Heat of Fusion and Vaporization Notes

Heat of Fusion and Vaporization

Intro to Heat of Fusion vs. Heat of Vaporization

• Substances can transition between solid, liquid, and gas states.
• Water exists as a liquid at room temperature, but can become ice (solid) or water vapor (gas).
• The amount of heat energy required for these transitions varies for different substances due to their specific heat capacity, melting points (MP), and boiling points (BP).

What is Heat of Fusion vs. Heat of Vaporization?

• Heat of Fusion: The energy needed to melt or freeze a substance at constant pressure.
  • Water's heat of fusion ($L_f$) is 334 J/g.
  • It requires 334J to melt 1g of ice and 334J to freeze 1g of water.
• Heat of Vaporization: The energy needed to evaporate or condense a substance at constant pressure.
  • Water's heat of vaporization ($L_v$) is 2257 J/g.
  • It requires 2257J to boil 1g of liquid water and 2257J to condense 1g of water vapor.

Time-Temperature Graph

• Shows phase changes (melting, freezing, evaporation, condensation) in relation to energy and temperature.
• Temperature remains constant during phase transitions.
• Heat of fusion for water: 6.01 kJ/mol
• Heat of vaporization for water: 40.65 kJ/mol

Quick Note: Temperature

• Temperature does not change during phase transitions (solid to liquid, liquid to gas).
• Energy is used to loosen bonds between molecules and overcome intermolecular forces.
• This energy is known as Latent Heat.

Molar Heat of Fusion and Molar Heat of Vaporization

• Molar Heat of Fusion: Heat absorbed by one mole of a substance during solid-to-liquid transition (or released during liquid-to-solid transition).
  • Water's molar heat of fusion is 6.02 kJ/mol.
• Molar Heat of Vaporization: Heat absorbed by one mole of a substance during liquid-to-gas transition (or released during gas-to-liquid transition).
  • Water's molar heat of vaporization is 40.67 kJ/mol.

How to Determine Heat of Fusion and Heat of Vaporization

• Adapt the specific heat capacity formula $Q = mC<em>p \Delta T$ to determine $\Delta H</em>{fus}$ and $\Delta H_{vap}$.
• Heat of Fusion: $L<em>f = \frac{\Delta H</em>{fus}}{Molar \, Mass}$
• Heat of Vaporization: $L<em>v = \frac{\Delta H</em>{vap}}{Molar \, Mass}$
• Molar Heat of Fusion: $\Delta H<em>{fus} = Molar \, Mass \times L</em>f$
• Molar Heat of Vaporization: $\Delta H<em>{vap} = Molar \, Mass \times L</em>v$

Consider the Following…

• The molar heat of fusion for Ethanol ($C<em>2H</em>5OH$) is approximately 5.02 kJ/mol.
• Determine the heat of fusion of ethanol in J/g.

Consider The Following #1 Answer:

• 109 J/g
• To determine the heat of fusion for ethanol, use the provided molar heat of fusion and molar mass.
• Molar heat of fusion for ethanol: 5.02 kJ/mol.
• Convert kJ to J: $5.02 \frac{kJ}{mol} = 5020 \frac{J}{mol}$
• Heat of fusion is equal to molar heat of fusion divided by molar mass.
• Molar mass of ethanol: 24.02 + 6.06 + 16.00 = 46.08 g/mol.
• $\frac{5020 \frac{J}{mol}}{46.08 \frac{g}{mol}} \approx 109 \frac{J}{g}$
• Melting/freezing one gram of ethanol requires the absorption/release of 109J of energy.

Triple Point and Critical Point

• Triple Point: The specific temperature and pressure at which a substance can exist in all three states of matter (solid, liquid, gas) in equilibrium.
• Critical Point: The specific temperature and pressure at which the distinction between liquid and gas phases disappears, resulting in equal densities.

Example Problem #1

• A 5.1g sample of solid gallium is at room temperature.
• Gallium's melting point is 30°C.
• Molar heat of fusion: 5.6 kJ/mol
• Specific heat of gallium: 0.37 J/g°C
• How much energy in joules is needed to melt the sample of gallium?

Example Problem #1 - Explained

• $Q = (5.1g) \times (0.37 \frac{J}{g°C}) \times (7°C) = 13.21 J$
  • Energy needed to heat the gallium to 30°C (melting point).
• $\frac{5600 \frac{J}{mol}}{69.72 \frac{g}{mol}} = 80.3 \frac{J}{g}$
  • Heat of fusion of gallium.
• $Q = (5.1g) \times (80.3 \frac{J}{g}) = 409.53 J$
  • Energy needed to melt the gallium.
• 13.21 J + 409.53 J = 422.74 J

Example Problem #2

• Consider a 11.7g sample of liquid water at room temperature.
• How much energy in kilojoules is needed to heat the sample of water from 23°C to 105°C?

Example Problem #2 - Explained

• $Q = (11.7g) \times (4.18 \frac{J}{g°C}) \times (77°C) = 3765.76 J$
  • Energy needed to heat the water to 100°C (boiling point).
• $Q = (11.7g) \times (2257 \frac{J}{g}) = 26406.9 J$
  • Energy needed to boil the water.
• $Q = (11.7g) \times (1.87 \frac{J}{g°C}) \times (5°C) = 109.40 J$
  • Energy needed to heat the water vapor to 105°C.
• 3765.76 J + 26406.9 J + 109.40 J = 30282.6 J = 30.28 kJ

Consider the following…

• If you have a 10g sample of ice at -3°C, what is the heat energy needed in kilojoules to get to the sample of water to room temperature (23°C)?

Consider The Following #2 Answer:

• 4.363 kJ
• This problem involves specific heat capacity and phase changes (Heat of Fusion equation).
• Since water will not reach its boiling point, we don't need to consider Heat of Vaporization for this problem.
• Given: a 10g sample of ice at -3°C.
• The ice absorbs energy to reach room temperature. The ice needs to completely melt first.
• $Q = mC_p\Delta T$
• $Q = (10g) \times (2.06 \frac{J}{g°C}) \times (3°C) = 61.8 J$
  • Energy needed to reach melting point.
• $Q = mL_f$
• $Q = (10g) \times (334 \frac{J}{g}) = 3340 J$
  • Energy needed to completely melt the sample.
• $Q = mC_p\Delta T$
• $Q = (10g) \times (4.18 \frac{J}{g°C}) \times (23°C) = 961.4 J$
  • Energy needed to reach 23°C after melting.
• 61.8 J + 3340 J + 961.4 J = 4363.2 J = 4.363 kJ