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).

• I apologize, but I honestly think this is basically a homework question. While I hope someone with the understanding to solve this gives you the answer in time, I have a feeling this is going to be closed as a homework question. Good luck. – heather Aug 2 '16 at 20:56
• I didn't know that yoghurt underwent an explosive thermonuclear reaction when it was cooled to less than 110 degrees. If it doesn't, it's not really important when it reaches that temperature, is it? :-) PS: Just put it in the fridge, it will be perfectly fine. – CuriousOne Aug 2 '16 at 21:01
• Heat to 185F and cool to 110F (reduction), 110F add starter. I could just store the reduced milk in the fridge, buy a new candy thermometer, heat directly to 110F then add starter, but hey, why not ask here? – user19087 Aug 2 '16 at 21:12
• Too many variables to solve from first principles. If you aren't actively stirring, the milk may or may not go through convection in the cylinder. Breezes/drafts will affect convection rate from the cylinder. Cylinder material and thickness will dramatically affect heat flow through the cylinder. – BowlOfRed Aug 2 '16 at 21:16
• 110 is hand warm. Just add the starter when it doesn't feel too hot. Lactobacillus will, by the way, survive and grow at slightly higher temperatures, too. :-) – CuriousOne Aug 2 '16 at 21:17

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.

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.

• It's easier to just get another thermometer and measure it. Factors that will invalidate the conditions provided by Gert: there will be convection in the container. There will also be evaporation from the surface, heat conduction through the walls of the container with an unknown heat transfer coefficient, natural or forced convection off the outside of the container, depending on whether a fan is blowing or not, etc. This really is a difficult problem to calculate from first principles. – David White Aug 2 '16 at 21:53
• @DavidWhite: You reckon? I mean have you seen the closest case on p.8 of my link? ;-) – Gert Aug 2 '16 at 21:55
• I didn't look that closely ... I'll go back and look again, but I am SURE that there is a LOT of physics and math going on in this problem. Update: I just looked again. Message to the OP: buy a new thermometer! – David White Aug 2 '16 at 21:58
• It's DEFINITELY a new thermometer or using NDSolve from Mathematica... – Gert Aug 2 '16 at 22:01

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.