= Solution
Let
$$
\kappa=\frac{k}{\rho c_p},
\qquad
\kappa_s=\frac{k_s}{\rho c_p},
\qquad
h(t)=2\lambda\sqrt{\kappa t}.
$$
The <Neumann solution of the Stefan problem> in the ice and substrate is
$$
T_i(z,t)=T_0+(T_m-T_0)
\frac{\operatorname{erf}(z/(2\sqrt{\kappa t}))}
{\operatorname{erf}\lambda}
\quad(0<z<h),
$$
and
$$
T_s(z,t)=T_s+(T_0-T_s)
\operatorname{erfc}\left(\frac{-z}{2\sqrt{\kappa_st}}\right)
\quad(z<0).
$$
Continuity of <heat flux> at the contact gives, with $r=\sqrt{k_s/k}$,
$$
\frac{T_m-T_0}{\operatorname{erf}\lambda}
=r(T_0-T_s),
$$
and therefore
$$
\boxed{
T_0=\frac{T_m+r\operatorname{erf}(\lambda)T_s}
{1+r\operatorname{erf}(\lambda)}}.
$$
At the ice–water interface, the water is isothermal at $T_m$. The <Stefan condition> $\rho L\dot h=kT_{i,z}(h^-)$ gives
$$
S\lambda
=\frac{T_m-T_0}{T_m-T_s}
\frac{e^{-\lambda^2}}
{\sqrt\pi\operatorname{erf}\lambda},
\qquad
S=\frac{L}{c_p(T_m-T_s)}.
$$
Eliminating $T_0$ yields the implicit equation
$$
\boxed{
\sqrt\pi S\lambda e^{\lambda^2}
\left(1+r\operatorname{erf}\lambda\right)=r},
$$
which determines $\lambda$ and hence $h(t)=2\lambda\sqrt{\kappa t}$.
If $k_s\gg k$, then $r\gg1$, $T_0\sim T_s$, and
$$
\boxed{S\lambda\sim
\frac{e^{-\lambda^2}}{\sqrt\pi\operatorname{erf}\lambda}}.
$$
The highly conducting substrate acts as a reservoir fixed near its initial cold temperature, giving the usual one-phase <Stefan problem>.
If $k_s\ll k$, then $r\ll1$, $T_0\sim T_m$, and $\lambda\sim r/(\sqrt\pi S)$. Consequently
$$
\boxed{
h(t)\sim\frac{2c_p(T_m-T_s)}{L}
\sqrt{\frac{\kappa_st}{\pi}}}.
$$
Here heat removal through the poorly conducting substrate is rate limiting, and only a small temperature drop is needed across the much more conducting ice.
Back to article page
