States of matter and phase transitions

Matter can exist as a solid, a liquid or a gas. When it changes from one state to another (melting, freezing, vaporisation, condensation) it absorbs or releases heat, yet the temperature stays constant during the change. This topic is the basis for calculating quantities of heat, writing the heat balance equation and understanding air humidity. Weather, cooking, engines and refrigerators all work by these laws.

§1. States of matter: the molecular picture

The state of a substance depends on how the attraction energy between molecules compares with the energy of their thermal motion.

• Solid: molecules are very close and vibrate about fixed equilibrium positions. Shape and volume are kept. In crystals the particles are arranged in order (a crystal lattice); in amorphous solids (glass, resin) there is no such order.
• Liquid: molecules are closely packed but now and then change places. The volume is kept, but the liquid takes the shape of its container.
• Gas: the distances between molecules are much larger than the molecules themselves and the interaction is very weak. A gas fills the whole container and is easily compressed.

Heating raises the average kinetic energy of the molecules; at some point it “overcomes” the bonds and the substance changes state. That is why a phase change needs extra energy.

§2. Quantity of heat and heat capacity

When a body is heated or cooled without changing state, the heat received or given out is

Here c is the specific heat capacity, J/(kg·K): the heat needed to warm 1 kg of the substance by 1 K. A temperature difference is the same in degrees Celsius and in kelvins.

• For water c = 4200 J/(kg·K), a very large value. So the sea warms slowly in summer, cools slowly in winter and softens the coastal climate.
• For ice c ≈ 2100 J/(kg·K), aluminium c ≈ 900 J/(kg·K), copper c ≈ 400 J/(kg·K).

The heat capacity of a body, C = cm (J/K), is the heat needed to warm the whole body by 1 K.

Heat released by burning fuel:

Worked example

What is the heat capacity of a 0.4 kg aluminium spoon (c = 900 J/(kg·K))?

1. Formula: C = cm.
2. C = 900·0.4 = 360 J/K, so 360 J warms the spoon by 1 K.

§3. Melting and freezing

A crystalline solid melts at a definite melting point. For ice it is 0 °C at normal pressure. While melting, the heat supplied goes into breaking the crystal lattice, so the temperature stays constant until all the ice has melted.

λ is the specific latent heat of fusion, J/kg. For ice λ = 3.3·10⁵ J/kg.
Freezing releases the same amount of heat. That is why melting snow cools the air in spring, while freezing lakes slightly warm their surroundings in autumn. Amorphous solids have no definite melting point; they soften gradually.

On a heating graph (temperature against time) melting appears as a horizontal segment: heat keeps arriving, but the temperature stands still.

§4. Evaporation, boiling and condensation

Evaporation is the escape of molecules from the surface of a liquid. It happens at any temperature and depends on the temperature, the surface area, wind (removal of vapour) and the kind of liquid. The fastest molecules leave, so the average energy of the rest falls and the liquid cools. That is why sweat cools the body.

Boiling is vaporisation throughout the whole volume: bubbles of vapour form inside the liquid. A liquid boils when the saturated vapour pressure inside a bubble equals the external pressure. Hence the boiling point depends on external pressure: in the mountains the pressure is lower and water boils below 100 °C; in a pressure cooker it is higher and water boils above 100 °C.

For water L = 2.3·10⁶ J/kg, almost 7 times the latent heat of fusion. Condensation (vapour turning into liquid) releases the same heat, which is why a steam burn is worse than a burn from boiling water.

Worked example

How much heat turns 1 kg of ice at −10 °C into steam at 100 °C (c_ice = 2100, c_water = 4200 J/(kg·K))?

1. Warming the ice to 0 °C: Q₁ = 2100·1·10 = 21 000 J.
2. Melting: Q₂ = 3.3·10⁵·1 = 330 000 J.
3. Warming the water from 0 °C to 100 °C: Q₃ = 4200·1·100 = 420 000 J.
4. Vaporisation: Q₄ = 2.3·10⁶·1 = 2 300 000 J.
5. Total: Q = 21 000 + 330 000 + 420 000 + 2 300 000 = 3 071 000 J ≈ 3.07 MJ.

