Enthalpy of fusion

The 'enthalpy' of fusion is a latent heat, because, while 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 substanceinmoles.

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.^[1] 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.^[2] In the case of ⁴He, this pressure range is between 24.992 and 25.00 atm (2,533 kPa).^[3]

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 (cal_th) = 4.184 joules rather than the International Steam Table calorie (cal_INT) = 4.1868 joules.

• To heat 1 kg of liquid water from 0 °C to 20 °C requires 83.6 kJ (see below). However, heating 0 °C ice to 20 °C requires additional energy to melt the ice. We can treat these two processes independently and using the specific heat capacity of water to be 4.18 J/(g⋅K); 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) × 20 K = 4.18 kJ/(kg⋅K) × 20 K = 83.6 kJ for 1 kg of water to increase in temperature by 20 K

  (1 + 2) 333.55 kJ + 83.6 kJ = 417.15 kJ for 1 kg of ice to increase in temperature by 20 K

  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.
• 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:

  50 kW = 50kJ/s = 180000kJ/h

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

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

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

Here, ${\displaystyle R}$  is the gas constant. For example, the solubility of paracetamol in water at 298 K is predicted to be:

${\displaystyle x_{2}=\exp {\left[-{\frac {28100~{\text{J mol}}^{-1}}{8.314~{\text{J K}}^{-1}~{\text{mol}}^{-1}}}\left({\frac {1}{298~{\text{K}}}}-{\frac {1}{442~{\text{K}}}}\right)\right]}=0.0248}$ 

Since the molar mass of water and paracetamol are 18.0153gmol⁻¹ and 151.17gmol⁻¹ and the density of the solution is 1000gL⁻¹, an estimate of the solubility in grams per liter is:

${\displaystyle {\frac {0.0248\times {\frac {1000~{\text{g L}}^{-1}}{18.0153~{\text{g mol}}^{-1}}}}{1-0.0248}}\times 151.17~{\text{g mol}}^{-1}=213.4~{\text{g L}}^{-1}}$ 

1000 g/L * (mol/18.0153g) is an estimate of the number of moles of molecules in 1L solution, using water density as a reference;

0.0248 * (1000 g/L * (mol/18.0153g)) is the molar fraction of substance in saturated solution with a unit of mol/L;

0.0248 * (1000 g/L * (mol/18.0153g)) * 151.17g/mol is the solute's molar fraction equivalent mass conversion;

1-0.0248 will be the fraction of the solution that is solvent.

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.^[5]

Atequilibrium the chemical potentials for the solute in the solution and pure solid are identical:

${\displaystyle \mu _{\text{solid}}^{\circ }=\mu _{\text{solute}}^{\circ }\,}$ 

or

${\displaystyle \mu _{\text{solid}}^{\circ }=\mu _{\text{liquid}}^{\circ }+RT\ln X_{2}\,}$ 

with ${\displaystyle R\,}$  the gas constant and ${\displaystyle T\,}$  the temperature.

Rearranging gives:

${\displaystyle RT\ln X_{2}=-\left(\mu _{\text{liquid}}^{\circ }-\mu _{\text{solid}}^{\circ }\right)\,}$ 

and since

${\displaystyle \Delta G_{\text{fus}}^{\circ }=\mu _{\text{liquid}}^{\circ }-\mu _{\text{solid}}^{\circ }\,}$ 

the heat of fusion being the difference in chemical potential between the pure liquid and the pure solid, it follows that

${\displaystyle RT\ln X_{2}=-\left(\Delta G_{\text{fus}}^{\circ }\right)\,}$ 

Application of the Gibbs–Helmholtz equation:

${\displaystyle \left({\frac {\partial \left({\frac {\Delta G_{\text{fus}}^{\circ }}{T}}\right)}{\partial T}}\right)_{p\,}=-{\frac {\Delta H_{\text{fus}}^{\circ }}{T^{2}}}}$ 

ultimately gives:

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

or:

${\displaystyle \partial \ln X_{2}={\frac {\Delta H_{\text{fus}}^{\circ }}{RT^{2}}}\times \delta T}$ 

and with integration:

${\displaystyle \int _{X_{2}=1}^{X_{2}=x_{2}}\delta \ln X_{2}=\ln x_{2}=\int _{T_{\text{fus}}}^{T}{\frac {\Delta H_{\text{fus}}^{\circ }}{RT^{2}}}\times \Delta T}$ 

the result is obtained:

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

• Enthalpy of vaporization
• Heat capacity
• Thermodynamic databases for pure substances
• Joback method (Estimation of the heat of fusion from molecular structure)
• Latent heat
• Lattice energy
• Heat of dilution

• Atkins, Peter; Jones, Loretta (2008), Chemical Principles: The Quest for Insight (4th ed.), W. H. Freeman and Company, p. 236, ISBN 978-0-7167-7355-9
• Ott, BJ. Bevan; Boerio-Goates, Juliana (2000), Chemical Thermodynamics: Advanced Applications, Academic Press, ISBN 0-12-530985-6