Cook like a physicist

Assume your turkey is a sphere of water ...

The holidays are upon us, and for many, that means one thing: cooking a turkey. While a real chef might rely on intuition, experience, and a good meat thermometer, the physicist sees this as a fascinating, real-world heat transfer problem.

At its heart, cooking a turkey is about getting heat from the hot oven into the cold, raw bird, raising its internal temperature to a safe (and delicious) level — a classic problem in heat conduction.

step 1: assume a sphere

A long running joke is that physicists always start out assuming that whatever they’re analyzing is a sphere.

Why? Because spheres are simple shapes that are easy to manipulate mathematically. In this case, while a turkey is a complex, irregularly shaped object with bones, cavities, varying densities of meat, a sphere is not a bad assumption.

Let’s further assume the turkey has the thermal properties of water.

Our spherical turkey starts cold (say, straight from the fridge at 5C). We then put it into a hot oven and assume the surface of the turkey quickly reaches 100C¹.

The goal? To get the very center of that “turkey-sphere” to a safe internal temperature (like 80C).

step 2: the physics

The governing equation for transient heat conduction in a sphere with constant thermal properties and no internal heat generation is:

\(\frac{1}{\alpha}\frac{\partial T}{\partial t} = \frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial T}{\partial r}\right)\)

This was originally derived by Joseph Fourier. To solve it, he also had to develop Fourier series. Smart guy!

The general solution is a complicated infinite sum. It’s the kind of problem that you slog through once, when you’re in grad school, and never again. If I need to solve this (like I did for this post), I looked up the solution and used a computer to work it out numerically.

The time it takes to cook is equal to:

\(t \approx \frac{R^2}{\alpha\pi^2}\ln \left[\frac{6}{\pi^2}\frac{(T_{init} - T_{hot})}{T_{center} - T_{hot}}\right]\)

where R is the radius, ⍺ is the thermal diffusivity of water, T_init is the initial temperature (5C), T_hot is the temperature of the surface (100C), and T_center is the temperature you want the center (80C).

step 3: the answer

The results are pretty reasonable. For a sphere of water that weighs 10 lbs (radius is about 4 inches), the time to cook until the center reaches 80C is 2.2 hours, or about 13 minutes per pound. That’s very close to the label on our turkey:

Interestingly, the equation above shows that time scales with the square of the radius. Because mass scales as the cube of the radius, the time to cook therefore scales with mass to the 2/3rds power, i.e., time ∝ mass^(2/3).

That’s why a 30-lb turkey doesn’t take three times as long to cook as a 10-lb turkey. To calculate how long it would take, we can scale the 2.2 hours for a 10-lb turkey to 30 lbs:

\(t_{\text{30 lbs}} = \frac{2.2}{10^{2/3}} \times {30^{2/3}} = 4.5\)

Again, this agrees well with the label.

When I originally worked this out 35 years ago, I asked my mother how long it took to cook a turkey and she said “12 minutes per pound”. That answer suggested cooking time was linear with mass, which is not what I calculated. But, after a pause, she said, “Unless it’s a really big turkey, in which case it’s 9 minutes per pound”.

“Ah hah,” I thought, “There’s your non-linearity.” Once again, physics FTW.

Happy thanksgiving everyone! Enjoy your approximately spherical turkey.

A few other things:

Today is the day everyone googles “how to cook a turkey”. This plot shows the relative frequency of the google search:

There’s a slight peak in the evening of the 26th and a big peak between 6 and 7 am CT on Thanksgiving day. Never let people tell you Americans don’t procrastinate. By 3 pm CT (when I’m writing this), it’s clearly too late to be googling.

On the Halcyon blog, there’s an interesting analysis of how energy consumption is different on Thanksgiving. People sleep late, so there’s no early peak, and they eat dinner early, so there’s one big mid-afternoon peak and no late afternoon peak:

Read the entire thing, it’s good.

1. I’m assuming here that water evaporating from the skin keeps the surface temperature from exceeding 100C.