Define the dimensionless stagnant-layer depth and time by
$$
\eta=\frac hH,
\qquad
\tau=\frac{\kappa t}{H^2},
$$
so $H^2/\kappa$ is the <thermal diffusion time>. The maximizing interface temperature gives
$$
T_h-T_s=\Delta T\left(1-\frac35\theta\right).
$$
Equating conductive and convective <heat fluxes> therefore gives the algebraic relation
$$
\boxed{
\frac{1-3\theta/5}{\eta}
=\mathcal F\theta^{5/3}}.
$$

Since $k=\rho c_p\kappa$, where $c_p$ is the <specific heat capacity>, the bulk heat balance in the convecting depth $H-h$ becomes
$$
\boxed{
(1-\eta)\frac{d\theta}{d\tau}
=-\mathcal F\theta^{5/3}}.
$$
Retaining the heat capacity of the thin stagnant layer changes this only by relative order $\eta$.

For $\mathcal F\gg1$ and $\theta=O(1)$, the flux relation gives $\eta=O(\mathcal F^{-1})$. The leading equation is consequently
$$
\frac{d\theta}{d\tau}
=-\mathcal F\theta^{5/3}.
$$
Separating variables and using $\theta(0)=1$ gives
$$
\theta^{-2/3}=1+\frac23\mathcal F\tau,
$$
hence
$$
\boxed{
\theta(\tau)=
\left(1+\frac23\mathcal F\tau\right)^{-3/2}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 332 / 2 / v / Solution

{title2=$L^2/\kappa$}

The thermal diffusion time across a distance $L$ is $L^2/\kappa$, where $\kappa$ is the <thermal diffusivity>. It is the time required for a temperature disturbance to spread over that distance by <thermal conduction>.

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 Incoming links: Thermal diffusion time