Enthalpy of fusion: Difference between revisions
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{{Short description|Enthalpy change when substance melts}} |
{{Short description|Enthalpy change when a substance melts}} |
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[[File:Enthalpies of melting and boiling for pure elements versus temperatures of transition.svg|right|thumb| |
[[File:Enthalpies of melting and boiling for pure elements versus temperatures of transition.svg|right|thumb|upright=1.35|alt=A log-log plot of the enthalpies of melting and boiling versus the melting and boiling temperatures for the pure elements. The linear relationship between the enthalpy of melting the temperature is known as ''Richard's rule''.|Enthalpies of melting and boiling for pure elements versus temperatures of transition, demonstrating [[Trouton's rule]]]] |
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⚫ | The |
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In [[thermodynamics]], the '''enthalpy of fusion''' of a [[Chemical substance|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 of matter|state]] from a [[solid]] to a [[liquid]], at [[Isobaric process|constant pressure]]. |
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⚫ | The enthalpy of fusion is the amount of energy required to convert one mole of solid into liquid. For example, when [[melting]] 1 kg of ice (at 0 °C under a [[c:File:Phase diagram of water.svg|wide range of pressures]]), 333.55 kJ of energy is absorbed with no [[temperature]] change. The '''heat of solidification''' (when a substance [[Freezing|changes from liquid to solid]]) is equal and opposite. |
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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 {{convert|1|atm|kPa|abbr=on|sigfig=6}} unless otherwise specified. |
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 {{convert|1|atm|kPa|abbr=on|sigfig=6}} unless otherwise specified. |
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| 163.93 |
| 163.93 |
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| [[sodium acetate]] |
| [[sodium acetate]]/H2O |
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| {{convert|264–289|J|cal|abbr=on|disp=number}} |
| {{convert|264–289|J|cal|abbr=on|disp=number}} |
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| 264–289<ref name=IDMA> |
| 264–289<ref name=IDMA>[{{GBurl|EsfcWE5lX40C|p=155|q=latent heat of fusion sodium acetate}} Page 155] in: {{cite book |doi=10.1002/9781119713173.ch3 |chapter=Thermal Energy Storage Methods |title=Thermal Energy Storage |date=2021 |pages=125–260 |isbn=978-1-119-71315-9 }}</ref> |
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⚫ | |||
| [[sodium sulfate]]/H2O |
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| {{convert|254|J|cal|abbr=on|disp=number}} |
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| 254<ref>{{cite journal |last1=Tao |first1=Wen |last2=Kong |first2=Xiangfa |last3=Bao |first3=Anyang |last4=Fan |first4=Chuangang |last5=Zhang |first5=Yi |title=Preparation and Phase Change Performance of Graphene Oxide and Silica Composite {{chem|Na|2|SO|4}}·10{{chem|H|2|O}} Phase Change Materials (PCMs) as Thermal Energy Storage Materials |journal=Materials |date=17 November 2020 |volume=13 |issue=22 |pages=5186 |doi=10.3390/ma13225186 |doi-access=free |pmid=33212870 |pmc=7698442 }}</ref> |
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|- |
|- |
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| [[stearic acid]] |
| [[stearic acid]] |
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| [[paraffin wax]] (C<sub>25</sub>H<sub>52</sub>) |
| [[paraffin wax]] (C<sub>25</sub>H<sub>52</sub>) |
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| 47. |
| 47.8–52.6 |
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| 200–220 |
| 200–220 |
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⚫ | |||
|} |
|} |
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These values are mostly from the [[CRC Press|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<sub>th</sub>) = 4.184 joules rather than the International Steam Table calorie (cal<sub>INT</sub>) = 4.1868 joules. |
These values are mostly from the [[CRC Press|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<sub>th</sub>) = 4.184 joules rather than the International Steam Table calorie (cal<sub>INT</sub>) = 4.1868 joules. |
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==Examples== |
==Examples== |
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{{bulleted list |
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|1= 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: |
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:(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 |
:(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 |
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:(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 |
:(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 |
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:(1+2) 333.55 kJ + 83.6 kJ = 417.15 kJ for 1 kg of ice 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 |
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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. |
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. |
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⚫ | |||
⚫ | |||
: 50 kW = {{gaps|50|kJ/s}} = {{gaps|180|000|kJ/h}} |
: 50 kW = {{gaps|50|kJ/s}} = {{gaps|180|000|kJ/h}} |
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⚫ | |||
