Change of state and latent heat

Heat without a rise in temperature

Put ice on a stove with a thermometer in it. The temperature climbs to $0\ ^\circ\text{C}$, then stops while the ice melts, though heat keeps flowing in. Then it climbs to $100\ ^\circ\text{C}$ and stops again while the water boils.

During a change of state, the heat goes into breaking the bonds between molecules, not into making them move faster. This heat is the latent heat:

• Ice melting: $L_f = 3.34 \times 10^{5}\ \text{J/kg}$.
• Water boiling: $L_v = 2.26 \times 10^{6}\ \text{J/kg}$, nearly seven times as much.

The same heat comes out again when water freezes or steam condenses.

You should find that boiling the water away takes far more heat than all the other stages together.

Pressure and boiling

Water boils when its vapour pressure equals the pressure on it. In a pressure cooker, at about two atmospheres, it boils near $120\ ^\circ\text{C}$ and food cooks faster; high in the mountains it boils below $100\ ^\circ\text{C}$ and food cooks slowly.

Evaporation cools

Water evaporates at any temperature, and the fastest molecules leave. Those left behind are cooler. Sweat, a wet cloth on a clay pot, and a dog's panting all use this.

Watch out. During melting or boiling, use $mL$, not $mc\,\Delta T$: the temperature does not change.

A worked example

How much heat turns $0.30\ \text{kg}$ of ice at $-10\ ^\circ\text{C}$ into water at $40\ ^\circ\text{C}$?

1. Warm the ice. $0.30\ \text{kg} \times 2100\ \text{J kg}^{-1}\text{K}^{-1} \times 10\ \text{K} = 6300\ \text{J}$.
2. Melt it. $0.30\ \text{kg} \times 3.34 \times 10^{5}\ \text{J/kg} = 100\,200\ \text{J}$.
3. Warm the water. $0.30\ \text{kg} \times 4186\ \text{J kg}^{-1}\text{K}^{-1} \times 40\ \text{K} = 50\,232\ \text{J}$.
4. Total. $1.57 \times 10^{5}\ \text{J}$; melting is the biggest part.

Practise with new numbers

Try it in Python

The program adds up the three stages of the example.

Remember these

• During a change of state the temperature stays steady: $Q = mL$.
• $L_f = 3.34 \times 10^{5}\ \text{J/kg}$, $L_v = 2.26 \times 10^{6}\ \text{J/kg}$.
• Higher pressure, higher boiling point; evaporation cools.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How much heat boils away $0.10\ \text{kg}$ of water already at $100\ ^\circ\text{C}$?

2. Why is a burn from steam worse than one from boiling water?

3. Why does food cook slowly high in the mountains?

4. Why does a clay pot keep water cool in summer?