 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-332/2/b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

η = \frac{z}{2 κ t}, h (t) = 2 λ κ t .

T_{i} (z, t) = T_{b} + (T_{m} - T_{b}) \frac{erf η}{erf λ},

T_{w} (z, t) = T_{\infty} + (T_{m} - T_{\infty}) \frac{erfc η}{erfc λ} .

These expressions satisfy all four thermal boundary conditions. At the interface,

T_{i, z} = \frac{T _{m} - T _{b}}{πκ t} \frac{e ^{- λ^{2}}}{erf λ}, T_{w, z} = \frac{T _{\infty} - T _{m}}{πκ t} \frac{e ^{- λ^{2}}}{erfc λ} .

Since

\dot{h} = λ κ / t

and

k = ρ c_{p} κ

\frac{L}{c _{p}} π λ e^{λ^{2}} = \frac{T _{m} - T _{b}}{erf λ} - \frac{T _{\infty} - T _{m}}{erfc λ} .

The second term represents heat supplied from the warm water and therefore reduces the freezing rate.

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