Date | November 2012 | Marks available | 8 | Reference code | 12N.2.SL.TZ0.7 |
Level | Standard level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Determine, Distinguish, Show that, and Suggest | Question number | 7 | Adapted from | N/A |
Question
Part 2 Melting of the Pobeda ice island
The Pobeda ice island forms regularly when icebergs run aground near the Antarctic ice shelf. The “island”, which consists of a slab of pure ice, breaks apart and melts over a period of decades. The following data are available.
Typical dimensions of surface of island = 70 km × 35 km
Typical height of island = 240 m
Average temperature of the island = –35°C
Density of sea ice = 920 kg m–3
Specific latent heat of fusion of ice = 3.3×105J kg–1
Specific heat capacity of ice = 2.1×10\(^3\)J kg–1K–1
(i) Distinguish, with reference to molecular motion and energy, between solid ice and liquid water.
(ii) Show that the energy required to melt the island to form water at 0°C is about 2×1020J. Assume that the top and bottom surfaces of the island are flat and that it has vertical sides.
(iii) The Sun supplies thermal energy at an average rate of 450 W m–2 to the surface of the island. The albedo of melting ice is 0.80. Determine an estimate of the time taken to melt the island assuming that the melted water is removed immediately and that no heat is lost to the surroundings.
Suggest the likely effect on the average albedo of the region in which the island was floating as a result of the melting of the Pobeda ice island.
Markscheme
(i) in water, molecules are able to move relative to other molecules, less movement possible in ice / in water, vibration and translation of molecules possible, in ice only vibration;
in liquid there is sufficient energy/vibration (from latent heat) to break and re-form inter-molecular bonds;
(ii) mass of ice=70000×35000×240×920(=5.4×1014kg);
energy to raise ice temperature to 0°C=5.4×1014×2.1×103×35(= 3.98×1019J);
energy to melt ice=5.4×1014×3.3×105(=1.8×1020J);
total= 2.2×1020J
(iii) energy incident=450×70000×35000(=1.1×1012Js–1m−2);
energy available for melting=1.1×1012×0.2(=2.2×1011J);
time \( = \left( {\frac{{2.2 \times {{10}^{20}}}}{{2.2 \times {{10}^{11}}}} = } \right)9.9 \times {10^8}{\rm{s}}\) or 32 years;
average albedo of ocean much smaller than (snow and) ice;
so average albedo (of Earth) is reduced;