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 Incoming links: Iceberg melt heat budget

Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 77 / 4 / b / Solution 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}

, so using the stated ice specific heat capacity,

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

The latent heat of fusion contribution is

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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Poleward heat transport by equatorward iceberg drift 2026-10-14

An iceberg moving toward the equator carries a thermal deficit into warmer water. Melting extracts heat there, equivalent to transport of heat toward the pole. For an annual number of similar bergs, multiply their individual iceberg melt heat budget by the rate per second. Freshwater effects on density stratification and ocean circulation are separate from this heat accounting.

 Read the full article