 Enthalpy of Fusion
Get Enthalpy of Fusion essential facts below. View Videos or join the Enthalpy of Fusion discussion. Add Enthalpy of Fusion to your PopFlock.com topic list for future reference or share this resource on social media.
Enthalpy of Fusion

The enthalpy of fusion of a substance, also known as (latent) heat of fusion, is the change in its enthalpy resulting from providing energy, typically heat, to a specific quantity of the substance to change its state from a solid to a liquid, at constant pressure. For example, when melting 1 kg of ice (at 0 °C under a wide range of pressures), 333.55 kJ of energy is absorbed with no temperature change. The heat of solidification (when a substance changes from liquid to solid) is equal and opposite.

This energy includes the contribution required to make room for any associated change in volume by displacing its environment against ambient pressure. The temperature at which the phase transition occurs is the melting point or the freezing point, according to context. By convention, the pressure is assumed to be 1 atm (101.325 kPa) unless otherwise specified.

## Overview

The 'enthalpy' of fusion is a latent heat, because during melting the heat energy needed to change the substance from solid to liquid at atmospheric pressure is latent heat of fusion, as the temperature remains constant during the process. The latent heat of fusion is the enthalpy change of any amount of substance when it melts. When the heat of fusion is referenced to a unit of mass, it is usually called the specific heat of fusion, while the molar heat of fusion refers to the enthalpy change per amount of substance in moles.

The liquid phase has a higher internal energy than the solid phase. This means energy must be supplied to a solid in order to melt it and energy is released from a liquid when it freezes, because the molecules in the liquid experience weaker intermolecular forces and so have a higher potential energy (a kind of bond-dissociation energy for intermolecular forces).

When liquid water is cooled, its temperature falls steadily until it drops just below the line of freezing point at 0 °C. The temperature then remains constant at the freezing point while the water crystallizes. Once the water is completely frozen, its temperature continues to fall.

The enthalpy of fusion is almost always a positive quantity; helium is the only known exception.Helium-3 has a negative enthalpy of fusion at temperatures below 0.3 K. Helium-4 also has a very slightly negative enthalpy of fusion below 0.77 K (-272.380 °C). This means that, at appropriate constant pressures, these substances freeze with the addition of heat. In the case of 4He, this pressure range is between 24.992 and 25.00 atm (2,533 kPa).

Substance Heat of fusion
(cal/g)
Heat of fusion
(J/g)
water 79.72 333.55
methane 13.96 58.99
propane 19.11 79.96
glycerol 47.95 200.62
formic acid 66.05 276.35
acetic acid 45.90 192.09
acetone 23.42 97.99
benzene 30.45 127.40
myristic acid 47.49 198.70
palmitic acid 39.18 163.93
sodium acetate 63-69 264-289
stearic acid 47.54 198.91
gallium 19.2 80.4
paraffin wax (C25H52) 47.8-52.6 200-220

These values are mostly from the CRC Handbook of Chemistry and Physics, 62nd edition. The conversion between cal/g and J/g in the above table uses the thermochemical calorie (calth) = 4.184 joules rather than the International Steam Table calorie (calINT) = 4.1868 joules.

## Examples

A) To heat 1 kg (1.00 liter) of water from 283.15 K to 303.15 K (10 °C to 30 °C) requires 83.6 kJ. However, to melt ice also requires energy. We can treat these two processes independently; thus, to heat 1 kg of ice from 273.15 K to water at 293.15 K (0 °C to 20 °C) requires:

(1) 333.55 J/g (heat of fusion of ice) = 333.55 kJ/kg = 333.55 kJ for 1 kg of ice to melt
PLUS
(2) 4.18 J/(g·K) × 20K = 4.18 kJ/(kg·K) × 20K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K
= 417.15 kJ

From these figures it can be seen that one part ice at 0 °C will cool almost exactly 4 parts water from 20 °C to 0 °C.

B) Silicon has a heat of fusion of 50.21 kJ/mol. 50 kW of power can supply the energy required to melt about 100 kg of silicon in one hour, after it is brought to the melting point temperature:

50 kW = 50kJ/s = 180000kJ/h

180000kJ/h * (1 mol Si)/50.21kJ * 28gSi/(mol Si) * 1kgSi/1000gSi = 100.4kg/h

## Solubility prediction

The heat of fusion can also be used to predict solubility for solids in liquids. Provided an ideal solution is obtained the mole fraction $(x_{2})$ of solute at saturation is a function of the heat of fusion, the melting point of the solid $(T_{\mathit {fus}})$ and the temperature (T) of the solution:

$\ln x_{2}=-{\frac {\Delta H_{\mathit {fus}}^{\circ }}{R}}\left({\frac {1}{T}}-{\frac {1}{T_{\mathit {fus}}}}\right)$ Here, R is the gas constant. For example, the solubility of paracetamol in water at 298 K is predicted to be:

$x_{2}=\exp {\left(-{\frac {28100{\mbox{ J mol}}^{-1}}{8.314{\mbox{ J K}}^{-1}{\mbox{ mol}}^{-1}}}\left({\frac {1}{298}}-{\frac {1}{442}}\right)\right)}=0.0248$ This equals to a solubility in grams per liter of:

${\frac {0.0248*{\frac {1000{\mbox{ g}}}{18.0153{\mbox{ mol}}^{-1}}}}{1-0.0248}}*151.17{\mbox{ mol}}^{-1}=213.4$ which is a deviation from the real solubility (240 g/L) of 11%. This error can be reduced when an additional heat capacity parameter is taken into account.

### Proof

At equilibrium the chemical potentials for the pure solvent and pure solid are identical:

$\mu _{solid}^{\circ }=\mu _{solution}^{\circ }\,$ or

$\mu _{solid}^{\circ }=\mu _{liquid}^{\circ }+RT\ln X_{2}\,$ with $R\,$ the gas constant and $T\,$ the temperature.

Rearranging gives:

$RT\ln X_{2}=-(\mu _{liquid}^{\circ }-\mu _{solid}^{\circ })\,$ and since

$\Delta G_{\mathit {fus}}^{\circ }=\mu _{liquid}^{\circ }-\mu _{solid}^{\circ }\,$ the heat of fusion being the difference in chemical potential between the pure liquid and the pure solid, it follows that

$RT\ln X_{2}=-(\Delta G_{\mathit {fus}}^{\circ })\,$ Application of the Gibbs-Helmholtz equation:

$\left({\frac {\partial ({\frac {\Delta G_{\mathit {fus}}^{\circ }}{T}})}{\partial T}}\right)_{p\,}=-{\frac {\Delta H_{\mathit {fus}}^{\circ }}{T^{2}}}$ ultimately gives:

$\left({\frac {\partial (\ln X_{2})}{\partial T}}\right)={\frac {\Delta H_{\mathit {fus}}^{\circ }}{RT^{2}}}$ or:

$\partial \ln X_{2}={\frac {\Delta H_{\mathit {fus}}^{\circ }}{RT^{2}}}*\delta T$ and with integration:

$\int _{X_{2}=1}^{X_{2}=x_{2}}\delta \ln X_{2}=\ln x_{2}=\int _{T_{\mathit {fus}}}^{T}{\frac {\Delta H_{\mathit {fus}}^{\circ }}{RT^{2}}}*\Delta T$ the end result is obtained:

$\ln x_{2}=-{\frac {\Delta H_{\mathit {fus}}^{\circ }}{R}}\left({\frac {1}{T}}-{\frac {1}{T_{\mathit {fus}}}}\right)$ 