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

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

functionduring 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.