Dam Son's problem: how long does it take to boil an ostrich egg?

The temperature distribution within each sphere as a function of time and radial position is governed by the transient heat conduction equation (in spherical coordinates): $$\frac{\partial T}{\partial t}=\alpha\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial T}{\partial r}\right)$$where $\alpha$ is the thermal diffusivity. Boundary and initial conditions are:

$T=T_0$ at t = 0, all r

$T=T_1$ all t, at r=R

$r^2\frac{\partial T}{\partial r}=0$, all t, r = 0 These equations can be reduced to dimensionless form by means of the following substitutions: $$\theta=\frac{T-T_0}{T_1-T_0}$$ $$\rho=\frac{r}{R}$$ $$\tau=\frac{\alpha t}{R^2}$$ With these substitutions, the equations become:$$\frac{\partial \theta}{\partial \tau}=\frac{1}{\rho^2}\frac{\partial}{\partial \rho}\left(\rho^2\frac{\partial \theta}{\partial \rho}\right)$$ $\theta = 0$ at $\tau=0$, all $\rho$

$\theta = 1$ all $\tau$, at $\rho=1$

$\rho^2\frac{\partial \theta}{\partial \rho}$, all $\tau$, $\rho = 0$

Note that there are no adjustable parameters in the dimensionless differential equation and boundary conditions. So, to achieve a specific value of the dimensionless temperature $\theta$ at the center of both spheres requires a specific value of dimensionless time $\tau=\tau^*$. In terms of the actual time, this would be $$t=\frac{R^2}{\alpha}\tau^*$$Assuming that the thermal diffusivities of the two egg materials are the same, this means that the time is proportional to the square of the egg radius R.


It looks like the assumption is being made that the time for the egg to cook is proportional to the square of the diameter.

$$T=pd^2$$

Where $p$ is the proportionality constant. Therefore

$$T_{chicken}=pd_{chicken}^2$$

$$T_{ostrich}=pd_{ostrich}^2$$

Or solving both for $p$

$$p=\frac{T_{chicken}}{d_{chicken}^2}=\frac{T_{ostrich}}{d_{ostrich}^2}$$

Therefore

$$T_{ostrich}=(\frac{d_{ostrich}^2}{d_{chicken}^2})*T_{chicken}=54\space min$$

So the question then is why are we assuming this proportionality? The easiest thing I can think of is that they are assuming this time is proportional to the surface area of the egg. So then $p=a*\pi$ where $a$ is some other factor dealing with the heat transfer from the water to the egg. This makes sense. More surface area allows for more heat transfer.

I would think that the volume should also come into play too though. It would take longer time to heat up something with more volume. If anyone can give guidance to this point I can adjust my answer accordingly.


Thanks to @BowlOfRed:

The time to cook is proportional to $\frac{energy\space needed\space to\space boil}{rate\space of\space heat\space exchange}$ (energy/(energy/time))->time).

The energy needed is going to proportional to the volume of the object (more stuff means more energy needed). This brings in a $d^3$ dependency.

The rate of heat exchange is proportional to both the area ($d^2$) of the surface as well as the temperature difference per unit length ($1/d$) (faster energy transfer if we have more energy and larger temperature differences over shorter lengths).

Therefore our time $T$ for heating is proportional to $\frac{d^3}{d^2/d}=d^2$

Then we can go through the above work to get to the final answer.


We want to show the time for cooking is proportional to $R^2$, where $R$ is the radius; then the ostrich egg will take $9$ times as long to cook.

You can get this intuitively by considering how the heat flow scales with the object size, but it also can be proven by dimensional analysis. The only things that can matter are $$\text{radius} \ R \sim [\text{m}], \quad \text{heat capacity } C \sim [\text{J/K}], \quad \text{thermal conductivity } \kappa \sim \left[ \frac{\text{J}}{\text{K} \cdot \text{m} \cdot \text{s}} \right].$$ I'm not including temperatures here, because the initial temperature, the temperature of the boiling water, and the "goal" temperature are the same for both eggs. Since the heat equation is linear that means any dependence on these temperatures will drop out when we take a time ratio.

The only way to get a time period $T$ is to have $$T \sim \frac{C}{R \kappa}$$ but since $C$ is proportional to the volume, $T$ is proportional to $R^2$ as desired.