GRADE 8 · CHANGE IN STATE OF AGGREGATION
Melting, solidification and graphs
Melting temperature and energy of a phase change.
Read the full lesson theory here without an account. In the interactive lesson, the same theory is presented as cards, followed by questions with feedback and practice problems. Your progress is saved.
Understanding step by step
We supply energy to ice at 0 °C. Less ice remains, but the thermometer still reads 0 °C. Why does energy enter while the familiar warming equation seems to give zero?
A change of state
Watch an ice cube: melting removes its fixed shape, but water does not become a different substance. Melting is the change from solid to liquid; solidification is the reverse. Particle arrangement and the ability to exchange neighbours change, not the molecules' composition. Thermal motion alone cannot explain this; we must also account for interaction energy.
The transition temperature
Pure ice and water can coexist near 0 °C at normal atmospheric pressure. This is the melting temperature: at that pressure, solid and liquid phases can be in equilibrium. It depends on the substance and pressure, so 0 °C is not a rule for every crystal. If some water freezes under the same equilibrium conditions, the change occurs at the same temperature.
Where does the energy go?
Where does energy go when the thermometer reading does not change? In a crystal, particles are held in ordered positions. Letting them exchange neighbours more freely requires changes in their arrangement and interactions. Energy goes into these changes during equilibrium melting. While both phases coexist at constant pressure, temperature does not rise. Once the last ice disappears, further energy transfer can warm the resulting water.
Read each graph segment
On a temperature-versus-supplied-heat graph, warming within one state gives a sloping segment: energy enters and temperature rises. During melting, energy keeps entering but the line is horizontal. Now imagine a temperature-versus-time graph T(t) after the heater is switched off: it can be horizontal without melting. Read a plateau together with the experimental conditions; time alone does not tell us how much energy was supplied.
Energy per kilogram
Consider one kilogram of a substance already at its melting temperature. The energy needed to turn it into liquid is its specific latent heat of fusion λ, measured in J/kg. A larger λ requires more energy to melt the same mass. This differs from specific heat capacity: c concerns a temperature change, while λ concerns a change of state at the transition temperature.
Calculate melting energy
If each kilogram needs λ joules, mass m needs Q = λm. Doubling mass doubles the heat; if only half the original substance melts, use that half's mass. At m = 0 no energy is needed for the transition itself. Solidifying the same mass under the same conditions releases the same energy magnitude, with negative Q for the substance. Units: Дж/кг · кг = Дж.
What if the ice starts colder?
What if the ice starts below 0 °C? Separate the process by what the thermometer and container show. First, the ice stays solid and warms, requiring cmΔT with the specific heat capacity of ice. Then water appears at the transition temperature, requiring λm. Add these contributions. The term λm does not also warm cold ice, and cmΔT cannot describe a segment where temperature stays constant while the state changes.
Limits of the school model
A clear plateau describes a pure crystalline substance undergoing a nearly equilibrium transition at constant pressure. Rapid uneven heating, impurities or changing pressure can alter the experiment. An amorphous material such as glass softens differently and need not have this plateau. The model usefully separates warming from a phase change, but its features cannot be applied unchecked to every material or every horizontal graph.
Melting without a temperature rise
Melt 0.5 kg of ice already at 0 °C. λ = 334 кДж/кг. How much heat is needed?
Worked example · 1
The ice is already at its melting temperature, so only the change of state is needed; there is no cmΔT warming stage here.
Worked example · 2
Each kilogram needs 334 kJ. For 0.5 kg, Q = λm = 334·0,5 = 167 кДж; kilograms cancel.
Worked example · 3
Half a kilogram needs half the energy needed for a kilogram. During the transition, the 167 kJ supplied changes the state while the mixture's temperature may remain 0 °C.
Check the result
167 kJ. This melts 0.5 kg of ice at 0 °C but does not include later warming of the water.
Key idea · 1
During a phase change, internal energy changes even without a temperature change.
Key idea · 2
Q = λm applies to the mass that actually melts at the transition temperature.
Key idea · 3
Cold ice is warmed first and melted afterwards; the graph helps separate these stages.
What comes next
A liquid can become gas without boiling. Next we explore evaporation and humidity.
Basic Concepts
Melting and solidification
Melting changes a solid into a liquid; solidification is the reverse change.
For a pure crystalline substance, both processes at a given pressure occur at the same temperature.
Melting point
Temperature of phase equilibrium of solid and liquid states at a given pressure.
During equilibrium melting of a pure crystalline substance at constant pressure, energy changes particle arrangement and interactions while temperature stays constant. The molecules themselves are not destroyed. Pressure and impurities can change the transition temperature.
Heating graph
The graph shows how temperature changes with time, allowing us to identify warming and phase-change segments.
A sloping segment indicates temperature change. A horizontal segment while heat is still supplied may indicate a phase change. Constant temperature alone does not prove this: thermal equilibrium without heating also gives a horizontal line.
Physical quantities
Specific latent heat of fusion
\lambdaEnergy needed to melt a unit mass of a substance at its melting temperature.
Unit: joule per kilogram · \frac{\mathrm{Дж}}{\mathrm{кг}}
Lesson formulas
Heat required for melting
Q = \lambda mAt constant temperature, the heat required for melting equals specific latent heat of fusion multiplied by mass.
Problems on this topic
Statements, answer options and the walkthrough.