• OBJECTIVES:
– Classify, by type, the heat changes that
occur during melting, freezing, boiling, and
condensing.
– Calculate heat changes that occur during
melting, freezing, boiling, and condensing.
1
Heats of Fusion and
Solidification
• Molar Heat of Fusion (Hfus) - the
heat absorbed by one mole of a
substance in melting from a solid to a
liquid
• Molar Heat of Solidification (Hsolid) heat lost when one mole of liquid
solidifies
2
Heats of Fusion and
Solidification
• Heat absorbed by a melting solid is
equal to heat lost when a liquid
solidifies
– Thus, Hfus = -Hsolid
3
Heats of Vaporization and
Condensation
• When liquids absorb heat at their
boiling points, they become vapors.
• Molar Heat of Vaporization (Hvap) the amount of heat necessary to
vaporize one mole of a given liquid.
4
Heats of Vaporization and
Condensation
• Condensation is the opposite of
vaporization.
• Molar Heat of Condensation (Hcond)
- amount of heat released when one
mole of vapor condenses
• Hvap = - Hcond
5
Heats of Vaporization and
Condensation
• H20(g)  H20(l)
Hcond = - 40.6kJ/mol
1) Vaporization (phase change from liquid to
gas) of the substance or
2) Fusion (phase change from solid to liquid)
of the substance.
6
Molar Enthalpy's of Vaporization and
Fusion (under standard conditions)
Substance
Formula
H vap
H fus
(kJ / mol ) (kJ/ mol )
23.3
5.66
Ammonia
NH3
Ethanol
C2H5OH
38.6
4.94
Methanol
CH3OH
35.2
3.22
Water
H2O
40.6
6.01
Calculate the total quantity of heat evolved when 10.0g of steam at
200 C is condensed, cooled , and frozen to ice at -50 C.
The specific heat capacity of ice and steam are 2.06J/gC and 1.87
J/gC respectively.
2
3
4
5
1
1. Cooling
steam
. 1. Cooling steam
Q = mcT
Q = (10.0g)(1.87J/gC)(100C)
Q = 1870 J = 1.870 kJ
1
2 n(H2O) = mass / molar mass
= 10.0g / 18g mol-1
= 0.56 mol H2O
Heat of vaporization
Hvap (H2O) = 40.6 kJ/mol
Hvap (H2O) = 0.56 mol  40.6 kJ/mol = 22.7 kJ
2
condensation
3
Q = mcT
Q = (10.0g)(4.18J/gC)(100C)
Q = 4180J = 4.180kJ
3
Cooling of water
4 n(H2O) = mass / molar mass
= 10.0g / 18g mol-1
= 0.56 mol H2O
Heat of fusion
 H fus (H2O) = 6.10 kJ/mol
 H fus (H2O) = 0.56 mol  6.01 kJ/mol = 3.37 kJ
4
Water freezing
5 Q = mcT
Q = (10.0g)(2.06J/gC)(50C)
Q = 1030J = 1.030kJ
Cooling ice
5
Calculate the total quantity of heat evolved when 10.0g of steam at
200 C is condensed, cooled , and frozen to ice at -50 C.
Total heat = 1 + 2 + 3 + 4 + 5
Total heat = 1.870 kJ + 22.7 kJ + 4.180 kJ + 3.37 kJ + 1.030 kJ
= 33.15 kJ
Question:
• How much energy is needed to raise the
temp of 25 grams of ice from -25°C to
105°C ?
• Answer: 90.4 kJ
• Solution:
• From ice to water
Q = mcT
Q = (25g)(2.06J/gC)(25C)
Q = 12,875 J = 12.9 kJ
Solid ice phase
n(H2O) = mass / molar mass
= 25 g / 18g mol-1
= 1.39 mol H2O
H fus (H2O) = 6.01 kJ/mol
H fus (H2O) = 1.39 mol  6.01 kJ/mol = 8.35 kJ
0 to 100
Q = mcT
Q = (25g)(4.184J/gC)(100C)
Q = 10,460 J = 10.4 kJ
Liquid to steam
n(H2O) = mass / molar mass
= 25g / 18g mol-1
= 1.39 mol H2O
H fus (H2O) = 40.6kJ/mol
H fus (H2O) = 1.39 mol  40.6 kJ/mol = 56.4 kJ
Above 100
Q = mcT
Q = (25g)(1.87J/gC)(5C)
Q = 2,337.5 J = 2.34 kJ
Ans: 12.9 kJ + 8.35 kJ + 10.4 kJ + 56.4 kJ + 2.34 kJ
= 90.4 kJ
Questions:
• 1. Find the amount of heat released
when 3.0 kg of water goes from 55ºC to
-35ºC.
• 2. Find the amount of heat absorbed
when 400. g ice goes from -20.0ºC to
10.0ºC.
• 3. Find the amount of heat absorbed
when 75.0 g ice goes from -40.ºC to
110ºC.