Calculation of temperature change in a refrigerated box Let’s say I have a refrigerated box held at temperature T, and I also have a door that opens on the front face for S seconds.  How does one go about calculating how long it takes for the temperature of the box to raise D degrees?  Or can someone suggest keywords to search on so I can learn how to do this?  All of the thermodynamics searches I have performed discuss temperature changes in plates, and I do not know how to extend this information to apply to an enclosed volume.
 A: A good place to start is the lumped-component model, which assumes that the entire system (here, the inside of the box) can be characterized by a single temperature $T$. That is, we're ignoring any temperature gradients within the material.
The typical approach is to perform an energy balance. We have a refrigeration mechanism that removes thermal energy at a constant rate, say, (call this $-Q$) and (with the door open) natural convection to the outside, which is at room temperature $T_\infty$. We could add up the rates of energy change and write something like
$$-Q-hA(T-T_\infty)=C\frac{dT}{dt}$$
where $h$ is a natural (or free) convection coefficient that characterizes the heat exchange with the room, $A$ is the exposed surface area, $C$ is the heat capacity of the cool area of the box and its refrigerated contents (a higher heat capacity means that the material is slower to change temperature), and $t$ is time. In other words, the speed at which the box changes temperature depends on the strength of cooling, the degree of convection, the exposed area, the temperature difference, and the heat capacity.
If we left the box open for a long time (so that steady state is achieved and $dT/dt$ approaches zero), we'd calculate that $T=T_\infty-Q/hA$, i.e., that the box is a little cooler than the room because of the refrigeration mechanism. So that makes sense.
Let's define the temperature difference $\Delta T=T-T_\infty$ to streamline  the equation:
$$-Q-hA\Delta T=C\frac{d\,\Delta T}{dt}$$
The solution is
$$\Delta T=-\frac{Q}{hA}+\left(T_0-T_\infty+\frac{Q}{hA}\right)\exp(-hAt/C)$$
where $T_0$ is the initial temperature inside the box. This model therefore tells us that the temperature difference asymptotically approaches the steady-state value in an exponential manner when the door is opened. The time constant, as discussed here, is $C/hA$; this is the time (in seconds) that it takes to erase a fair amount of the original temperature difference. After a few time constants, we're nearly at the steady-state value. So estimating the time constant is the key to answering your question regarding quantitative temperature changes.
$A$ is the total exposed cooled area, and $C$ is the heat capacity of all of the cooled material. Estimating these values involves a whole additional set of assumptions, of course. Intuitively, areas within crevices are not going to participate as much in the heat transfer process. In addition, it may be challenging to estimate the amount of cooled material because in reality, temperature gradients will exist between the refrigerated part of the box and the outer surfaces and cooling mechanism. Estimates here may vary within an order of magnitude. 
We have a similar degree of uncertainty regarding the natural convection coefficient $h$, which  is notoriously challenging to estimate from first principles, as it depends on the gas density, viscosity, temperature, and flow rate and the system geometry. We often end up applying complex empirical relations that involve nondimensional relations such as the Rayleigh, Prandtl, and Reynolds numbers. As a first pass, I'd plug in $1-10\, \mathrm{Wm^{-2}K^{-1}}$ as an order-of-magnitude estimate for natural convection in air. However, $h$ could change substantially depending on whether the opening is facing upwards, to the side, or downwards. It's likely going to be necessary to calibrate these predictions with an actual system. However, this approach at least yields the insight that the characteristic time should scale up with $C$ and inversely with $hA$, which should match our intuition that a stocked fridge maintains its temperature better, for example.
