Properties of Matter
Chapter 4
Hein and Arena
Version 2.0
12th Edition
Eugene Passer
Chemistry Department
Bronx Community
1 College
© John Wiley and Sons, Inc
Chapter Outline
4.1 Properties of Substances
4.6 Heat: Quantitative Measurement
4.2 Physical Changes
4.7 Energy in Chemical Changes
4.3 Chemical Changes
4.8 Conservation of Energy
4.4 Conservation of Mass
4.5 Energy
2
4.1
Properties of
Substances
3
Properties of a Substance
• A property is a characteristic of a
substance.
• Each substance has a set of properties
that are characteristic of that substance
and give it a unique identity.
4
Physical Properties
5
• The inherent characteristics of a
substance that are determined without
changing its composition.
• Examples:
 taste
 color
 physical state
 melting point
 boiling point
6
Physical Properties of Chlorine
•
•
•
•
•
2.4 times heavier than air
color is yellowish-green
odor is disagreeable
melting point –101oC
boiling point –34.6oC
7
Chemical Properties
8
Describe the ability of a substance
to form new substances, either by
reaction with other substances or by
decomposition.
9
Chemical Properties of Chlorine
• It will not burn in oxygen.
• It will support the combustion of
certain other substances.
• It can be used as a bleaching agent.
• It can be used as a water
disinfectant.
• It can combine with sodium to
form sodium chloride.
10
11
4.2
Physical Changes
12
Physical Changes
• Changes in physical properties (such as
size, shape, and density) or changes in
the state of matter without an
accompanying change in composition.
• Examples:
 tearing of paper
 change of ice into water
 change of water into steam
 heating platinum wire
• No new substances are formed.
13
4.3
Chemical Changes
14
In a chemical change new substances are formed
that have different properties and composition
from the original material.
15
Formation of Copper(II) Oxide
Heating
copper
insubstance
a Bunsen
burner
material
iswire
ainnew
called
The black
formation
ofwire
copper(II)
oxide
from
copper
Heating
a acopper
a Bunsen
burner
causes
the copper
copper
tolose
loseits
its
originalappearance
appearance
copper(II)
oxide.
and
oxygen
is a chemical
change.
causes
the
to
original
and
become
a100%
black
material.
The
copperisa(II)
oxide
is aby
new
substance with
Copper
copper
mass.
and
become
black
material.
properties
that
are
different
from
copper.
Copper (II) oxide is: 79.94% copper by mass
20.1% oxygen by mass.
16
Formation of Copper(II) Oxide
2+
2- 22+ or
Neither
A chemical
Cu norischange
O
contains
has
occurred.
Cu
O
Copper(II)
oxide
made
up
of
and
O
2
17
4.2
Decomposition of Water
The
composition
hydrogen
explodes
and
physical
with
a appearance
pop upon
of
Water
They
But
the
are
is decomposed
burning
both colorless
splint
into
gases.
ishydrogen
extinguished
and the
when
hydrogen
addition
ofand
athe
burning
oxygen
splint.
are different
from
oxygeninto
placed
by
passing
water
electricity
sample.
through
it. water.
The oxygen causes the flame of a burning
splint to intensify.
18
Chemical Equations
19
Water decomposes into hydrogen and
oxygen when electrolyzed.
reactant
yields
products
20
Chemical symbols can be used to
express chemical reactions
21
Water decomposes into hydrogen and
oxygen when electrolyzed.
2H2O
reactant
2H2
yields
products
O2
22
Copper plus oxygen yields copper(II)
oxide.
heat
reactants
yield
product
23
Copper plus oxygen yields copper(II)
oxide.
heat
2Cu
reactants
O2
2Cu2O
yield
product
24
25
4.4
Conservation
of Mass
26
No change is observed in the total
mass of the substances involved in a
chemical change.
27
sodium + sulfur  sodium sulfide
46.0 g
32.1 g
78.1 g
78.1 g reactant → 78.1 g product
mass reactants = mass products
28
4.5
Energy
29
Energy is the capacity to do work
30
Types of Energy
•
•
•
•
•
•
mechanical
chemical
electrical
heat
nuclear
radiant
31
Potential Energy
Energy that an object possesses
due to its relative position.
32
The potential energy of the ball increases
with increasing height.
increasing
potential energy
increasing
potential energy
50 ft
20 ft
33
Potential Energy
Stored energy
34
• Gasoline is a source of chemical
potential energy.
• The heat released when gasoline burns
is associated with a decrease in its
chemical potential energy.
• The new substances formed by burning
have less chemical potential energy
than the gasoline and oxygen.
35
Kinetic Energy
Energy matter possesses due to
its motion.
36
Moving bodies possess kinetic energy.
• The flag waving in
the wind.
37
Moving bodies possess kinetic energy.
• A bouncing ball.
• The running man.
38
Moving bodies possess kinetic energy.
• The runner
39
Moving bodies possess kinetic energy.
• The soccer player.
40
4.6
Heat:
Quantitative Measurement
41
Heat
• A form of energy associated with
small particles of matter.
Temperature • A measure of the intensity of heat, or
of how hot or cold a system is.
42
Units of Heat Energy
43
• The SI unit for heat
energy is the joule
(pronounced
“jool”).
• Another unit is
the calorie.
(exactly)
4.184 Joules = 1 calorie
4.184 J = 1 cal
This amount of heat energy will raise the
temperature of 1 gram of water 1oC.
44
An Example of the Difference
Between Heat and Temperature
A form of energy
associated with
small particles of
matter.
A measure of the
intensity of heat,
or of how hot or
cold a system is.
45
Twice as much
heat energy is
required to raise
the temperature
of 200 g of
water 10oC as
compared
to
100 g of water.
temperature
heat beakers
rises 10oC
A
B
100 g water
200 g water
30oC
20
30oC
20
4184 J
8368 J
46
Specific Heat
47
The specific heat of a substance is the quantity
of heat required to change the temperature of 1 g
of that substance by 1oC.
48
49
The units of
specific heat in
joules are:
Joules


