We shall answer this question in three easy steps:
Given our little potato experiment, we’d like to estimate the thermal diffusivity constant \(\kappa\). To do the computation properly, we would derive the heat equation, think about the boundary conditions, and solve messy partial differential equations in the case of a uniform sphere. This all seems rather painful. Hence, we instead recall that the process of thermal diffusion is a random walk.
In each step of the walk, a phonon travels distance \(\lambda\) (the mean free path) before scattering in a random direction. Suppose we restrict the phonon to travel in only one dimension, moving either \(+1\) or \(-1\) at each step. After \(N\) steps \(s_i\), the average distance from the origin is $$\begin{aligned} \sqrt{\mathbb{E}\left[\left(\sum_i s_i\right)^2 \right]} = \sqrt{\sum_i s_i^2} = O\left(\sqrt{N}\right). \end{aligned}$$ In 3 dimensions, we can consider random choices of displacement vectors \((\pm1,0,0)\), \((0,\pm1,0)\), or \((0,0,\pm 1)\), still giving \(O(\sqrt{N})\) distance after \(N\) steps by Euclidean distance. Since each step takes a fixed amount of time (set by the speed of sound in the material), we conclude that a phonon takes \(O(N)\) time to travel distance \(\sqrt{N}\).
Writing in terms of a length scale \(L\), the time to diffuse distance \(L\) is \(\tau \sim L^2 / \kappa\). Plugging in our numbers for a potato, we find $$\begin{aligned} \tau \sim 10^3~\mathrm{s} \sim \frac{(0.1~\mathrm{m})^2}{\kappa} \implies \kappa \sim 10^{-5}~\mathrm{m^2~s^{-1}}. \end{aligned}$$ One convenient aspect of this approach is it gives the time \(\tau\) for a body at any initial temperature to reach a final “room temperature” regardless of the room temperature. Since the initial temperature moves towards the final temperature following exponential decay in time (due to Newton’s law of cooling), it suffices to estimate the characteristic timescale of the decay, which is independent of the initial temperature difference between the potato and the room.
To verify that we can use the same thermal diffusivity value for a potato and Mars, let’s try to compute \(\kappa\) from first principles using the phonon picture. The time between scattering events of the phonon is \(\lambda/c\), where \(c\) is the speed of sound in the material. From the random walk argument, it requires \(L^2/\lambda^2\) steps to travel distance \(L\). Hence, we obtain characteristic time $$\begin{aligned} \tau \sim \frac{L^2}{\lambda^2} \cdot \frac{\lambda}{c} \sim \frac{L^2}{\lambda c} \implies \kappa \sim \lambda c. \end{aligned}$$ We proceed to estimate both the mean free path \(\lambda\) — i.e., the average distance between atoms in the solid — and the speed of sound \(c\).
We’ll try to recover \(\lambda\) in street-fighting style based on the interatomic spacing \(a\) of bonded atoms. For simplicity, we shall evaluate \(a\) for an ionic or covalent bond, where the bonded atoms are directly next to each other; for weaker intermolecular forces, we will introduce a fudge factor \(\lambda = 3a\). The radius of the hydrogen atom is known to be the Bohr radius \(a_0 = 0.5~\mathrm{Å}\). Roughly speaking, the radius is constant with atomic number \(Z\): electron shielding causes the new electron to see only one proton. (In reality, the larger number of protons reduces the radius due to Coulomb attraction, while starting a new shell of electrons will increase the radius due to the Pauli exclusion principle.) Classically, an orbiting system has angular momentum like \(mvr\sim\sqrt{r}\); quantum mechanically, the Bohr model gives angular momentum like \(n\) for the \(n\)th energy level or shell. Hence, we might expect the radius to go like the shell number squared. For most solids with two or three shells, we thus expect \(a \sim 6a_0 \sim 3~\mathrm{Å}\), which corresponds to the typical interatomic distance for a covalent bond. Since bonds such as van der Waals bonds are typically larger, we include a fudge factor to estimate \(\lambda \sim 3a \sim 10~\mathrm{Å}\).
To estimate the speed of sound, we already did most of the hard work when calculating the pitch of squeaking chalk. By analogy from a wave on a string, we estimated the speed of sound \(c\) in a solid of density \(\rho\) in terms of Young’s elastic modulus \(M\): $$\begin{aligned} c \sim \sqrt\frac{M}{\rho}. \end{aligned}$$ A microscopic argument of elastic modulus as a normalized spring constant \(k/l\) between bonds of length \(l\) suggested that we can rewrite things in terms of the spring energy \(E \sim kl^2\): $$\begin{aligned} c \sim \sqrt\frac{E}{\rho l^3}. \end{aligned}$$ Since we’re looking at the stretching of bonds, \(E\) is the cohesive energy required to break the bond. We can estimate this from the Rydberg energy of the hydrogen atom, scaling the Coulomb potential from the Bohr radius \(a_0\) to our bond distance \(l\): $$\begin{aligned} c \sim \sqrt\frac{\mathrm{Ry} a_0}{\rho l^4}. \end{aligned}$$ Consequently, the thermal diffusivity is given by $$\begin{aligned} \kappa \sim \lambda c \sim \frac{\lambda}{l^2}\sqrt\frac{\mathrm{Ry} a_0}{\rho}. \end{aligned}$$ Sure, the density of a potato is perhaps a quarter of that of a rock; but that only introduces an error of a factor of two, which is small in an order-of-magnitude estimate. Indeed, the bond lengths of rock (such as SiO\(_2\)) form a tighter lattice than potato bonds, so we might pick up a factor of two that cancels our density error. Hence, we may rest assured that our potato does scale up to Mars.
We can also verify our formula against the boiled potato experiment above. If we take \(\rho=1~\mathrm{g~cm^{-3}}\) (since the potato is mostly water) and assume the relevant springs occur between molecules separated by distance \(l = \lambda = 10~\mathrm{Å}\), we obtain from first principles a potato thermal diffusivity of $$\begin{aligned} \kappa \sim \frac{1}{10~\mathrm{Å}}\sqrt{\frac{(13.6~\mathrm{eV})(0.5~\mathrm{Å})}{1~\mathrm{g~cm^{-3}}}} \sim 3 \times 10^{-6}~\mathrm{m^2~s^{-1}}, \end{aligned}$$ which is reasonably close to the result of our potato experiment, and which we expect to be similar to the thermal diffusivity of Mars.
Plugging in the numbers for Mars, we obtain \(\tau \sim (10^6~\mathrm{m})^2 / 10^{-5}~\mathrm{m^2~s^{-1}} \sim\) 3 billion years. Despite the proximity of our answer to the official estimate of 4.5 billion years, it is interesting to note that Mars actually cooled down around an order of magnitude faster than we anticipated. Two reasons may explain the discrepancy: first, solid-state convection and evaporative cooling physically transported hot material faster than thermal diffusion at the molecular scale; and secondly, the rate of radiative cooling via infrared radiation scales like the difference between the fourth powers of the planet’s temperature and that of the adjacent environment (due to the Stefan-Boltzmann law), unlike the exponential decay we assumed (i.e., a linear difference of temperatures). On the flip side, the core of Mars is also considerably more radioactive than a potato, providing a heat source unaccounted for in our computations. Despite these shortcomings of our approach, the order-of-magnitude result remains reasonably accurate!