1. 3.03 Phase Changes of Matter
Water crystal
Heating-Cooling Curve
Dr. Fred Omega Garces
Chemistry 100
Miramar College

2. Phases of Matter: Terminology
Energy is required for phase
changes to occur.
Solid-Liquid-Gas
Triangle

3. Intermolecular Forces
At the molecular level:
Molecules or matter is held together by “glue” called intermolecular forces
Energy added (K.E. increases)

4. Keeping Matter together
Intramolecular Forces -
Force which keeps molecule together, i.e., bonds.
Intermolecular Forces -
Attractive force between molecules. Responsible for keeping matter in solid or liquid phase.

5. Heating Cooling Curve From Ice to Steam and Vice-versa 6.01 kJ mol
Stage1
Stage2
Stage3
Stage4
Stage5
Heat Addition
What is the energy needed to take 1g H2O at 0°C to 100°C ?
=720cal
1.84 J
g °
0.43 cal g
6.01 kJ mol
540 cal g J
g °
1 cal
g o
40.7 kJ
mol
80 cal g
2.09 J
g °
0.50 cal g

6. Heating / Cooling Curves
From Steam to Ice and Vice-versa
2.09 J
g °
0.50 cal g
1.84 J
g °
0.43 cal g
.334 kJ g
80 cal J
g °
1 cal
g o
2.26 kJ g
540 cal g
What is the energy needed to take 1g H2O at 0°C to 100°C ?
=720cal

7. What is the energy needed to take 1g H2O ice at 0°C to steam 100°C?
Heating Cooling Curve
From Ice to Steam and Vice-versa
Stage1
Stage2
Stage3
Stage4
Stage5
80 cal g
60.2 kJ
mol
540 cal g
40.1 kJ
mol
1 cal
g o J
g °
What is the energy needed to take 1g H2O ice at 0°C to steam 100°C?
=720cal
Units
Stage1
S.h. ice
Stage2
Heat fusion
Stage3
specific heat (s.h.) water
Stage4
Heat vaporization
Stage5
S.h. gas
cal /g
0.50
cal / g K 80
cal/g
1.0
540
cal / g
J g-1 K-1
2.11
J / g K
334
J/g
4,18
2257
2.08
cal mol-1
9.01
cal / mol K
1441
18.102cal / mol K
9728
J mol-1 K-1
38.1
J / mol K
6.02
kJ / mol K
75.33
40.1
kJ/mol
37.5
J /mol K

8. SOLUTION STUDY CHECK SAMPLE PROBLEM Heat of Fusion
Ice cubes at 0 °C with a mass of 26 g are added to your soft drink.
a. How much heat (joules) must be added to melt all the ice at 0 °C?
b. What happens to the temperature of your soft drink? Why?
calorie
STEP 1 Given 26 g of H2O(s) Need joules to melt ice
SOLUTION
cal
STEP 2
calorie
STEP 3 Equalities/Conversion Factors
80 cal
80 cal
80 cal
STEP 4 Set Up Problem
80 cal
2080 calorie
b. The soft drink will be colder because heat from the soft drink is providing the energy to melt the ice.
STUDY CHECK
In a freezer, 150. g water at 0 °C is placed in an ice cube tray. How much heat, in kilojoules, must be removed to
form ice cubes at 0 °C?

9. SOLUTION STUDY CHECK SAMPLE PROBLEM Heat removed from water in sauna.
In a sauna, 150 g of water is converted to steam at 100 °C. How many kilocalories of heat are needed?
STEP 1 Given 150 g of H2O(l) to H2O(g) Need kilocalories of heat to change state
SOLUTION
STEP 2
STEP 3 Equalities/Conversion Factors
STEP 4 Set Up Problem
STUDY CHECK
When steam from a pan of boiling water reaches a cool window, it condenses. How much heat, in kilocalories
(kcal), is released when 25 g of steam condenses at 100 °C?

10. SAMPLE PROBLEM
2.76
Raising temperature of water
A pitcher containing 0.75 L (750 g) of water at 4°C is removed from the refrigerator. How many Kcal (and kJ) are needed to warm the water to a room temperature of 22 °C
First calculate for 1 g water, then multiply by 750 gram H2O
22°C – 4°C = 18 °Temp change
Therefore it will cost 18 cal for 1 gram
For 750 grams :
750 g * (18cal/g) = cal or 13.5 Cal or 58.2 KJ
Stage1
Stage2
Stage3
Stage4
Stage5 J
g °
1 cal
g o
- 22°C
- 4°C
Heat Addition

11. Answer: Calculate all for 1 g then multiply final answer by mass
SAMPLE PROBLEM
Combining Heat Calculations
Calculate in calories and Joule the following:
a) Energy required to heat 25.0 g of water from 12.5°C to 25.7°C
b) Energy required to heat 38.0 g of water from 1°C to 100°C steam.
c) Energy lose when 15.0 g of water cools from 75 °C to 0°C solid ice.
d) Final temperature when 600. cal is added to 6.00 g ice at starting at 0°C.
Answer: Calculate all for 1 g then multiply final answer by mass
13. 2 cal / g x g = 330 cal = 1420 J
(99 cal cal ) x 38.0 g = cal = 24.3 kcal =105 kJ
(75 cal cal ) x 15.0 g = 2330 cal = 2.33 kcal = 10.0 kJ
Cost (80 cal) x 6 g = 480 cal. There fore 600 – 480 = 120 cal
120 cal / 6 g = 20° Final temperature = 20° C
Double Check: (80 cal cal ) x 6.00 g = 600. cal. This checsks

12. SOLUTION Answer: SAMPLE PROBLEM 2.14 Combining Heat Calculations
Using Table 2.13 and the specific heat of ethanol (2.46 J/g °C), calculate the total heat, in joules, needed to
convert 15.0 g of ethanol at 25.0 °C to gas at 78.0 °C.
STEP 1 Given g of ethanol at 25.0 °C
Need heat (J) needed to warm the ethanol and change it to gas
SOLUTION
STEP 2 When several changes occur, draw a diagram of heating and changes of state.
Total heat = joules needed to warm ethanol from 25.0 °C to 78.0 °C
+ joules to change liquid to gas at 78.0 °C
Answer:
PART 1: Energy to Warm Ethanol 25° to 88°
Temperature change = 78°– 25° = 53°
for 1 gram ethanol, cost 2.46 J for every °
So 2.46 J/° x 53° = J for 1 gram
15.0 g x ( J / g) = J
Part 2: Energy to convert ethanol liquid to gas
Need heat vaporization ethanol = 841 J for 1 g
Therefore for 15.0 g, that is
841 J x 15.0 g = J
Part 1 + part 2 = = 14.6 kJ = 3.38 Kcal

13. Summary
The equilibria between the three phase of a substance is a function of the pressure and the temperature as shown in a phase diagram. Equilibria between any two phase are indicated by a line. The line through the melting point usually slopes slightly to the right as pressure increases (solid is more dense than liquid). For water this slope is to the left. Features of the phase diagram is the normal melting point (freezing point) and the normal boiling point. The triple point is the condition in which all three phases of matter exist and the critical point is the condition in which the liquid and the gas phase are indistinguishable.