I have an interesting problem here. I have a box of certain dimensions. Inside of the box is some insulation and another box, from the same material as the outer box. I know everything possible about the materials possible. Is there an equation to calculate that if there are for example -4 degrees Celsius outside, it will take XY minutes to cool the inside of the inner box by 1 degree? I really don’t know where to start. Thanks for any help.


The insulation will have a certain calculable thermal conductivity, in Watts per Kelvin (or the unit of measure of your choice).

The inner box will have a certain calculable thermal capacity, in Joules per Kelvin. That can be turned into Kelvins per Joule.

Calculate the heat flux (in watts) into the inner box, note that a Watt is a Joule / second, and do some math.

In the real world, the insulating material will have its own heat capacity, which will complicate things, and a rectilinear box will complicate things more -- a spherically symmetrical box would let you turn this into a 1-D problem.

Also in the real world, if there are any fluids involved (i.e., air), then there's a possibility of convection, at which point things go well over my head, and you'll find me in the lab with a bazillion thermocouples and an oven, figuring out answers experimentally.


There is also another way to do this. But of course, like all engineering exercises, there is a lot of things I have to assume (read: making an ass out of u and me)

So first, I assume that the situation is 1D (easier to work with), then I ignore the heat capacity of the walls (boxes and insulation layer), I also assume the shape of the box is not important, no heat is lost from the inside (as in, if heat goes in, it is in). Along the way, if I make any more assumption, I will write it out.

Technically speaking, the flow of heat is "out" (from hot to cold), or in this case, the "cold stuff" is going in the box (and the "hot stuff" is going out).

The heat flux (ie heat loss rate) is $\dot Q = k*A*\Delta T$ with $\dot Q$ as heat loss rate, k is the heat transfer coefficient, and A is the area. Here, we are working with heat flux (heat loss per unit of area), so $\dot q=k*\Delta T$ shall be used.

We have $\frac{1}{k}=\sum \frac{1}{\alpha} + \sum \frac{\delta}{\lambda}$

With $\alpha $ as heat transfer coefficient (hot to cold place, and vice versa, W/m^2/K

$\delta$ is the thickness of the material layer

$\lambda$ is thermal conductivity (W/m/K)

In your case You plug in the data for three layers of materials (2 times of the boxes, and once for the insulation layer), you would have \frac{1}{k} and then, k. You know the original temperature of (air - ?) inside the inner box, so you can calculate the $\Delta T$. From here you can calculate the heat flux $\dot q$.

But you also ask for the time to drop by 1 degree... I'm pretty sure that this one is a long formula with logarithm, exponent and stuff like that. Think about it, the colder you get, the less difference between you and the target, making flushing the heat out a bit harder. However, being engineers, we can "approximate" this my taking the average.

Making the next assumption that the system is static and the only energy considered is thermal energy, we calculate the internal energy of the gas

$U=\frac{3}{2}nRT$ with n as the number of mol of air, T is the beginning temperature in Kelvin, and R is 8.31 (8.3144 if you want more decimal). Do that again for the final temperaure (T=-4C), you would have the change in internal energy $\Delta U$. Because of my assumption of averaging above (or constant heat loss rate), we can calculate the time as

$t=\Delta U/ \dot q$

Divide this by the temperature difference $\Delta T$ would give you the (super) time-average needed to decrease the temperature of the system by 1C or 1K.

Any criticism is welcome


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