 14, 2021 at 7:57

Relationships (3) and (4) are true in general since heat and work depend on the path and are therefore treated using inexact differentials. But in this case for heat you know the path by relationship (1), so you can treat the heat as an exact differential as in relationship (2).


I'm the original commenter.

Process function vs state function.

My argument is simply that heat is a process function, and not a state function.

A system does not have heat. It does not make sense to talk about variation of heat for a system, so it's for example forbidden to write $\Delta Q$, which would mean $Q_2 - Q_1$. Sadly, this notation is used many times over the Internet.

What is defined, though, is: "At the end of a process, how much energy has been transferred from one body to another, due to temperature difference?" This is heat, and it's simply written $Q$.

Exact differential vs inexact differential

Writing $dQ$ would basically mean "a very small $\Delta Q$". But $\Delta Q$ isn't defined, so another notation needs to be used. That's why $\delta Q$, an inexact differential can be used instead of $dQ$, which would be an exact differential.

From "Fundamentals of Engineering Thermodynamics":

heat is not a property

Note that exact differentials are mathematically well-defined, and come with a number of nice properties, which heat or work do not have.

For example, integrating a state function $U$ over a cycle, $$\oint \mathrm{d} U = 0 \, ,$$ while for a path function $Q$ $$\oint \delta Q \neq 0 \, .$$

If $Q$ and $W$ were state functions, engines would be useless : they would absorb no heat and do no work at all, since $Q$ and $W$ would be reset at each cycle.

$\delta Q$ is basically a "tread lightly" sign, indicating that not every operation is allowed or even defined. It's not a differential, it is just a "small bit of heat transferred to the system".

Your question

As mentioned in J.Murray's excellent answer, it is possible to define a new $q$ function for your specific case, and say that $\delta Q = \dot q \mathrm dt$. It's also mentioned in the "Fundamentals of Engineering Thermodynamics":

rate of heat transfer

As far as I can tell, it's never wrong to write $\delta Q$, but it can be wrong to write $dQ$ (e.g. in $dU = dQ + dW$) so I find it easier to stick to $\delta Q$.

In your specific case, $\dot{Q}$ might be the easiest way to avoid confusion.

  • $\begingroup$ A system does not have heat. It does not make sense to talk about variation of heat for a system, so it's for example forbidden to write $ΔQ$, which would mean $Q_2−Q_1$ I'll probably go to my grave not understanding what that means. The kettle I use for my tea contains heat energy. Then I add heat energy to bring the water to the boil. I've added $\Delta Q=Q_2-Q_1$ heat energy to the water. Simples, Aleksei! $\endgroup$
    – Gert
    Feb 13, 2021 at 23:55
  • $\begingroup$ @Gert: I guess we found the crux of the matter, then. You can talk about the internal energy of your kettle, because it's a state function, and basically describes how much "thermal energy" is inside your kettle. During a process (e.g. "preparing your tea"), you can transfer energy to your kettle, as heat, for example with a stove. But heat is just energy being transferred. It's not inside the stove or inside the tea. If no work is performed on your tea, you could say that $\Delta U = Q + 0$, so $U2 - U1 = Q$. $\endgroup$ Feb 14, 2021 at 0:13
  • $\begingroup$ @Gert: If you only remember one thing from this whole thread, it is that "The kettle I use for my tea contains heat energy." is nonsensical and should never be written. This isn't a matter of taste, this isn't about notation or about pedantry. This is about the definition of heat, and it's basically the foundation on which thermodynamics is built. I strongly advise you to read the first chapters of "Thermodynamics and an Introduction to Thermostatistics", by Herbert B. Callen or "Fundamentals of Engineering Thermodynamics" by Moran. Don't go to your grave before you read those! $\endgroup$ Feb 14, 2021 at 13:26
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    $\begingroup$ @Gert Consider two identical boxes filled with the same ideal gas at $300$ K. I adiabiatically compress the first box to 1/2 its original volume, raising its temperature to $300\cdot 2^{2/3}\approx 476$ K. I then isothermally compress the first box to 1/2 its original volume, and then subsequently put it on a hot plate to raise its temperature to match the first box. The initial and final states of both boxes are identical, but heat was delivered to the second box and not the first. Therefore, heat (or more suggestively, heat transfer) is a quantity which is associated [...] $\endgroup$
    – J. Murray
    Feb 15, 2021 at 2:05
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    $\begingroup$ [...] to a process, not to a state. It makes perfect sense to talk about the amount of heat which has been added to a system during whatever process is under consideration; it also makes sense to talk about how much heat has been delivered to a box since some initial time. But unlike dU (an infinitesimal change in the internal energy due to a change in S,V, or N), there is no function Q of the thermodynamical variables which would make the notation dQ justified in the same way. $\endgroup$
    – J. Murray
    Feb 15, 2021 at 2:06

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