}} |
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⚫ | |||
==Solubility prediction== |
==Solubility prediction== |
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The heat of fusion can also be used to predict [[solubility]] for solids in liquids. Provided an [[ideal solution]] is obtained the [[mole fraction]] <math>(x_2)</math> of solute at saturation is a function of the heat of fusion, the [[melting point]] of the solid <math>(T_\text{fus})</math> and the [[temperature]] <math>(T)</math> of the solution: |
The heat of fusion can also be used to predict [[solubility]] for solids in liquids. Provided an [[ideal solution]] is obtained the [[mole fraction]] <math>(x_2)</math> of solute at saturation is a function of the heat of fusion, the [[melting point]] of the solid <math>(T_\text{fus})</math> and the [[temperature]] <math>(T)</math> of the solution: |
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:<math> |
:<math>\ln x_2 = - \frac {\Delta H^\circ_\text{fus}}{R} \left(\frac{1}{T}- \frac{1}{T_\text{fus}}\right)</math> |
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Here, <math>R</math> is the [[gas constant]]. For example, the solubility of [[paracetamol]] in water at 298 [[Kelvin (unit)|K]] is predicted to be: |
Here, <math>R</math> is the [[gas constant]]. For example, the solubility of [[paracetamol]] in water at 298 [[Kelvin (unit)|K]] is predicted to be: |
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:<math> |
:<math>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</math> |
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Since the molar mass of water and paracetamol are {{gaps|18.0153|g|mol<sup>−1</sup>}} and {{gaps|151.17|g|mol<sup>−1</sup>}} and the density of the solution is {{gaps|1000|g|L<sup>−1</sup>}}, an estimate of the solubility in grams per liter is: |
Since the molar mass of water and paracetamol are {{gaps|18.0153|g|mol<sup>−1</sup>}} and {{gaps|151.17|g|mol<sup>−1</sup>}} and the density of the solution is {{gaps|1000|g|L<sup>−1</sup>}}, an estimate of the solubility in grams per liter is: |
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:<math> |
:<math>\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}</math> |
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: |
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: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; |
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:0.0248 * (1000 g/L * (mol/18.0153g)) is the molar fraction of substance in saturated solution with a unit of mol/L; |
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:0.0248 * (1000 g/L * (mol/18.0153g)) * 151.17g/mol is the solute's molar fraction equivalent mass conversion; |
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:1-0.0248 will be the fraction of the solution that is solvent. |
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which is a deviation from the real solubility (240 |
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.<ref>{{cite journal |last1=Hojjati |first1=H. |last2=Rohani |first2=S. |title=Measurement and Prediction of Solubility of Paracetamol in Water−Isopropanol Solution. Part 2. Prediction |journal=Organic Process Research & Development |date=November 2006 |volume=10 |issue=6 |pages=1110–1118 |doi=10.1021/op060074g }}</ref> |
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===Proof=== |
===Proof=== |
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At [[Chemical equilibrium|equilibrium]] the [[chemical potential]]s for the |
At [[Chemical equilibrium|equilibrium]] the [[chemical potential]]s for the solute in the solution and pure solid are identical: |
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:<math>\mu^\circ_\text{solid} = \mu^\circ_\text{ |
:<math>\mu^\circ_\text{solid} = \mu^\circ_\text{solute}\,</math> |
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oder |
oder |
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Rearranging gives: |
Rearranging gives: |
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:<math>RT\ln X_2 = - |
:<math>RT\ln X_2 = -\left(\mu^\circ_\text{liquid} - \mu^\circ_\text{solid}\right)\,</math> |
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and since |
and since |
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the heat of fusion being the difference in chemical potential between the pure liquid and the pure solid, it follows that |
the heat of fusion being the difference in chemical potential between the pure liquid and the pure solid, it follows that |
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:<math>RT\ln X_2 = - |
:<math>RT\ln X_2 = -\left(\Delta G^\circ_\text{fus}\right)\,</math> |
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Application of the [[Gibbs–Helmholtz equation]]: |
Application of the [[Gibbs–Helmholtz equation]]: |
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:<math>\left( \frac{\partial ( \frac{\Delta G^\circ_\text{fus} } {T} ) } {\partial T} \right)_{p\,} = - |
:<math>\left( \frac{\partial \left( \frac{\Delta G^\circ_\text{fus} } {T} \right) } {\partial T} \right)_{p\,} = -\frac {\Delta H^\circ_\text{fus}} {T^2} </math> |
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ultimately gives: |
ultimately gives: |
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:<math>\left( \frac{\partial ( \ln X_2 ) } {\partial T} \right) |
:<math>\left( \frac{\partial \left( \ln X_2 \right) } {\partial T} \right) = \frac {\Delta H^\circ_\text{fus}} {RT^2}</math> |
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or: |
