A lake freezes

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The temperature of the air is \(T_a=263\;K\) and the temperature of the liquid water is \(T_F=273\;K\).

We give the massic latent heat of fusion of ice \(L_f\), its density \(\rho\) and its thermal conductivity \(\lambda\).

The thermal capacity of ice is supposed to be negligible.

Let \(h(t)\) be the height of ice at time \(t\). At \(t=0\), \(h(0)=0\).

Question

  • Establish the differential equations which have \(T(z)\) and \(h(t)\) as solutions.

  • Solve them and determine \(h(t)\).

Solution

The heat equation in the ice :

\(\rho c\frac{{\partial T}}{{\partial t}} = \lambda \frac{{{\partial ^2}T}}{{\partial {z^2}}}\)

So if we suppose that \(c\) is almost \(0\) :

\(\frac{{{\partial ^2}T}}{{\partial {z^2}}} \approx 0\)

So :

\(T(z,t) = {T_a} + \frac{{{T_F}-{T_a}}}{{h(t)}}z\)

We make the approximation of the quasi-stationary mode.

The heat flow in the ice is due to its solidification, that is to say, between \(t\) and \(t+dt\) :

\(j(z,t) = - \lambda \frac{{\partial T}}{{\partial z}} = - \frac{{\lambda ({T_F}-{T_a})}}{{h(t)}}\)

So :

\(- \frac{{\lambda ({T_F}-{T_a})}}{{h(t)}}Sdt = - {L_F}dm = - \rho Sdh{L_F}\)

That is to say :

\(\frac{{\lambda ({T_F}-{T_a})}}{{h(t)}}Sdt = \rho Sdh{L_F}\)

And :

\(hdh = \frac{{\lambda ({T_F}-{T_a})}}{{\rho {L_F}}}dt\)

After integration :

\(h = \sqrt {\frac{{2\lambda ({T_F}-{T_a})}}{{\rho {L_F}}}\;t}\)