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Past exam of the mathematics course of the University of Cambridge

2026

Paper 332

The diffusion length suggests the <similarity variable>
$$
\eta=\frac{z}{2\sqrt{\kappa t}},
\qquad
h(t)=2\lambda\sqrt{\kappa t}.
$$
The <Neumann solution of the Stefan problem>, written using the <error function>, is
$$
\boxed{
T_i(z,t)=T_b+(T_m-T_b)
\frac{\operatorname{erf}\eta}{\operatorname{erf}\lambda}},
$$
$$
\boxed{
T_w(z,t)=T_\infty+(T_m-T_\infty)
\frac{\operatorname{erfc}\eta}{\operatorname{erfc}\lambda}}.
$$
These expressions satisfy all four thermal <boundary conditions>. At the interface,
$$
T_{i,z}
=\frac{T_m-T_b}{\sqrt{\pi\kappa t}}
\frac{e^{-\lambda^2}}{\operatorname{erf}\lambda},
\qquad
T_{w,z}
=\frac{T_\infty-T_m}{\sqrt{\pi\kappa t}}
\frac{e^{-\lambda^2}}{\operatorname{erfc}\lambda}.
$$
Since $\dot h=\lambda\sqrt{\kappa/t}$ and $k=\rho c_p\kappa$, the <Stefan condition> becomes
$$
\boxed{
\frac{L}{c_p}\sqrt{\pi}\lambda e^{\lambda^2}
=\frac{T_m-T_b}{\operatorname{erf}\lambda}
-\frac{T_\infty-T_m}{\operatorname{erfc}\lambda}}.
$$
The second term represents heat supplied from the warm water and therefore reduces the freezing rate.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 332 / 2 / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 332 / 2 / b / Solution

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Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 332 / 2 / b / Solution

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