Latent Heat of Fusion

Latent Heat of Fusion

The Latent Heat Of Fusion Formula, commonly known as the enthalpy of fusion, describes how much energy must be given to a solid substance (often in the form of heat) to produce a change in its physical state and turn it into a liquid (when the pressure of the environment is kept constant). For instance, the latent heat of fusion of one kilogram of water is 333.55 kilojoules, which is the amount of heat energy required to transform one kilogram of ice while maintaining a constant ambient temperature of 0 degrees Celsius. Learn more about latent heat of fusion, its formula and examples

What is Latent Heat of Fusion?

The Latent Heat of Fusion refers to the amount of heat energy required to change a substance from a solid state to a liquid state at its melting point, without changing its temperature. It is a specific type of latent heat, which is the heat absorbed or released during a phase change (such as melting or freezing) that occurs at a constant temperature.

When a substance reaches its melting point, the particles within the solid structure begin to overcome the forces holding them in place. Energy is needed to break these intermolecular bonds and transition the substance into a liquid state. This energy, known as the latent heat of fusion (L_f​), is absorbed from the surroundings by the substance undergoing fusion.

Latent Heat of Fusion Formula

The formula for Latent heat of fusion is

Q = m × L

Wherein

L = specific latent heat of fusion of substance.

The heat that the substance absorbs or releases is stated as when the temperature changes from t₁ (low temperature) to t₂ (high temperature).

Q = mc Δt  

Q = mc (t₂ – t₁)

The total amount of heat absorbed or liberated by the material is

Q = mL + mc Δt

Unit of Latent Heat of Fusion

The formula of Latent heat of fusion is J/kg

Specific Heat of Fusion and Molar Heat of Fusion

The terms “Specific Heat of Fusion” and “Molar Heat of Fusion” both relate to the amount of heat energy required for a substance to undergo a phase change from solid to liquid, specifically during the process of fusion (melting). Here’s an explanation of each term:

Specific Heat of Fusion

The Specific Heat of Fusion (\( \Delta H_{fus} \)) is the amount of heat energy required to change one unit of mass (often one kilogram or one gram) of a substance from solid to liquid at its melting point, without changing its temperature.

Formula:
\[ \Delta H_{fus} = \frac{Q}{m} \]

Where:
\( Q \) is the amount of heat energy absorbed or released during fusion (in joules),
\( m \) is the mass of the substance undergoing fusion (in kilograms or grams).

Units: The Specific Heat of Fusion is typically expressed in joules per kilogram (J/kg) or joules per gram (J/g).

Example:
For water, the Specific Heat of Fusion is approximately 334,000 J/kg. This means that 334,000 joules of energy are required to melt 1 kilogram of ice at 0°C into liquid water at 0°C.

Molar Heat of Fusion

The Molar Heat of Fusion (\( \Delta H_{fus}^{\circ} \)) is the amount of heat energy required to change one mole of a substance from solid to liquid at its melting point, without changing its temperature.

Formula:
\[ \Delta H_{fus}^{\circ} = \frac{Q}{n} \]

Where:
\( Q \) is the amount of heat energy absorbed or released during fusion (in joules),
\( n \) is the number of moles of the substance undergoing fusion.

Units: The Molar Heat of Fusion is expressed in joules per mole (J/mol).

Example:
For water, the Molar Heat of Fusion is approximately 6.01 kJ/mol (or 6,010 J/mol). This means that 6,010 joules of energy are required to melt one mole of ice at 0°C into liquid water at 0°C.

Relationship Between Specific Heat of Fusion and Molar Heat of Fusion

The Specific Heat of Fusion and Molar Heat of Fusion are related by the substance’s molar mass (\( M \)):

\[ \Delta H_{fus}^{\circ} = M \cdot \Delta H_{fus} \]

Where:
\( M \) is the molar mass of the substance (in grams per mole).

