 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/iii/paper-77/4/b/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

The horizontal area is

A = 1 0^{6} m^{2}

and the total mass from part (a) is

m = 2.55 \times 1 0^{11} kg

. The iceberg melt heat budget includes warming to

0^{\circ} C

followed by fusion. With temperature

T (z) = - 0.6 z

in the upper

50

m and

- 3 0^{\circ} C

below,

Q_{sensible} = A c_{i} [\int_{0}^{50} (300 + 12 z) (0.6 z) d z + \int_{50}^{300} 900 (30) d z] .

The two integrals are

525000

and

6750000 kg K m^{- 2}

Q_{sensible} = 1 0^{6} (2100) (7275000) = 1.52775 \times 1 0^{16} J .

Q_{latent} = m L_{f} = (2.55 \times 1 0^{11}) (336000) = 8.568 \times 1 0^{16} J .

Therefore

Q = Q_{sensible} + Q_{latent} = 1.009575 \times 1 0^{17} J ≃ 1.01 \times 1 0^{17} J .

This produces liquid water at

0^{\circ} C

; warming the meltwater further would require additional heat.

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