Transcript Properties of Matter Chapter 4 Chapter 4 - Properties of Matter 4.1 Properties of Substances 4.6 Heat: Quantitative Measurement 4.2 Physical Changes 4.7 Energy in.

Properties of Matter
Chapter 4
1
Chapter 4 - Properties of Matter
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
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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.
 Can be classified as either physical or
chemical
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Properties of a Substance
Physical Properties
 The inherent characteristics of a substance
that are determined without changing its
composition.
 Examples:
 taste
 color & odor
 physical state
 melting point
 boiling point
 density
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Physical Properties
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Chemical Properties
Describe the ability of a substance to form
new substances, either by reaction with
other substances or by decomposition.
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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.
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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.
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Chemical Changes
In a chemical change new substances are
formed that have different properties and
composition from the original material.
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Formation of Copper(II) Oxide
Heating
copperofwire
wire
ina Bunsen
asubstance
Bunsen
burner
black
iscopper(II)
a in
new
called
The
formation
oxide
from
Heating
aa material
copper
burner
causes
theoxide.
copper
lose
its
original
copper(II)
copper
oxygen
atochemical
change.
causes and
the
copper
toislose
its
original
appearance
aa black
material.
The
copper
(II)become
oxide
is
a new
substance
Copper
isand
100%
copper
by
mass.
appearance
and
become
black
material.
with
properties
that
are
different
from
Copper (II) oxide is: 79.94% copper by
copper.
mass 20.1% oxygen by mass.
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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
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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.
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Chemical Equations
Water decomposes into hydrogen and
oxygen when electrolyzed.
reactant
yields
products
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Chemical symbols can be used to
express chemical reactions
Water decomposes into hydrogen and oxygen
when electrolyzed.
2H2O
reactant
2H2
yields
products
O2
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Copper plus oxygen yields copper(II)
oxide.
heat
reactants
yield
product
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Copper plus oxygen yields copper(II)
oxide.
heat
2Cu
reactants
O2
2Cu2O
yield
product
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Conservation of Mass
No change is observed in the total
mass of the substances involved in
a chemical change.
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Conservation of Mass
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
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Energy
Energy is the capacity to do work
Potential Energy
Energy that an object possesses due
to its relative position.
Kinetic Energy
Energy matter possesses due to its
motion.
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Types of Energy
 mechanical
 chemical
 electrical
 heat
 nuclear
 radiant
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The potential energy of the ball increases
with increasing height.
increasing
potential energy
increasing
potential energy
50 ft
20 ft
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Potential Energy
Stored energy
• 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.
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Moving bodies possess kinetic energy.
 The flag waving in the
wind.
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Moving bodies possess kinetic energy.
 A bouncing ball.
 The running man.
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Units of Heat Energy
 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.
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Heat vs. Temperature
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.
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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
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4184 J
8368 J
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Specific Heat
The specific heat of a substance is the quantity of
heat required to change the temperature of 1 g of
that substance by 1oC.
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Units of Specific Heat
 J 
 g oC 


The
units
of
specific heat in
joules are:
Joules


 gram oCelcius 


The
units
of
specific heat in
calories are:
 calories   cal 
 gram oCelcius   g oC 

 

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General Equation - Specific Heat
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
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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 = 

mass = 125 g
 g x Δt 
Δt = 52.6oC – 25.0oC = 27.6oC
1638 J

 0.475 J
=
specific heat = 
o
o 
g
C
 125 g x 27.6 C 
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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 
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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?
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.
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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?
 Calculate the heat gained by the water.
 Calculate the final temperature of the metal.
 Calculate the specific heat of the metal.
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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 
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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
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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
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Energy in Chemical Changes
In all chemical changes, matter either
absorbs or releases energy.
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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.
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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.
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Conservation of Energy
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.
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Conservation of Energy
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
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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.
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Concepts
1. Physical Properties
2. Distinguish Chemical from Physical Properties
3. Classify changes – Chemical or Physical
4. Kinetic vs. Potential Energy
5. Law of Conservation of Mass
6. Law of Conservation of Energy
7. Heat vs. Temperature
8. Use equation:
(mass) (specific heat) (Δt) = heat
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