or: |
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:<math> |
:<math>\partial \ln X_2 = \frac {\Delta H^\circ_\text{fus}} {RT^2} \times \delta T</math> |
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and with [[ |
and with [[integral|integration]]: |
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:<math> |
:<math>\int^{X_2=x_2}_{X_2 = 1} \delta \ln X_2 = \ln x_2 = \int_{T_\text{fus}}^T \frac {\Delta H^\circ_\text{fus}} {RT^2} \times \Delta T</math> |
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the |
the result is obtained: |
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:<math> |
:<math>\ln x_2 = - \frac {\Delta H^\circ_\text{fus}} {R}\left(\frac{1}{T}- \frac{1}{T_\text{fus}}\right)</math> |
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==See also== |
==See also== |
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*[[ |
* [[Enthalpy of vaporization]] |
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*[[Heat capacity]] |
* [[Heat capacity]] |
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*[[Thermodynamic databases for pure substances]] |
* [[Thermodynamic databases for pure substances]] |
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*[[Joback method]] (Estimation of the heat of fusion from molecular structure) |
* [[Joback method]] (Estimation of the heat of fusion from molecular structure) |
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*[[Latent heat]] |
* [[Latent heat]] |
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*[[Lattice energy]] |
* [[Lattice energy]] |
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*[[Heat of dilution]] |
* [[Heat of dilution]] |
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==Notes== |
==Notes== |
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{{DEFAULTSORT:Enthalpy Of Fusion}} |
{{DEFAULTSORT:Enthalpy Of Fusion}} |
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[[Category:Thermodynamic properties]] |
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[[Category:Enthalpy]] |
[[Category:Enthalpy]] |
Revision as of 18:24, 2 July 2024
![A log-log plot of the enthalpies of melting and boiling versus the melting and boiling temperatures for the pure elements. The linear relationship between the enthalpy of melting the temperature is known as Richard's rule.](http://upload.wikimedia.org/wikipedia/commons/thumb/8/89/Enthalpies_of_melting_and_boiling_for_pure_elements_versus_temperatures_of_transition.svg/300px-Enthalpies_of_melting_and_boiling_for_pure_elements_versus_temperatures_of_transition.svg.png)
In thermodynamics, 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.
The enthalpy of fusion is the amount of energy required to convert one mole of solid into liquid. 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.
Übersicht
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 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.[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]
![](http://upload.wikimedia.org/wikipedia/commons/thumb/7/70/Molar_heat_of_fusion_period_two.png/150px-Molar_heat_of_fusion_period_two.png)
Substance | Heat of fusion | |
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(cal/g) | (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/H2O | 63–69 | 264–289[4] |
sodium sulfate/H2O | 61 | 254[5] |
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
- 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
- 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
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 of solute at saturation is a function of the heat of fusion, the melting point of the solid and the temperature of the solution:
Here, is the gas constant. For example, the solubility of paracetamol in water at 298 K is predicted to be:
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:
- 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.[6]
Proof
At equilibrium the chemical potentials for the solute in the solution and pure solid are identical:
oder
with the gas constant and the temperature.
Rearranging gives:
and since
the heat of fusion being the difference in chemical potential between the pure liquid and the pure solid, it follows that
Application of the Gibbs–Helmholtz equation:
ultimately gives:
or:
and with integration:
the result is obtained:
See also
- 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
Notes
- ^ Atkins & Jones 2008, p. 236.
- ^ Ott & Boerio-Goates 2000, pp. 92–93.
- ^ Hoffer, J. K.; Gardner, W. R.; Waterfield, C. G.; Phillips, N. E. (April 1976). "Thermodynamic properties of 4He. II. The bcc phase and the P-T and VT phase diagrams below 2 K". Journal of Low Temperature Physics. 23 (1): 63–102. Bibcode:1976JLTP...23...63H. doi:10.1007/BF00117245. S2CID 120473493.
- ^ Page 155 in: "Thermal Energy Storage Methods". Thermal Energy Storage. 2021. pp. 125–260. doi:10.1002/9781119713173.ch3. ISBN 978-1-119-71315-9.
- ^ Tao, Wen; Kong, Xiangfa; Bao, Anyang; Fan, Chuangang; Zhang, Yi (17 November 2020). "Preparation and Phase Change Performance of Graphene Oxide and Silica Composite Na
2SO
4·10H
2O Phase Change Materials (PCMs) as Thermal Energy Storage Materials". Materials. 13 (22): 5186. doi:10.3390/ma13225186. PMC 7698442. PMID 33212870. - ^ Hojjati, H.; Rohani, S. (November 2006). "Measurement and Prediction of Solubility of Paracetamol in Water−Isopropanol Solution. Part 2. Prediction". Organic Process Research & Development. 10 (6): 1110–1118. doi:10.1021/op060074g.
References
- 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