Application of Latent Heat of Fusion

The latent heat of fusion finds application in various fields due to its role in phase transitions, particularly in the conversion of solids to liquids. Here are some significant applications:

• Freezing: Preserving perishable foods by removing heat from them to freeze them solid, preserving freshness and extending shelf life.
• Thawing: Allowing foods to thaw by absorbing heat from the surroundings, ensuring safe and controlled thawing without cooking.
• Ice Formation: Cooling systems use the latent heat of fusion to convert refrigerants from liquid to solid form (ice) in ice-making processes, contributing to efficient cooling.
• Heat Removal: During the phase change from liquid to solid, significant heat absorption occurs, enhancing the efficiency of cooling cycles.
• Cryogenic Cooling: Using phase change materials to achieve extremely low temperatures for scientific experiments, medical applications (like MRI machines), and space exploration.
• Superconductors: Cooling superconducting materials to their critical temperature using phase change materials, enabling zero-resistance electrical conductivity and other unique properties.

Solved Examples of Latent Heat of Fusion Formula

Example 1: Calculate the amount of heat energy required to melt 500 grams of ice at 0°C.

Solution:

Given:
Mass (\( m \)) = 500 grams = 0.5 kg (since 1 kg = 1000 grams)
Latent heat of fusion (\( L_f \)) for water = 334,000 J/kg

\[ Q = m \cdot L_f \]

\[ Q = 0.5 \text{ kg} \times 334,000 \text{ J/kg} \]
\[ Q = 167,000 \text{ joules} \]

Answer: The amount of heat energy required to melt 500 grams of ice at 0°C is 167,000 joules.

Example 2: A substance has a molar heat of fusion (\( \Delta H_{fus}^{\circ} \)) of 6.01 kJ/mol. Calculate the latent heat of fusion in joules per kilogram.

Solution:

Given:
Molar heat of fusion (\( \Delta H_{fus}^{\circ} \)) = 6.01 kJ/mol = 6.01 × 10³ J/mol
Molar mass of the substance (\( M \)) = 18.02 g/mol (for water)

\[ L_f = \frac{\Delta H_{fus}^{\circ}}{M} \]

Convert molar mass to kilograms per mole:
\[ M = 18.02 \text{ g/mol} = 0.01802 \text{ kg/mol} \]

Now, calculate latent heat of fusion:
\[ L_f = \frac{6.01 \times 10^3 \text{ J/mol}}{0.01802 \text{ kg/mol}} \]
\[ L_f \approx 333,407 \text{ J/kg} \]

Answer: The latent heat of fusion of the substance is approximately 333,407 joules per kilogram.

Example 3: How much energy is required to heat 2 kg of ice at -10°C to water at 20°C?

Solution:

Mass (\( m \)) = 2 kg
Initial temperature (\( T_{initial} \)) = -10°C
Final temperature (\( T_{final} \)) = 20°C
Latent heat of fusion (\( L_f \)) for water = 334,000 J/kg

1. Calculate energy to melt ice at -10°C to 0°C:
\[ Q_1 = m \cdot L_f \]
\[ Q_1 = 2 \text{ kg} \times 334,000 \text{ J/kg} \]
\[ Q_1 = 668,000 \text{ joules} \]

2. Calculate energy to heat water from 0°C to 20°C:
– Heat capacity of water (\( C \)) = 4,186 J/kg°C (specific heat capacity of water)

\[ Q_2 = m \cdot C \cdot \Delta T \]
\[ \Delta T = T_{final} – T_{initial} = 20°C – 0°C = 20°C \]

\[ Q_2 = 2 \text{ kg} \times 4,186 \text{ J/kg°C} \times 20°C \]
\[ Q_2 = 167,440 \text{ joules} \]

3. Total energy required:
\[ Q_{total} = Q_1 + Q_2 \]
\[ Q_{total} = 668,000 \text{ J} + 167,440 \text{ J} \]
\[ Q_{total} = 835,440 \text{ joules} \]

Answer: The total energy required to heat 2 kg of ice at -10°C to water at 20°C is 835,440 joules.