This worked example problem demonstrates how to calculate the energy required to raise the temperature of a sample that includes changes in phase. This problem finds the energy required to turn cold ice into hot steam. What is the heat in Joules required to convert 25 grams of -10 °C ice into 150 °C steam?Useful information:heat of fusion of water = 334 J/gheat of vaporization of water = 2257 J/gspecific heat of ice = 2.09 J/g·°Cspecific heat of water = 4.18 J/g·°Cspecific heat of steam = 2.09 J/g·°C The total energy required is the sum of the energy to heat the -10 °C ice to 0 °C ice, melting the 0 °C ice into 0 °C water, heating the water to 100 °C, converting 100 °C water to 100 °C steam and heating the steam to 150 °C. To get the final value, first calculate the individual energy values and then add them up. Step 1: Find the heat required to raise the temperature of ice from -10 °C to 0 °C. Use the formula: q = mcΔT where
In this problem:
Plug in the values and solve for q: q = (25 g)x(2.09 J/g·°C)[(0 °C - -10 °C)]q = (25 g)x(2.09 J/g·°C)x(10 °C) q = 522.5 J
Find the heat required to convert 0 °C ice to 0 °C water.
q = m·ΔHf where
For this problem:
Plugging in the values gives the value for q:
q = (25 g)x(334 J/g) The heat required to convert 0 °C ice to 0 °C water = 8350 J
Find the heat required to raise the temperature of 0 °C water to 100 °C water.q = mcΔTq = (25 g)x(4.18 J/g·°C)[(100 °C - 0 °C)]q = (25 g)x(4.18 J/g·°C)x(100 °C)q = 10450 JThe heat required to raise the temperature of 0 °C water to 100 °C water = 10450 J Step 4:
Find the heat required to convert 100 °C water to 100 °C steam. q = heat energy m = massΔHv = heat of vaporization q = (25 g)x(2257 J/g)q = 56425 JThe heat required to convert 100 °C water to 100 °C steam = 56425 Step 5: Find the heat required to convert 100 °C steam to 150 °C steamq = mcΔTq = (25 g)x(2.09 J/g·°C)[(150 °C - 100 °C)]q = (25 g)x(2.09 J/g·°C)x(50 °C)q = 2612.5 J The heat required to convert 100 °C steam to 150 °C steam = 2612.5 Step 6: Find total heat energy. In this final step, put together all of the answers from the previous calculations to cover the entire temperature range.
Answer: The heat required to convert 25 grams of -10 °C ice into 150 °C steam is 78360 J or 78.36 kJ.
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. 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. 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] Standard enthalpy change of fusion of period three Standard enthalpy change of fusion of period two of the periodic table of elements
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.
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 ) {\displaystyle (x_{2})} of solute at saturation is a function of the heat of fusion, the melting point of the solid ( T fus ) {\displaystyle (T_{\text{fus}})} and the temperature ( T ) {\displaystyle (T)} of the solution: ln x 2 = − Δ H fus ∘ R ( 1 T − 1 T fus ) {\displaystyle \ln x_{2}=-{\frac {\Delta H_{\text{fus}}^{\circ }}{R}}\left({\frac {1}{T}}-{\frac {1}{T_{\text{fus}}}}\right)}Here, R {\displaystyle R} is the gas constant. For example, the solubility of paracetamol in water at 298 K is predicted to be: x 2 = exp [ − 28100 J mol − 1 8.314 J K − 1 mol − 1 ( 1 298 K − 1 442 K ) ] = 0.0248 {\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: 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] ProofAt equilibrium the chemical potentials for the pure solvent and pure solid are identical: μ solid ∘ = μ solution ∘ {\displaystyle \mu _{\text{solid}}^{\circ }=\mu _{\text{solution}}^{\circ }\,}or μ solid ∘ = μ liquid ∘ + R T ln X 2 {\displaystyle \mu _{\text{solid}}^{\circ }=\mu _{\text{liquid}}^{\circ }+RT\ln X_{2}\,}with R {\displaystyle R\,} the gas constant and T {\displaystyle T\,} the temperature. Rearranging gives: R T ln X 2 = − ( μ liquid ∘ − μ solid ∘ ) {\displaystyle RT\ln X_{2}=-\left(\mu _{\text{liquid}}^{\circ }-\mu _{\text{solid}}^{\circ }\right)\,}and since Δ G fus ∘ = μ liquid ∘ − μ solid ∘ {\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 R T ln X 2 = − ( Δ G fus ∘ ) {\displaystyle RT\ln X_{2}=-\left(\Delta G_{\text{fus}}^{\circ }\right)\,}Application of the Gibbs–Helmholtz equation: ( ∂ ( Δ G fus ∘ T ) ∂ T ) p = − Δ H fus ∘ T 2 {\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: ( ∂ ( ln X 2 ) ∂ T ) = Δ H fus ∘ R T 2 {\displaystyle \left({\frac {\partial \left(\ln X_{2}\right)}{\partial T}}\right)={\frac {\Delta H_{\text{fus}}^{\circ }}{RT^{2}}}}or: ∂ ln X 2 = Δ H fus ∘ R T 2 × δ T {\displaystyle \partial \ln X_{2}={\frac {\Delta H_{\text{fus}}^{\circ }}{RT^{2}}}\times \delta T}and with integration: ∫ X 2 = 1 X 2 = x 2 δ ln X 2 = ln x 2 = ∫ T fus T Δ H fus ∘ R T 2 × Δ T {\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 end result is obtained: ln x 2 = − Δ H fus ∘ R ( 1 T − 1 T fus ) {\displaystyle \ln x_{2}=-{\frac {\Delta H_{\text{fus}}^{\circ }}{R}}\left({\frac {1}{T}}-{\frac {1}{T_{\text{fus}}}}\right)}
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What is the energy required to melt 1kg of ice at?
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