3.2.3a Changes of state and specific latent heat
Latent heat: energy that changes state, not temperature
Latent heat
Latent heat is the energy needed to change the state of a substance, transferred without changing its temperature.
- When a substance changes state, energy is transferred to it when it melts or boils, and away from it when it condenses or freezes.
- This energy changes the internal energy of the substance.
- During the change of state the temperature stays constant, because the energy is used to change the state rather than to raise the temperature.
- For example, while ice is melting, energy is transferred to it, but its temperature stays at the melting point until all the ice has melted.
During a change of state, energy is transferred and the internal energy changes, but the temperature stays the same.
Specific latent heat and the equation E = mL
Specific latent heat
The specific latent heat of a substance is the energy needed to change the state of 1 kg1\ \text{kg}1 kg of it with no change in temperature, measured in joules per kilogram, J/kg\text{J/kg}J/kg.
- The word “specific” means “per kilogram”, so specific latent heat is the energy needed for each 1 kg1\ \text{kg}1 kg of a substance to change state.
- Different substances have different specific latent heats, so changing the state of 1 kg1\ \text{kg}1 kg of one may need a different amount of energy from another.
- The energy transferred during a change of state is E=mLE = mLE=mL.
- EEE is the energy transferred in joules, J\text{J}J; mmm is the mass in kilograms, kg\text{kg}kg; LLL is the specific latent heat in joules per kilogram, J/kg\text{J/kg}J/kg.
- The mass must be in kilograms; if a question gives grams, convert first using 1000 g=1 kg1000\ \text{g} = 1\ \text{kg}1000 g=1 kg.
A student melts 0.50 kg0.50\ \text{kg}0.50 kg of a solid whose specific latent heat of fusion is 200 000 J/kg200\,000\ \text{J/kg}200000 J/kg. Calculate the energy needed to melt it.
Write the equation:
E=mL E = mL E=mLSubstitute and calculate:
E=0.50×200 000=100 000 J E = 0.50 \times 200\,000 = 100\,000\ \text{J} E=0.50×200000=100000 JThe energy needed is 100 000 J100\,000\ \text{J}100000 J.
Fusion and vaporisation
Specific latent heat of fusion
The energy needed to change 1 kg1\ \text{kg}1 kg of a substance from solid to liquid with no change in temperature (melting or freezing).
Specific latent heat of vaporisation
The energy needed to change 1 kg1\ \text{kg}1 kg of a substance from liquid to vapour with no change in temperature (boiling or condensing).
- Use the specific latent heat of fusion when a solid melts into a liquid, or a liquid freezes into a solid.
- Use the specific latent heat of vaporisation when a liquid boils into a vapour, or a vapour condenses into a liquid.
- The exam clue is the change named: melting or freezing points to fusion, while boiling or condensing points to vaporisation.
- Do not say the temperature rises during melting or boiling just because energy is supplied; the temperature stays constant while the state changes.
- Use fusion for solid to liquid and vaporisation for liquid to vapour; do not mix them up.
- Convert the mass to kilograms before using E=mLE = mLE=mL.
- For a calculation, show E=mLE = mLE=mL, the mass in kg\text{kg}kg, the correct value of LLL, and the final answer in J\text{J}J.
- For a written answer, say the temperature does not change, then explain that the energy supplied changes the internal energy and the state of the substance.
- What is latent heat?
- Why does the temperature stay constant during a change of state?
- Define specific latent heat and give its unit.
- State the equation for the energy transferred during a change of state.
- When do you use the specific latent heat of fusion rather than of vaporisation?
3.2.3b Heating and cooling graphs
Reading heating and cooling graphs
- A sloping section shows the temperature changing while the substance stays in one state.
- A flat section, or plateau, shows a change of state happening at constant temperature.
- A heating or cooling graph plots temperature on the vertical axis against time or energy transferred on the horizontal axis.
- On a heating graph the order is: the solid warms, the solid melts at constant temperature, the liquid warms, the liquid boils at constant temperature, then the gas warms.
- A cooling graph is the reverse: the gas cools, it condenses at constant temperature, the liquid cools, it freezes at constant temperature, then the solid cools.
- The temperature of the melting or boiling plateau gives the melting point or boiling point of the substance.
Why the temperature is constant during a change of state
- On a sloping section, the energy transferred changes the average kinetic energy of the particles, so the temperature changes.
- On a flat section, energy is still being transferred, but it changes the particles’ potential energy as their arrangement changes, not their kinetic energy, so the temperature stays constant.
- During melting or boiling, energy is transferred to the substance to free the particles from their positions; during freezing or condensing, energy is transferred away as the particles come closer together.
- If the horizontal axis shows time, the length of a section only compares energy transfers when energy is supplied or removed at a constant rate.
Which equation to use
Specific heat capacity
The energy needed to raise the temperature of 1 kg1\ \text{kg}1 kg of a substance by 1 ∘C1\ ^\circ\text{C}1 ∘C, measured in J/kg ∘C\text{J/kg}\,^\circ\text{C}J/kg∘C.
Specific latent heat
The energy needed to change the state of 1 kg1\ \text{kg}1 kg of a substance with no change in temperature, measured in J/kg\text{J/kg}J/kg.
- On a sloping section the temperature changes, so use ΔE=mcΔθ\Delta E = mc\Delta\thetaΔE=mcΔθ with the specific heat capacity.
- On a flat section the state changes at constant temperature, so use E=mLE = mLE=mL with the specific latent heat.
- Specific heat capacity is measured in J/kg ∘C\text{J/kg}\,^\circ\text{C}J/kg∘C, while specific latent heat is measured in J/kg\text{J/kg}J/kg.
A 0.50 kg0.50\ \text{kg}0.50 kg substance is heated. Its temperature first rises from 20 ∘C20\ ^\circ\text{C}20 ∘C to 50 ∘C50\ ^\circ\text{C}50 ∘C, and it then melts at a constant temperature. Its specific heat capacity is 800 J/kg ∘C800\ \text{J/kg}\,^\circ\text{C}800 J/kg∘C and its specific latent heat of fusion is 40 000 J/kg40\,000\ \text{J/kg}40000 J/kg.
Sloping section (temperature rise):
ΔE=mcΔθ=0.50×800×(50−20)=12 000 J \Delta E = mc\Delta\theta = 0.50 \times 800 \times (50 - 20) = 12\,000\ \text{J} ΔE=mcΔθ=0.50×800×(50−20)=12000 JFlat section (melting):
E=mL=0.50×40 000=20 000 J E = mL = 0.50 \times 40\,000 = 20\,000\ \text{J} E=mL=0.50×40000=20000 JThe first step uses specific heat capacity because the temperature changes; the second uses specific latent heat because the state changes at constant temperature.
- A flat section does not mean no energy is being transferred; energy is transferred, but it changes the state rather than the temperature.
- Do not confuse the units: specific heat capacity is J/kg ∘C\text{J/kg}\,^\circ\text{C}J/kg∘C, while specific latent heat is J/kg\text{J/kg}J/kg.
- Check the axes first, then read each sloping section as a temperature change and each flat section as a change of state.
- For a plateau, write that the temperature is constant because the energy transferred changes the particles’ potential energy rather than their kinetic energy.
- Then decide whether the calculation needs ΔE=mcΔθ\Delta E = mc\Delta\thetaΔE=mcΔθ or E=mLE = mLE=mL.
- What does a sloping section of a heating or cooling graph represent?
- What does a flat section represent?
- Why does the temperature stay constant during a change of state?
- When do you use specific heat capacity, and when do you use specific latent heat?
- What are the units of specific heat capacity and specific latent heat?
