Last time, we modeled the egg as a rod, then irresponsibly reduced it to a cone by dividing by 3. Now finally getting to the realm of a closer estimation: a sphere. As it turns out, the differential equation for heat equation in spherical coordinates is a fraction harder than the rod case, so I spent some time trying to figure out how it works. Only if we can naively bring out the r squared, the formulas look the same.

\[\frac{\partial u(r, t)}{\partial t} = \frac{\alpha}{r^2} \frac{\partial}{\partial r} \left(r^2 \frac{\partial u(r, t)}{\partial r}\right)\]

As a reminder, u(r, t) is temperature as a function of radius r and time t, \(\alpha\) is a constant. r = 0 means the center of the egg and r = R is located at the shell of the egg.

The boundary conditions are:

The center of the egg is as cold as the fridge: \(u(0, 0) = 4\)

No heat flow at the center of the egg: \(\frac{\partial u(0, t)}{\partial r} = 0\)

The shell of the egg is always boiling: \(u(R, t) = 100\)

Eventually, the egg reaches thermal equilibrium with the water: \(u(r, \infty) = 100\)

Let’s split \(u(r, t)\) into \(u(r, t) = \Gamma(r) T(t)\)

I was gonna use \(R\) instead of \(\Gamma\), then I realized I’ve already used \(R\) to denote the radius of the egg.

Fast forwarding the computations, a similar pattern to the rod case emerges

\[\frac{T'(t)}{\alpha T(t)} = \frac{X''(r)}{X(r)} = -\lambda^2\] \[X(r) = r\Gamma(r)\] \[T(t) = C \exp(-\lambda^2 \alpha t)\]

After some rigorous internet research, if we skip all the rest of the smaller terms, a generic functional form that works most of the time is a sinc function

\[\Gamma(r) = A sinc(\lambda r)\]

\(\lambda_k = \frac{k \pi}{R}, k = 1, 2, 3, ...\), picking k = 1

As always, applying this technique of rewriting the boundary conditions

\[u(r, t) = u(r, \infty) + v(r, t)\]

New boundary conditions:

\[v(R, t) = 0\] \[\frac{\partial v(0, t)}{\partial r} = 0\] \[v(0, 0) = -96\]

So conveniently, \(A = -96\)

\[u(0, t) = 100 - 96 sinc\left(\frac{\pi}{R} r\right)\exp\left(-\frac{\pi^2}{R^2} \alpha t\right) = 70\]

Note \(sinc(0) = 1\)

\[u(0, t) = 100 - 96 \exp\left(-\frac{\pi^2}{R^2} \alpha t\right) = 70\] \[\frac{30}{96} = \exp\left(-\frac{\pi^2}{R^2} \alpha t\right)\] \[t = \frac{R^2}{\pi^2 \alpha} \ln\frac{96}{30}\] \[R = \left(\frac{3M}{4000\pi}\right)^{1/3}\] \[t = \frac{\left(\frac{3M}{4000\pi}\right)^{2/3}}{\pi^2 \alpha} \ln\frac{96}{30}\]

Using \(\alpha = \frac{k}{\rho c}\), \(c = 3120 \frac{J}{kg C}\), \(\rho = 1000 \frac{kg}{m^3}\), \(k = 0.4 \frac{J}{s\ m\ C}\)

This gives us:

\[t = 3500 M^{2/3}\]

So another inaccurate set of computations for the hard boiled egg, note that there is a chance I’ve made a mistake somewhere.

\(t = 3500 M^{2/3}\)

  • Small: 42g -> 0.042kg -> 422s = 7min
  • Medium: 49g -> 0.049kg -> 469s = 7.8min
  • Large: 56g -> 0.056kg -> 512s = 8.5min
  • Extra Large: 63g -> 0.063kg -> 554s = 9.2min
  • Jumbo: 70g -> 0.070kg -> 595s = 10min