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.

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


*

*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;

*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.
