cylinder of milk: time required to reach given temperature This is a very practical question, I've looked at Wikipedia's heat equation page but it is too complicated. I am in the middle of making some yogurt but I've broken my thermometer. The temperature of the milk (post-stirring) was approximately 145 degrees Fahrenheit at 16:30. The milk is contained in a pot/cylinder of 9" diameter and 4" high (1 gallon of milk before evaporation). The air temperature is 73 degrees Fahrenheit. How long till the temperature reaches 110 degrees Fahrenheit?
I'll accept an actual time but would prefer an annotated simple (non-differential) equation pre-solved for a cylindrical shape including a reference to milk's heat capacity.
I can't believe how much this sounds like one of those inane homework questions, but I need an answer soon (before 110 degrees).
 A: There is no quick and easy way to calculate this and certainly not before your yoghurt reaches $110\:\mathrm{F}$ (that's the most unusual 'ultimatum' I've come across on this site, ever!)
Nor is it possible to derive the expressions for that temperature (temperature v. time function) without using differential equations.
We can distinguish two main cases:
1. Temperature inside the pot is uniform:
For stirred liquids that's a reasonable assumption. We can then derive a cooling curve by simple application of Newton's cooling law. But a differential equation is needed for that too.
The cooling curve takes on the general form:
$$\Large{T_2=T_{\infty}+(T_1-T_{\infty})e^{-\frac{hA}{mc_p}t}}$$
Where $T_1$ is the initial temperature, $T_2$ the temperature after an amount of time $t$ has elapsed, $T_{\infty}$ the ambient temperature, $h$ the convection heat transfer coefficient, $A$ the total surface area of the object, $m$ its mass and $c_p$ the specific heat capacity of the object's material. 
2. There are radial temperature gradients:
For non-stirred liquids or solids we know the temperature will vary from the outermost layer to the core:
$$\frac{\partial T}{\partial r} < 0$$
(For cooling).
The consequence of this temperature gradient is that, all other things being equal, this second mode of cooling is slower than the first case because core heat has to conduct (diffuse) through the various layers between core and surface.
In this case Fourier's heat equation for a cylinder can be used. That's a second order, linear, partial differential equation in $r$, $T$ and $t$ ($T(r,t)$).
This excellent paper should give an idea of the mathematical complexity involved in solving this equation. Most engineers/physicists would seek practical, numerical solutions using Wolfram's NDSolve feature or similar.
A: While you have your "answer" already, I would like to offer this alternative:
When you need to reach a certain temperature - not higher, not lower - the best thing you can do is create an environment with that temperature. In this case you might create a large bath at 110F and leave your soon-to-be-yoghurt in there for a while; the key is you don't want it to get too cold.
So how do you go about creating a bath of 110F? There are two points that are easily attained: boiling water, and melting ice. Assuming you still have a working scale, boil a large quantity of water (weigh the pot, weigh pot plus water) and put a reasonable amount of ice in cold water. Then calculate (this is easy) how much ice to add to boiling water to get to 110F (take account of both the latent heat of fusion, and the heat capacity of the melted ice). Put the entire mix in a big pot that is inside an even bigger pot - with insulation (air, towels, ...) in between. Put the yoghurt-to-be inside the liquid (which is at 110 F and has very large heat capacity$ and cover the whole thing to avoid evaporation.
Wait a while. Make yoghurt. Enjoy.
The key is that a sufficient quantity of water at known temperature will barely budge - if you prevent evaporation and limit conductive heat loss.
