Key Takeaways
- →Heat of fusion (enthalpy of fusion, ΔH_fus) is the energy required to convert one mole or one gram of a solid into a liquid at constant temperature and pressure.
- →The total heat needed to melt a mass m is Q = m · ΔH_fus, where ΔH_fus is expressed in J/g or kJ/mol.
- →Heat of fusion is a latent heat: it changes the phase of a substance without changing its temperature.
- →Water has a high heat of fusion (334 J/g), which moderates climate by absorbing or releasing large amounts of energy during freezing and melting.
- →The same magnitude of heat is released when a liquid freezes; the sign of Q changes, but ΔH_fus remains the same physical quantity.
Heat of Fusion Calculator: Energy for Solid-to-Liquid Phase Transitions
In 1761, the Scottish chemist Joseph Black distinguished between sensible heat and latent heat, coining the term "latent" to describe heat that changes the state of a substance without raising its temperature. His experiments with ice showed that a surprisingly large amount of heat is required to melt ice into water — far more than the heat required to warm the same water by a few degrees. This latent heat of fusion is now called the enthalpy of fusion (ΔH_fus). It is a fundamental thermodynamic property that governs melting, freezing, climate, refrigeration, and materials processing. Ice's heat of fusion of 334 joules per gram means that melting one gram of ice at 0°C absorbs enough energy to cool 80 grams of water by 1°C — a property that stabilizes Earth's climate and protects ecosystems from rapid temperature swings.
- What heat of fusion measures
- How to use the heat of fusion calculator
- The heat of fusion formula
- Heat of fusion values for common substances
- Sensible heat vs. latent heat
- Applications in nature and industry
- Worked examples
- Frequently Asked Questions
What heat of fusion measures
Heat of fusion, or enthalpy of fusion (ΔH_fus), is the amount of heat energy that must be supplied to one mole (or one gram) of a substance at its melting point to convert it from the solid phase to the liquid phase. The process occurs at constant temperature, because the absorbed energy goes into breaking intermolecular bonds in the solid rather than increasing kinetic energy.
Key points:
- ΔH_fus is always positive for melting (endothermic).
- The same amount of heat is released during freezing (exothermic), so ΔH_freeze = −ΔH_fus.
- Heat of fusion is a latent heat because it is hidden in the phase change and does not change temperature.
- Substances with strong intermolecular forces in the solid generally have higher heats of fusion.
- Heat of fusion is measured calorimetrically, often by observing the temperature plateau during melting.
The physical interpretation: at the melting point, molecules in the solid are vibrating but still locked in their lattice positions. The heat of fusion provides exactly enough energy to overcome the lattice binding energy and allow molecules to slide past one another as a liquid. No kinetic energy is added — the temperature stays constant — but the potential energy of the system increases as the ordered lattice breaks down into the disordered liquid.
How to use the heat of fusion calculator
The heat of fusion calculator on this page computes the total energy Q required to melt a given mass of a substance using the formula Q = m × ΔH_fus.
Inputs:
- Mass — Enter the mass of the substance in grams. The calculator accepts any positive value.
- Substance (preset ΔH_fus) — Select from a dropdown of common substances with pre-loaded heat of fusion values:
- Water / Ice — 334 J/g
- Ethanol — 108 J/g
- Mercury — 11.4 J/g
- Aluminum — 396 J/g
- Gold — 63.7 J/g
- Copper — 201 J/g
- Iron — 272 J/g
- Sodium Chloride — 492 J/g
- Custom — enter your own value
- Custom ΔH_fus — If you select "Custom," enter the specific heat of fusion in J/g for your substance.
The calculator returns:
- Heat Required (J) — the total energy in joules
- Heat Required (kJ) — the same value in kilojoules
- Heat Required (kcal) — the same value in food calories
- ΔH_fus Used — confirms the value applied in the calculation
- Formula — displays Q = m × ΔH_fus for reference
Practical use cases:
- Refrigeration design: Sizing ice packs and phase-change thermal storage for transport cooling.
- Metallurgy: Calculating furnace energy requirements to melt metal ingots.
- Climate science: Estimating the energy absorbed when polar ice melts.
- Food science: Determining cooling capacity needed for frozen food transport.
The heat of fusion formula
The total energy required to melt a mass m of a substance is:
Q = m · ΔH_fus
where:
- Q is the heat absorbed (J or kJ)
- m is the mass of the substance (g)
- ΔH_fus is the specific heat of fusion (J/g)
For molar calculations:
Q = n · ΔH_fus_molar
where n is the number of moles and ΔH_fus_molar is the molar heat of fusion (kJ/mol or J/mol).
When a substance must first be heated to its melting point and then melted, the total energy is the sum of sensible heat and latent heat:
Q_total = m · c_solid · ΔT + m · ΔH_fus
where c_solid is the specific heat capacity of the solid and ΔT is the temperature rise to the melting point.
Unit conversions:
- 1 cal = 4.184 J
- 1 kcal = 4,184 J
- 1 kJ = 1,000 J
- 1 BTU = 1,055 J
- To convert from kJ/mol to J/g: divide by molar mass and multiply by 1,000. For water: 6.01 kJ/mol ÷ 18.015 g/mol × 1,000 = 333.6 J/g ≈ 334 J/g.
