Latent heat during a phase change

Heat here means energy transferred across the system boundary. Positive material value $L$ specifies the joules required per kilogram for the selected transition. Heat into the substance is positive: melting and vaporisation require $Q=+mL$, whereas freezing and condensation release $Q=-mL$. If only part melts, $m$ includes only that part. Remaining ice is not mass whose melting energy has already been supplied.[1]

Choose the value after identifying the process: $L_f$ refers to fusion and $L_v$ to vaporisation. Both depend on substance and state. Warming liquid water does not melt it again. Ice below its melting temperature must first reach that temperature. These require different calculation stages even though heat enters in each case.

The relation follows from the meaning of specific transition energy. Two equivalent portions together require the sum of their individual energies. A small converted mass $dm$ therefore requires $\delta Q=L\,dm$. If state and transition conditions remain fixed, $L$ is constant and can be taken outside the integral when adding contributions. Proportionality to mass does not mean all substances need the same energy.

Where does this energy go? The new phase has different molecular interactions and structure. Vaporisation also expands volume against external pressure. At constant pressure, with only pressure–volume work and negligible macroscopic kinetic and potential changes, supplied heat therefore equals $\Delta H$, not generally $\Delta U$. A temperature plateau means equilibrium transition conditions are maintained; it does not mean energy remains unchanged.[2][3]

For a pure substance at fixed pressure, two phases coexist in equilibrium at a definite transition temperature. Additional heat changes their fractions: less ice and more water, initially at the same temperature. After the last ice melts, the liquid can warm. A horizontal segment in a temperature–energy plot therefore represents an energy interval, not an absence of supplied energy. Mixtures or varying pressure need not produce this simple shape.[1]

Near normal atmospheric pressure, ice melts at about 0 °C and pure water boils at about 100 °C. Pressure is part of these approximate statements. Evaporation can occur below boiling temperature and requires energy without the whole liquid boiling. Supercooling, superheating or delayed nucleation can make the actual path depart from the equilibrium picture.[1][3]

Use matching mass and property units: 333 kJ/kg equals 333000 J/kg or 333 J/g. For 150 g, use either 0.150 kg with J/kg or 150 g with J/g. Magnitude $|Q|$ and sign answer different questions: how much energy is transferred, and in which direction? Check the phase after each stage as well. A negative calculated water temperature after presumed complete melting can reveal that insufficient energy was available for the assumed final phase.

Heat capacity belongs to the separate heating stages before or after the transition. First subtract independently determined preheating energy from the available net energy budget; only the remainder can convert mass at the transition. If energy has not yet reached the transition temperature, the remaining budget cannot justify a positive converted mass.[1]

Zero converted mass gives zero transition heat. These two zero inputs do not determine L. Nonzero transition heat with zero converted mass does not fit $Q=\pm mL$. For a positive converted mass, heat sign must agree with the selected transition direction.[1]

Check the budget before calculating a final temperature. Warm water cooling to the melting point supplies only the energy it releases during that cooling. Compare this amount with the energy needed for all the ice. If insufficient, ice remains alongside water. Converted mass is then $m_{\mathrm{melt}}=\frac{Q_{\mathrm{available}}}{L_f}$, and equilibrium temperature stays at the melting point. Simply averaging initial temperatures misses the phase-change contribution.

With exactly sufficient energy, conversion finishes at the transition temperature. Only a remaining surplus can warm the new phase. A vessel and additional bodies may absorb or release energy and belong in the balance. If ice initially lies below melting temperature, its preheating must be supplied first; that energy is no longer available for melting.

A time calculation becomes possible only after useful power is specified. It does not replace the phase check. In the following model, all net power reaches the sample and heat transfer and geometry introduce no additional limitation.[1]

The first law of thermodynamics places transition heat within the complete energy balance. The limits below explain when pressure–volume work and departing mass also need to be included.[2]

For water, $L_v$ is the enthalpy difference between saturated liquid and saturated vapour at the relevant saturation state. Temperature and corresponding pressure change this value; IAPWS describes that state dependence. Rounded values in the examples are explicitly specified model values, not a table for arbitrary hot or compressed states. At the critical point the liquid–vapour phase distinction ends.[3][1]

The formula gives energy, not duration directly. A time estimate additionally needs the net power actually entering the transition. Conduction and heat transfer can limit that power. During open evaporation mass also leaves the vessel, so a complete energy balance includes its enthalpy flow. A fixed-mass calculation is appropriate only when the chosen boundary and final states account for those masses correctly.[2]

First ask: which phase is present, which mass changes phase, and is enough energy available? Calculate temperature stages and transitions separately. Sign describes energy direction; a plateau describes equilibrium between two phases. A duration requires an additional power or heat-transfer specification.