Heating and cooling curve for water heated at a constant rates.
A-B = Solid ice, temperature is increasing.
Particles gain kinetic energy, vibration of particles increases.
Ice
B-C = Solid starts to change state from solid to liquid.
Temperature remains constant as energy is used to break intermolecular bonds.
H2O (s)  H2O () energy required  6 kJ/mol
0ºC
C-D = temperature starts to rise once all the solid has melted.
Particles gain kinetic energy.
Liquid
water
D-E = Liquid starts to vaporize, turning from liquid to gas. The
temperature remains constant as energy is used to break intermolecular forces.
H2O ()  H2O (g)
energy required  41 kJ/mol
100ºC
E-F = temperature starts to rise once all liquid is vaporized. Gas
particles gain kinetic energy.
steam
Heating and Cooling Curve of Water
vaporization
condensation
freezing
Endothermic and Exothermic
Processes
Sublimation and Deposition
• Sublimation is when a solid phase goes
directly into a gaseous phase without first
becoming a liquid.
• Deposition is when a gaseous phase goes
directly to the solid phase without first
becoming a liquid.
Sublimation and Deposition
Phase Changes
Energy Changes Accompanying Phase
Changes
–
–
–
–
–
–
Sublimation: Hsub > 0 (endothermic).
Vaporization: Hvap > 0 (endothermic).
Melting or Fusion: Hfus > 0 (endothermic).
Deposition: Hdep < 0 (exothermic).
Condensation: Hcon < 0 (exothermic).
Freezing: Hfre < 0 (exothermic).
• Generally heat of fusion (enthalpy of fusion) is
less than heat of vaporization:
– it takes more energy to completely separate molecules,
Ice molecules are locked in
fixed positions, held by
intermolecular-bonds.
Ice is less dense than liquid
water because the molecules are
further apart than in liquid
water.
Energy Requirements for changing state:
In ice the water molecules are held together by strong intermolecular
forces.
The energy required to melt 1 mole of a substance is called the
molar heat of fusion
(fus H) For ice it is 6.02 kJmol-1
The energy required to change 1 mole of a liquid to its vapor is
called the molar heat of vaporization
(vap H) For water it is 40.6 kJmol-1
H (delta H) is the change in energy or heat content.
It takes more energy to vaporize water than to melt it.
This is because in melting you weaken the intermolecular forces.
Here about 1/6 of the hydrogen bonds are broken.
In vaporization you totally break them.
All the hydrogen bonds are broken
vap H is always greater than fus H.
Fusion is when a solid melts to form a liquid
Vaporization is when a liquid evaporates to form a gas.
Calculating Energy Changes: Solid to liquid
How much energy is required to melt 8.5 g of ice at 0C?
The molar heat of fusion for ice is 6.02 kJmol-1
Step 1: How many moles of ice do we have?
n = m/M
n = 8.5g / 18gmol-1
= 0.47 mol H2O
Step 2: Use the equivalence statement to work the energy (6.02 kJ is
required for 1 mol H2O)
kJ = 0.47 mol H2O  6.02 kJ / mol H2O
= 2.8kJ
What is specific heat capacity?
The amount of energy required to change the temperature of one
gram of a substance by 1C .
10 C
11 C
Another name for specific heat is a calorie
(1 calorie = 4.184 Joules)
Specific heat capacity of liquid water (H2O (L) ) is 4.18 J g-1C–1.
Water (s) = 2.03 J g-1 C –1
Water (g) = 2.0 J g-1 C –1
 0.5 cal/g to break up ice
Calculating the energy to increase the temperature of liquid water.
Calculating specific heat using the equation:
Q = ms (tf  ti)
or
Q = energy (heat) required
Q = ms T
or
s = specific heat capacity
Heat (H) = ms (tf  ti)
m = mass of the sample
T = change in temperature in C
EXAMPLE:
How much energy does it take to heat 10g of water from 50 to 100 C ?
Specific heat capacity of water = 4.184 J g-1C–1
Q = m  s  T
Q=
(10g)  (4.184 J g-1 C -1)  (50 C) = 2.1  10 3 J
Problem
How much energy is required to heat 25 g of liquid water from 25C
to 100C and change it to steam?
Step 1: Calculate the energy needed to heat the water from 25C to
100C
Q = m  s  T
Q = 25g

4.184 J g-1 C -1  75 C = 7.8  10 3 J
Step 2: Vaporization: Use the vap H to calculate the energy
required to vaporize 25g of water at 100C
Molar heat of vaporization of water is 40.6 kJmolThe vap H is per mole, not per gram. So first you have to convert
the grams into moles.
.25g  1mol H2O / 18g mol-1 H2O = 1.4 mol H2O
vap H (H2O) = 1.4 mol H2O  40.6kJ/mol = 57 kJ
n(H2O) = mass / molar mass
= 25g / 18g mol-1
= 1.4 mol H2O
vap H (H2O) = 40.6kJ/mol
vap H (H2O) = 1.4 mol  40.6kJ/mol = 57 kJ
Total energy change is:
7.8kJ + 57kJ = 65kJ