# Is cooling really exponential?

I did the following experiment: I took boiling water, put a thermomenter (from Vernier hooked up to a TInspire calculator) into it and pulled it out. Then we started measuring the temperature it read with time. We waved the thermometer so that the air aorund it was not heating up. I did an exponential regression on the data (well, appropriately shifted data). The results show a clear non-exponential heat loss. So what is this curve? Or what have I done wrong?

From a comment I have added a further plot: where the y-axis is logged. Again, it is clear that the data is not linear.

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How do you fit the data? –  qfzklm Feb 25 at 8:13
I mean, what is your expression to fit? –  qfzklm Feb 25 at 8:14
@qfzklm The function indicated is y=231*(0.97)^x. –  Geoff Feb 25 at 8:16
Exactly as @JarosławKomar says: it is hard to fit exponential functions to data otherwise because it becomes a nonlinear fit, i.e. you're not simply seeking an optimal superposition. So, depending on the fitter you used, the fact that it didn't fit too well may not be the final answer! You should be able to coax MS-Excel Solver to do the appropriate nonlinear fit, but Jarosław's suggestion is likely to give better results. –  WetSavannaAnimal aka Rod Vance Feb 25 at 8:34
@WetSavannaAnimalakaRodVance ohmigod please don't suggest using Excel's Solver! It's an ugly misbehaving beast. There are many great tools out there with solvers that will do a far better job. –  Carl Witthoft Feb 25 at 13:11

Given your description, you clearly have non-exponential behaviour. However, there are two possible reasons for this behaviour:

1. Some materials in the system are nonlinear and do not follow Newton's law of cooling, which is that the heat flux at a given point is proportional to the temperature gradient vector. I should think this is the least likely of the two reasons;
2. General solutions of the Heat Equation do not have a simple exponential time dependence; the general solution for materials that fulfil Newton's law of cooling (i.e. heat flux proportional to temperature gradient) is a superposition of functions of the form $f_n(x,\,y,\,z) \,\exp(-\alpha_n t)$, so you will only get a good fit to an exponential dependence when the boundary conditions / initial conditions are such that only one of these terms is dominant. See the discussion under the heading "Solving the heat equation using Fourier series" on the the Heat Equation Wiki Page to see how the superposition arises.

As I said, the second is the most likely reason for your behaviour. The superosition weights are set by initial and boundary conditions. Indeed a particular "prototypical" solution to the heat equation is the behaviour we see if we have a "hot spot" of very high temperature in a one-dimensional system, and we let this hot spot diffuse. Our idealised description of this case is:

$$\begin{cases} u_t(x,t) - k u_{xx}(x,t) = 0& (x, t) \in \mathbf{R} \times (0, \infty)\\ u(x,0)=\delta(x)& \end{cases}$$

and its solution, with a highly non-exponential time dependence, is the heat kernel:

$$\Phi(x,t)=\frac{1}{\sqrt{4\pi kt}}\exp\left(-\frac{x^2}{4kt}\right)$$

whence we can build solutions to arbitrary initial temperature distributions in the bar by linear superposition. Here, of course, $\delta$ stands for the Dirac delta. Again, see the discussion under the heading "Fundamental Solutions" on the the Heat Equation Wiki Page.

So you likely need a more sophisticated model of your cooling thermometer. As an aside, I am surprised you got a curve with a simple dependence at all: I should not have thought "waving the thermometer around" to make it cool would set up particularly repeatable cooling conditions needed for proper experimental investigation.

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I needed a bit of time to work through your solution (as I am a non-physicist). I am though a maths teacher and was expecting that this simple experiment would lead to an exponential decay graph (which was my original intention). All the same, many thanks. –  Geoff Mar 3 at 15:51