4.3 Latent heat and enthalpy of phase transitions

Latent Heat and Phase Transitions

Role of latent heat in transitions

Latent heat is the energy absorbed or released when a substance changes phase without changing temperature. During a phase transition (melting, vaporization, sublimation), the temperature stays constant until the entire substance has converted to the new phase.

Why doesn't the temperature change? All the energy goes toward breaking or forming intermolecular bonds rather than increasing the kinetic energy of particles. On a heating curve, you can see this as the flat horizontal segments where temperature plateaus even though heat is still being added.

Fusion vs. vaporization latent heat

These are the two most common types of latent heat you'll work with:

• Latent heat of fusion ($L_f$): the energy needed to convert a substance from solid to liquid (melting) or released when liquid becomes solid (freezing), all at constant temperature. This energy overcomes enough intermolecular forces to break the rigid structure of the solid.
• Latent heat of vaporization ($L_v$): the energy needed to convert a substance from liquid to gas (vaporization) or released when gas becomes liquid (condensation), at constant temperature. This energy must completely separate molecules from each other.

The latent heat of vaporization is always significantly larger than the latent heat of fusion for a given substance. For water, $L_v = 2260 \text{ kJ/kg}$ while $L_f = 334 \text{ kJ/kg}$. That's roughly 6.8 times more energy. The reason: going from liquid to gas requires fully overcoming intermolecular attractions so molecules can move independently, whereas melting only loosens the solid structure without fully separating molecules.

Calculations with the latent heat equation

The equation for energy involved in a phase change is:

$Q = mL$

where $Q$ is energy (J), $m$ is mass (kg), and $L$ is the specific latent heat (J/kg).

Example: How much energy is needed to melt 2 kg of ice at 0°C?

1. Identify the correct latent heat value: $L_f = 334 \text{ kJ/kg}$ for water.
2. Plug into the equation: $Q = 2 \text{ kg} \times 334 \text{ kJ/kg}$
3. Solve: $Q = 668 \text{ kJ}$

Note that this only covers the phase change itself. If you also need to raise the temperature of the resulting water, you'd use $Q = mc\Delta T$ for that portion and add the two energies together. Multi-step heating problems almost always combine both equations.

Enthalpy and Phase Transitions

Enthalpy in phase transitions

Enthalpy ($H$) is a thermodynamic property representing the total heat content of a system, defined as:

$H = U + PV$

where $U$ is internal energy, $P$ is pressure, and $V$ is volume.

During a phase transition at constant pressure, the change in enthalpy ($\Delta H$) equals the latent heat of that transition. This is why you'll often see the terms used interchangeably in practice:

• $\Delta H_{fus}$ = latent heat of fusion
• $\Delta H_{vap}$ = latent heat of vaporization
• $\Delta H_{sub}$ = latent heat of sublimation

For endothermic transitions (melting, vaporization, sublimation), $\Delta H > 0$ because the system absorbs energy. For exothermic transitions (freezing, condensation, deposition), $\Delta H < 0$ because the system releases energy. The magnitudes are the same for a transition and its reverse; only the sign changes.

Enthalpy analysis using phase diagrams

Phase diagrams plot the phases of a substance as a function of pressure and temperature. They contain three key boundary lines:

• Solid-liquid line: melting/freezing points at various pressures
• Liquid-gas line: vaporization/condensation points at various pressures
• Solid-gas line: sublimation/deposition points at various pressures

Along each line, two phases coexist in equilibrium.

The Clausius-Clapeyron equation connects the slope of these boundary lines to the enthalpy and volume changes during the transition:

$\frac{dP}{dT} = \frac{\Delta H}{T \Delta V}$

where $T$ is the transition temperature and $\Delta V$ is the change in molar volume between the two phases. (The simpler form $\frac{dP}{dT} = \frac{\Delta H}{\Delta V}$ omits the temperature term and is sometimes called the Clapeyron equation, but the full Clausius-Clapeyron form is more commonly used.)

By comparing slopes on the phase diagram, you can draw conclusions about relative enthalpy changes. A steeper liquid-gas line compared to the solid-liquid line tells you that the enthalpy change during vaporization is larger than during melting, which is consistent with $L_v > L_f$.