Heat of fusion values for common substances
| Substance | Melting Point (°C) | ΔH_fus (J/g) | ΔH_fus (kJ/mol) | Crystal Type |
|---|---|---|---|---|
| Water (ice) | 0.0 | 334 | 6.01 | Molecular (H-bond) |
| Ethanol | −114 | 108 | 4.93 | Molecular |
| Mercury | −38.8 | 11.4 | 2.30 | Metallic |
| Sodium chloride | 801 | 492 | 28.8 | Ionic |
| Aluminum | 660 | 396 | 10.7 | Metallic |
| Gold | 1,064 | 63.7 | 12.5 | Metallic |
| Copper | 1,085 | 201 | 13.0 | Metallic |
| Iron | 1,538 | 272 | 15.2 | Metallic |
| Silver | 962 | 105 | 11.3 | Metallic |
| Lead | 328 | 23.0 | 4.77 | Metallic |
| Paraffin wax | 46–68 | 200–250 | — | Molecular |
| Naphthalene | 80 | 147 | 18.8 | Molecular |
| Gallium | 29.8 | 80.1 | 5.59 | Metallic |
Ionic and metallic solids generally have high molar heats of fusion because strong bonds must be broken. Molecular substances tend to have lower values. Water's heat of fusion is unusually high for a small molecule because of extensive hydrogen bonding in ice — each water molecule forms four hydrogen bonds in a tetrahedral lattice that requires significant energy to disrupt.
Sensible heat vs. latent heat
Sensible heat changes the temperature of a substance. Latent heat changes its phase. The distinction is essential in calorimetry and engineering.
For example, to convert 1 kg of ice at −10°C to water at 20°C, three steps are required:
- Warm ice from −10°C to 0°C: Q = 1,000 g × 2.09 J/g·K × 10 K = 20,900 J
- Melt ice at 0°C: Q = 1,000 g × 334 J/g = 334,000 J
- Warm water from 0°C to 20°C: Q = 1,000 g × 4.18 J/g·K × 20 K = 83,600 J
Total = 20,900 + 334,000 + 83,600 = 438,500 J ≈ 438.5 kJ
The latent heat of fusion (step 2) accounts for about 76% of the total energy, illustrating its importance. This is why ice is so effective at cooling drinks — the phase change absorbs vastly more energy than simple warming of the resulting water.
The same principle applies in reverse during freezing. When water turns to ice, it releases 334 J/g — the same magnitude but opposite in sign. This is why coastal areas near large bodies of water experience milder winters: the latent heat released during freezing moderates temperature drops.
Applications in nature and industry
Climate regulation: Ice on Earth's polar regions reflects sunlight and absorbs large amounts of heat when melting. The heat of fusion of water creates a massive thermal buffer. Melting the Greenland ice sheet (approximately 2.85 million km³ of ice) would absorb roughly 9.5 × 10²³ joules — equivalent to about 1,000 years of global energy consumption at current rates. Without this latent heat buffer, global temperatures would rise much faster.
Refrigeration and thermal storage: Phase-change materials (PCMs) such as paraffin wax and salt hydrates are used in building materials and packaging to store and release thermal energy at nearly constant temperature. A wallboard infused with paraffin PCM absorbs heat as it melts during the day and releases it as it refreezes at night, passively regulating indoor temperature without electricity.
Metallurgy: Foundries must supply the heat of fusion to melt metals. The energy required determines furnace size, fuel consumption, and processing time. Melting 1 ton of aluminum requires approximately 396 MJ just for the phase change — before accounting for heating the solid to 660°C.
Food preservation: Ice packs and frozen water bottles rely on the high heat of fusion of water to keep food cold during transport. A 500-gram ice pack absorbs 167,000 J (167 kJ) as it melts — enough to keep a cooler cold for hours.
Cryosurgery and cryopreservation: Controlled freezing and thawing of tissues and biological samples depends on understanding and managing latent heat effects. Vitrification — an alternative to crystalline freezing — avoids the heat of fusion entirely by forming a glass instead of ice.
Geology: The heat of fusion of silicate minerals governs magma formation and volcanic processes. Partial melting of rock in the Earth's mantle releases latent heat that drives plate tectonics.
Worked examples
Example 1: Melting ice
How much heat is required to melt 250 g of ice at 0°C?
Q = m · ΔH_fus = 250 g × 334 J/g = 83,500 J = 83.5 kJ
This is the calculator's default operation. Select "Water / Ice" and enter 250 g.
Example 2: Cooling water by melting ice
How many grams of ice at 0°C must melt to absorb 100 kJ of heat?
m = Q / ΔH_fus = 100,000 J / 334 J/g = 299 g
About 299 g of ice must melt. This is the calculation behind ice bath cooling in chemistry labs.
Example 3: Heating and melting aluminum
How much heat is needed to convert 500 g of aluminum at 25°C to liquid aluminum at its melting point (660°C)? The specific heat of solid aluminum is 0.90 J/g·K, and its heat of fusion is 396 J/g.
Step 1: Heat solid from 25°C to 660°C. Q1 = 500 g × 0.90 J/g·K × (660 − 25) K = 500 × 0.90 × 635 = 285,750 J
Step 2: Melt aluminum at 660°C. Q2 = 500 g × 396 J/g = 198,000 J
Total = 285,750 + 198,000 = 483,750 J = 483.8 kJ
The latent heat (Q2) accounts for 41% of the total — a significant fraction even at these high temperatures.
Example 4: Phase-change material for building insulation
A wallboard contains 5 kg of paraffin wax (ΔH_fus ≈ 210 J/g). How much heat can it absorb during the day as the wax melts?
Q = 5,000 g × 210 J/g = 1,050,000 J = 1,050 kJ
This is equivalent to the cooling capacity of a small air conditioner running for about 15 minutes — passive, free, and silent.
Example 5: Freezing releases heat
A lake freezes and forms 10,000 kg of ice overnight. How much heat is released into the environment?
Q = 10,000,000 g × 334 J/g = 3,340,000,000 J = 3,340 MJ
The freezing process releases 3,340 megajoules of heat — which warms the surrounding air and water, slowing further freezing. This is why lakes freeze gradually rather than all at once.