§5. The heat balance equation

In a thermally insulated system energy is conserved: the heat given out by hot bodies equals the heat taken in by cold ones.

Method:

1. Decide which bodies cool (give heat) and which warm up (receive heat).
2. Check whether there is a phase change (does ice melt, does steam condense?).
3. Write a separate Q for each process and form the equation.
4. Check the answer for sense: the final temperature must lie between the initial ones.

When two portions of the same substance are mixed, the formula simplifies to t = (m₁t₁ + m₂t₂)/(m₁ + m₂).

Worked example

0.2 kg of ice at 0 °C is put into 1 kg of water at 50 °C. What is the final temperature?

1. Melting the ice needs λm = 3.3·10⁵·0.2 = 66 000 J. Cooling to 0 °C, the water could give 4200·1·50 = 210 000 J, which is enough, so all the ice melts.
2. Balance: 4200·1·(50 − t) = 66 000 + 4200·0.2·t.
3. 210 000 − 4200t = 66 000 + 840t → 144 000 = 5040t.
4. t ≈ 28.6 °C.

Worked example

How much gas must be burnt on a 40 % efficient cooker to heat 3 kg of water from 20 °C to 100 °C (q = 4.4·10⁷ J/kg)?

1. The water needs Q = 4200·3·80 = 1 008 000 J.
2. The gas must release Q/η = 1 008 000/0.4 = 2 520 000 J.
3. m = 2 520 000/(4.4·10⁷) ≈ 0.057 kg = 57 g.

§6. Saturated vapour and humidity

In a closed vessel above a liquid the rates of evaporation and condensation become equal: a dynamic equilibrium is reached. Such vapour is called saturated. Its pressure depends only on temperature, not on volume: if the volume is reduced, some vapour condenses and the pressure stays the same.

• Absolute humidity ρ is the mass of water vapour in 1 m³ of air (g/m³).
• Relative humidity shows how far the vapour is from saturation:

• The dew point is the temperature to which air must be cooled for its vapour to become saturated. Below it, dew and fog form.

The density of saturated vapour follows from the ideal gas equation:

Humidity is measured with a psychrometer or a hygrometer. A comfortable relative humidity for people is roughly 40–60 %.

Worked example

5 m³ of air at 19 °C contains 34.5 g of water vapour; the saturated vapour density is 16.3 g/m³. Find the relative humidity and how much more water must evaporate to saturate the air.

1. Absolute humidity: ρ = 34.5/5 = 6.9 g/m³.
2. φ = 6.9/16.3·100 % ≈ 42 %.
3. Missing vapour: (16.3 − 6.9)·5 = 47 g.

Worked example

A closed 400 l vessel holds air at 27 °C; evaporating 2 g of water in it makes the vapour saturated. p_sat = 3.6 kPa. What was the initial relative humidity?

1. ρ_sat = 3600·0.018/(8.31·300) ≈ 0.026 kg/m³ = 26 g/m³.
2. Saturated vapour in the vessel: 26·0.4 = 10.4 g; initially there was 10.4 − 2 = 8.4 g, so ρ = 8.4/0.4 = 21 g/m³.
3. φ = 21/26 ≈ 81 %.

Check yourself

1. How much heat warms 2 kg of water from 10 °C to 60 °C?

2. How much heat turns 0.5 kg of water at 100 °C into steam?

3. 0.3 kg of water at 90 °C is mixed with 0.7 kg at 10 °C. Final temperature?

4. Absolute humidity is 8 g/m³, saturated vapour density 16 g/m³. Relative humidity?

5. How much heat melts 0.5 kg of ice at 0 °C and warms the water to 20 °C?

6. How much heat is released when 50 g of petrol burns completely (q = 4.6·10⁷ J/kg)?

7. How does the temperature of ice change while it melts?

8. In the mountains, is the boiling point of water above or below 100 °C?

Common mistakes

• Forgetting the latent heat (λm, Lm) and writing only cmΔt.
• Thinking that the temperature rises during melting or boiling.
• Not checking first whether all the ice melts: a final temperature below 0 °C means an error.
• Taking relative humidity relative to atmospheric pressure instead of the saturated vapour pressure.
• Leaving mass in grams and multiplying by J/(kg·K).