define and use specific latent heat and distinguish between specific latent heat of fusion and specific latent heat of vaporisation

Specific Latent Heat

When a substance changes phase without a change in temperature, the energy required (or released) is called the latent heat. The amount of energy per unit mass is the specific latent heat, denoted by \$L\$.

Definition

The specific latent heat \$L\$ of a substance is the energy required to change the phase of 1 kg of the substance at constant temperature and pressure.

Mathematical Form

\$Q = mL\$

where

• \$Q\$ = heat energy transferred (J)
• \$m\$ = mass of the substance (kg)
• \$L\$ = specific latent heat (J kg\(^{-1}\))

Types of Specific Latent Heat

1. Specific latent heat of fusion (\$L_f\$) – energy required to melt a solid into a liquid.
2. Specific latent heat of vaporisation (\$L_v\$) – energy required to convert a liquid into a gas.

Distinguishing \$Lf\$ and \$Lv\$

Although both are forms of latent heat, they differ in the phase transition involved and in magnitude:

• Fusion occurs at the melting point; the molecular arrangement changes from ordered (solid) to less ordered (liquid) while the intermolecular forces are partially broken.
• Vaporisation occurs at the boiling point; the intermolecular forces are largely overcome, producing a gaseous state.

Typical \cdot alues (at 1 atm)

Worked Example – Melting Ice

Calculate the heat required to melt 250 g of ice at 0 °C.

Given \$L_f\$(ice) = \$3.34 \times 10^5\ \text{J kg}^{-1}\$.

Solution:

\$m = 0.250\ \text{kg}\$

\$Q = mL_f = 0.250 \times 3.34 \times 10^5 = 8.35 \times 10^4\ \text{J}\$

Thus, \$8.35 \times 10^4\ \text{J}\$ of heat is needed.

Worked Example – Boiling Water

How much energy is required to convert 1.5 kg of water at 100 °C to steam at 100 °C?

Given \$L_v\$(water) = \$2.26 \times 10^6\ \text{J kg}^{-1}\$.

Solution:

\$Q = mL_v = 1.5 \times 2.26 \times 10^6 = 3.39 \times 10^6\ \text{J}\$

The required energy is \$3.39 \times 10^6\ \text{J}\$.

Key Points to Remember

• Latent heat is absorbed or released at constant temperature.
• \$Lf\$ is always smaller than \$Lv\$ for the same substance because breaking all intermolecular bonds (vaporisation) requires more energy than merely loosening them (fusion).
• Use \$Q = mL\$ for any phase change, substituting the appropriate \$Lf\$ or \$Lv\$.