# 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.

-
How do you fit the data? – qfzklm Feb 25 '14 at 8:13
I mean, what is your expression to fit? – qfzklm Feb 25 '14 at 8:14
@qfzklm The function indicated is y=231*(0.97)^x. – Geoff Feb 25 '14 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 '14 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 '14 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.

-
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 '14 at 15:51

the cooling process depends on many factors: - convection, radiation and conduction inside the body.

Exponential decay is good only if the lumped capacitance model is appropiate and if the relation with radiation is linear. I would say that in your case you also had convection inside the fluid. And ontop of that the relation with radiation is not linear it is to the fourth power, so your pretty much had a non-linear dependency and off course far away from being exponential, the limits of the exponential decay assumption have to be tested before application.

I hope this helps.

-

Water evaporates if it has a higher temperature and or if the humidity of the air isn't at 100%. This is a nonlinear process as the vapour pressure of water is nonlinear with temperature and the rate of evaporation is linear with partial pressure difference (the partial pressure in the water is equal to the vapour pressure). The evaporation reduces the temperature of the remaining liquid as it requires energy to turn water into water vapour. You would therefore expect that the cooling initially is faster as a lot of water evaporates (which sometimes condenses and becomes visible as steam right above the vessel).

Three solutions to redo your experiment: 1) Put a lid on the pan as this reduces the rate of evaporation because the air gets saturated (this is also the reason why you would want to put a lid on a pan when cooking as to consume less energy) 2) Use a liquid that has a much lower rate of evaporation at those temperatures. 3) Include the rate of evaporation in the equation with which you do the curve fit.

-

## protected by Qmechanic♦Jun 23 at 15:26

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).