Newton's Law of Cooling: $\delta Q$ or $\mathrm{d}Q$?

Question

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?

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

Answer 1

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" $q$.

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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.

Answer 2

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).

Answer 3

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.

Answer 4

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":

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":

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.