Transcript Phase Changes and their Calculations

Phase Changes
and their Calculations
J. Flint Baumwirt
Granada Hills Charter High School
Symbols used in heat and phase
change calculations:
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 - The Greek letter "delta" represents a “change in.”
H - Represents enthalpy, which we think of as heat.
Q (or q) - Represents heat energy gained or lost.
m - Represents mass.
T - Represents temperature.
Cp - Represents specific heat.
NOTE: Cp will differ depending on state (solid, liquid or gas)
Enthalpy of fusion: Hfus
• As energy is added to a solid at its melting point,
all the energy is used to increase the kinetic
energy of the molecules during the phase
change.
• Because of this, the temperature of the melting
system remains constant until all of the solid has
become liquid. If energy is still being added to
the system, the temperature will begin to climb
when all of the solid becomes liquid.
Enthalpy of vaporization: Hvap
• As energy is added to a liquid at its boiling point,
all the energy is used to increase the kinetic
energy of the molecules during the phase
change.
• Because of this, the temperature of the boiling
system remains constant until all of the liquid
has become vapor. If energy is still being added
to the system, the temperature will begin to
climb when all of the liquid has been
vaporized.
Heat transfer
• The law of conservation of energy tells us that
heat lost by one quantity of matter must be
gained by another.
• As a quantity of matter gains heat energy, its
temperature will increase based on its individual
specific heat.
• Specific heat is defined as: the heat required to
raise the temperature of one gram of a
substance by one Celsius degree.
Overview
• Transitions between solid, liquid, and gaseous
phases typically involve large amounts of energy
compared to the specific heat. If heat were
added at a constant rate to a mass of ice to take
it through its phase changes to liquid water and
then to steam, the energies required to
accomplish the phase changes (called the latent
heat of fusion and latent heat of vaporization )
would lead to plateaus in the temperature vs
time graph.
Latent Heat of Fusion and
Latent Heat of Vaporization:
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A Table of Heats of Fusion and Vaporization are provided below for a
variety of substances:
Substance
Hfus (kJ/mol)
Hvap (kJ/mol)
Water
5.98
40.6
Benzene
9.92
30.7
Chloroform
12.4
31.9
Diethel Ether
6.86
26.0
Ethanol
7.61
38.6
Notice that the latent heats of vaporization are considerably higher in
energy than the latent heats of fusion. This means that the amount of
energy required to overcome the attractive forces between the particles of
a solid to form the particles of a liquid is small compared to the energy
required to pull apart liquid molecules to allow them to form a gas. In other
words, it takes MORE energy to turn a liquid to a gas than to melt a solid to
form a liquid.
• From this information we can now
calculate how much energy it takes to melt
an ice cube. Notice that the values of the
heats of fusion and the heats of
vaporization are in kilojoules per mole.
• Notice also that specific heat is usually
given in the unit of Joules/mol
Calculations
• Example: Calculate the energy, in kilojoules
necessary to melt 1.00 gram of ice.
– This is a phase change calculation
– Where a solid changes to a liquid
– The equation used here is Q = nHfus
– Where Q is the energy in Joules
– n is the number of moles and
– Hfus is the change in energy for this phase
change
This quantity will be provided for you in the
problem or in a table or chart
The latent heat of fusion of water = 5.98 kJ/mol.
1.00 gram H2O
1 mole
5.98 kJ
18.0 grams
1 mole
1.00 x 5.98 = 0.332
18.00
= 0.332 kJ
Calculating Energy Changes: Liquid to Gas
• Example: Calculate the energy required to heat 25.0
grams of liquid water from 25C to 100C and change it
to steam at 100 C.
• The specific heat of water is 4.184 J/gC, and the molar
heat of vaporization of water is 40.6 kJ/mol.
• In calculating the energy required to change a substance
from one state to another in this problem two different
energies must be considered;
– the energy required to change the substance to a different
state AND
– the energy required to raise the temperature to the point of
change.
• Here you must do two problems and then add the energies
together. One is a simple energy equation calculation
Q=mCpT and the other is like our first example multiplying
mass (in the proper units) times the molar heat of
vaporization, Q = n Hvap
Step #1:
Q=mCT
m = 25.0 grams
25.0 grams
C=4.184 J/gC
4.184 J
75 C
T=25C to 100C = 75C
= 7.8 x 103 J = 7.8 kJ
gC
Step #2:
Vaporization
Q = n Hvap
25.0 g H2O
1 mol H2O
40.6 kJ
18.0 grams
1 mol H2O
Step #3:
Add the two energies together (make sure units match):
7.8 kJ + 57 kJ = 65 kJ
= 57 kJ
Sample Heat Calculation Problem:
How much heat is needed to convert 250 g of ice at -30 oC to steam at
150 oC?
Molar Heat of Fusion
Equations:
H2O(s) = 2.06 J/gC
H2O = 5.98 kJ/mol
H2O(l) = 4.184 J/gC
Molar Heat of Vaporization
H2O(g) = 2.02 J/gC
H2O = 40.6 kJ/mol
Q= mCpT
Q=n Hfus
Q=n Hvap
Specific Heat Values
Draw a Phase Diagram showing all possible changes for this problem
first.
150 oC
How much heat is needed
to convert 250 g of ice at
- 30C to steam at 150C?
0 oC
- 30 oC
1
ice  water
melting
2
100 oC
water  vapor (steam)
vaporizing (boiling)
4
5
100 oC
3
0 oC
1-Ice temp rising 
2-melting 
3-water temp rising 
4-vaporizing 
5-vapor temp rising
Specific Heat Values
H2O(s) Cp solid = 2.06 J/gC
H2O(l) Cp liquid = 4.184 J/gC
H2O(g) Cp gas = 2.02 J/gC
Molar Heat of Fusion
Hfus of H2O = 5.98 kJ/mol
Molar Heat of Vaporization
Hvap of H2O = 40.6 kJ/mol
This problem requires six individual
calculations. There are three in which the
phase does not change (yellow), two in
which the phase changes (green), and a
final calculation to total the energy of the
problem. (pink)
How much heat is needed to convert 250 g of ice at -30C to steam at 150C?
This problem requires six individual calculations. There are three in which the phase
does not change (yellow), two in which the phase changes (green), and a final
calculation to total the energies. (pink)
1. Raise the temperature of ice from -30 oC to 0 oC - no phase change
Q = m T Cp solid = (250g) (30 Co) (2.06 J/g . Co) = 15,450 J/1000J kJ-1 =15.45 kJ
2. Melt the ice - phase change occurs - temperature does not change
Q = m  Hfus = (250g/18.0g mol-1) (5.98 kJ/mol) = 0.831 kJ
3. Raise the temperature of liquid water from 0 oC to 100 oC - no phase change
q = m  T Cp liquid= (250g) (100 Co) (4.184 J/g . Co) = 104,600 J/1000J kJ-1 = 104.6 kJ
4. Vaporize the liquid water - phase change occurs - temperature does not change
Q = m  Hvap = (250g/18.0g mol-1) (40.6 kJ/mol) = 564 kJ
5. Raise the temperature of the water vapor from 100 oC to 150 oC - no phase change
Q = m  T Cp gas = (250g) (50 Co) (2.02 J/g . Co) = 25,250 J/1000J kJ-1 = 25.25 kJ
6. Find the total heat needed for the conversion - Add the Q 's from steps 1 - 5.
15.45 kJ + 0.831 kJ + 104.6 kJ + 564 kJ + 25.25 kJ = Total Q = 710 kJ