 gram oCelcius 


 J 
 g oC 


50
The units of
specific heat in
calories are:
 calories 
 gram oCelcius 


 cal 
 g oC 


51
The relation of mass, specific heat,
temperature change (Δt), and quantity of
heat lost or gained is expressed by the
general equation:
(
)(
specific heat
of substance
)
mass of
Δt = heat
substance
52
Example 1
53
Calculate the specific heat of a solid in J/goC and in
cal/ goC if 1638 J raise the temperature of 125 g of the
solid from 25.0oC to 52.6oC.
(mass of substance)(specific heat of substance)Δt = heat
(g)(specific heat of substance)Δt = heat
heat = 1638 J
 heat 
specific heat = 

 g x Δt 
mass = 125 g
Δt = 52.6oC – 25.0oC = 27.6oC
1638 J

 0.475 J
=
specific heat = 
o
o 
g
C 54
 125 g x 27.6 C 
Calculate the specific heat of a solid in J/goC and in
cal/ goC if 1638 J raise the temperature of 125 g of the
solid from 25.0oC to 52.6oC.
Convert joules to calories using 1.000 cal/4.184 J
 0.475 J   1.000 cal  0.114 cal
specific heat =  o

 = g oC

 g C   4.184 J 
55
Example 2
56
A sample of a metal with a mass of 212 g is heated to
125.0oC and then dropped into 375 g of water at
24.0oC. If the final temperature of the water is 34.2oC,
what is the specific heat of the metal?
When the metal enters the water, it begins to cool,
losing heat to the water. At the same time, the
temperature of the water rises. This process continues
until the temperature of the metal and the temperature of
the water are equal, at which point (34.2oC) no net flow
of heat occurs.
57
A sample of a metal with a mass of 212 g is heated to
125.0oC and then dropped into 375 g of water at
24.0oC. If the final temperature of the water is 34.2oC,
what is the specific heat of the metal?
• Calculate the heat gained by the water.
• Calculate the final temperature of the metal.
• Calculate the specific heat of the metal.
58
A sample of a metal with a mass of 212 g is heated to
125.0oC and then dropped into 375 g of water at
240.0oC. If the final temperature of the water is
34.2oC, what is the specific heat of the metal?
Heat Gained by the Water
temperature rise
of the water
Δt = 34.2oC – 24.0oC = 10.2oC
heat gained (375g ) 4.184 J 
o
4
(10.2
C)
=
=
1.60 x 10 J


o
by the water
 gC 
59
A sample of a metal with a mass of 212 g is heated to
125.0oC and then dropped into 375 g of water at
240.0oC. If the final temperature of the water is
34.2oC, what is the specific heat of the metal?
Heat Lost by the Metal
Once the metal is dropped into the water, its temperature
will drop until it reaches the same temperature as the
water (34.2oC).
temperature drop
of the metal
Δt = 125.0oC – 34.2oC = 90.8oC
heat lost
heat gained
=
= 1.60 x 104 J
by the metal
by the water
60
A sample of a metal with a mass of 212 g is heated to
125.0oC and then dropped into 375 g of water at
240.0oC. If the final temperature of the water is
34.2oC, what is the specific heat of the metal?
The heat lost or gained by the system is given by:
(mass) (specific heat) (Δt) = energy change
rearrange
heat 

specific heat = 

 mass x Δt 
specific heat
=
of the metal
 1.60 x 10 J   0.831 J 
 (212g)(90.8oC)    g oC) 


 
4
61
4.7
Energy in
Chemical Changes
62
In all chemical changes, matter either
absorbs or releases energy.
63
Energy Release From
Chemical Sources
Type of
Energy
Energy Source
Electrical
Storage batteries
Light
A lightstick. Fuel combustion.
Heat and Light
Combustion of fuels.
Body
Chemical changes occurring within
body cells.
64
Chemical Changes Caused by
Absorption of Energy
Type of
Energy
Chemical Change
Electrical
Electroplating of metals.
Decomposition of water into
hydrogen and oxygen
Light
Photosynthesis in green plants.
65
4.8
Conservation
of Energy
66
An energy transformation occurs
whenever a chemical change occurs.
•
If energy is absorbed during a chemical
change, the products will have more
chemical potential energy than the
reactants.
•
If energy is given off in a chemical
change, the products will have less
chemical potential energy than the
reactants.
67
H2 + O2 have higher
potential energy than H2O
higher
energy
potential
is absorbed
energy
Electrolysis of Water
lower
energy
potential
is given
energy
off
Burning of
Hydrogen in Air
68
4.3
Law of Conservation of Energy
Energy can be neither created nor destroyed,
though it can be transformed from one form of
energy to another form of energy.
69
70