# IB DP Physics 3.1 – Thermal concepts Question Bank SL Paper 2 Part B

Question

This question is about thermal energy transfer.

A hot piece of iron is placed into a container of cold water. After a time the iron and water reach thermal equilibrium. The heat capacity of the container is negligible.

a

Define specific heat capacity.[2]

b.

The following data are available.

Mass of water = 0.35 kg
Mass of iron = 0.58 kg
Specific heat capacity of water = 4200 J kg–1K–1
Initial temperature of water = 20°C
Final temperature of water = 44°C
Initial temperature of iron = 180°C

(i) Determine the specific heat capacity of iron.

(ii) Explain why the value calculated in (b)(i) is likely to be different from the accepted value.[5]

## Markscheme

a

the energy required to change the temperature (of a substance) by 1K/°C/unit degree;
of mass 1 kg / per unit mass;

b.

(i) use of mcΔT;
0.58×c×[180-44]=0.35×4200×[44-20];
c=447Jkg-1K-1≈450Jkg-1K-1;

(ii) energy would be given off to surroundings/environment / energy would be absorbed by container / energy would be given off through vaporization of water;
hence final temperature would be less;
hence measured value of (specific) heat capacity (of iron) would be higher;

Question

Part 2 Internal energy

Humans generate internal energy when moving, while their core temperature remains approximately constant.

a

Distinguish between the concepts of internal energy and temperature.[3]

b.

An athlete loses 1.8 kg of water from her body through sweating during a training session that lasts one hour.

Estimate the rate of energy loss by the athlete due to sweating. The specific latent heat of evaporation of water is 2.3×106 J kg–1.[2]

## Markscheme

a

internal energy:
total energy of component particles (in the human);
comprises potential energy + (random) kinetic energy;

temperature:
measure of average kinetic energy of particles;
indicates direction of (natural) flow of thermal energy;
internal energy measured in J and temperature measured
in K/°C ; (both needed)
(accept alternative suitable units)

b.

total energy lost=2.3×106×1.8(=4.14×106J);
1.2 kW;

Question

Part 2 Melting of the Pobeda ice island

a

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.[8]

b.

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.

[2]

### Markscheme

a

(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;

b.

average albedo of ocean much smaller than (snow and) ice;
so average albedo (of Earth) is reduced;