Learning Check

Chapter 3 Matter and Energy
3.1
Classification of Matter
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1
Matter
Matter
• Is the material that
makes up all things.
• Has mass and
occupies space.
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2
Pure Substances
A pure substance is classified as
• Matter with a specific composition.
• An element when composed of one type of
atom.
• A compound when composed of two or
more elements combined in a definite ratio.
3
Elements
Elements are
• Pure substances that
contain only one type
of material.
• Examples are:
Copper Cu
Lead Pb
Aluminum Al
4
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Compounds
Compounds
• Contain two or more
elements in a definite ratio.
Salt (NaCl)
Table sugar (C12H22O11)
Water (H2O)
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5
Elements in a Compound
“Table salt” is a compound that contains the
elements sodium and chlorine.
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6
Mixtures
A mixture is a type of matter that consists of
• Two or more substances that are physically
mixed, not chemically combined.
• Two or more substances in different
proportions.
• Substances that can be separated by physical
methods.
7
Mixtures
Example of a mixture:
• Pasta and water be separated by using a
strainer.
• Uses a physical method to separate.
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8
Homogeneous Mixtures
In a homogeneous mixture,
• The composition is
uniform throughout.
• The different parts of the
mixture are not visible.
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9
Heterogeneous Mixtures
In a heterogeneous mixture,
• The composition of
substances is not
uniform.
• The composition varies
from one part of the
mixture to another.
• The different parts of the
mixture are visible.
10
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Classification of Matter
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11
Learning Check
Identify each of the following as a pure substance or a
mixture.
A. pasta and tomato sauce
B. aluminum foil
C. helium
D. air
12
Solution
Identify each of the following as a pure substance
or a mixture.
A. pasta and tomato sauce
mixture
B. aluminum foil
pure substance
C. helium
pure substance
D. air
mixture
13
Learning Check
Identify each of the following as a homogeneous or
heterogeneous mixture:
A. hot fudge sundae
B. shampoo
C. sugar water
D. peach pie
14
Solution
Identify each of the following as a homogeneous
or heterogeneous mixture:
A. hot fudge sundae
heterogeneous mixture
B. shampoo
homogeneous mixture
C. sugar water
homogeneous mixture
D. peach pie
heterogeneous mixture
15
Chapter 3 Matter and
Energy
3.2 Properties of Matter
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16
Matter
Matter
• Has characteristics called physical
and chemical properties.
17
Physical Properties
Physical properties are
• Characteristics observed or measured
without changing the identify of a
substance.
• Shape, physical state, odor, boiling and
freezing points, density, and color of
that substance.
18
Physical Properties of
Copper
Copper has the following
physical properties:
• Reddish-orange
• Very shiny
• Excellent conductor of
heat and electricity
• Solid at 25C
• Melting point 1083C
• Boiling point 2567 C
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19
States of Matter
The states of matter are
• Solid
Definite volume and shape
• Liquid
Definite volume, but take the
shape of its container
• Gas
No definite volume or shape
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20
Examples of States of
Matter
Solids
• Rocks, shells, baseballs, tennis
racquets, books
Liquids
• Lakes, rain, melted gold, mercury
in a thermometer
Gases
• Air, helium in a balloon, neon in a
21 neon tube
Learning Check
Identify the state of matter for each of
the following:
A.
B.
C.
D.
E.
22
Vitamin tablets
Eye drops
Vegetable oil
A candle
Air in a tire
Solution
Identify the state of matter for each of
the following:
A.
B.
C.
D.
E.
23
Vitamin tablets
Eye drops
Vegetable oil
A candle
Air in a tire
solid
liquid
liquid
solid
gas
Physical Change
A physical change occurs in a
substance if there is
• A change in the state.
• A change in the physical shape.
• No change in the identity and
composition of the substance.
24
Examples of Changes of
State
Some changes of state for water:
• Solid water (ice) melts and
forms liquid water.
• Liquid water boils and forms
gaseous water (steam).
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25
Examples of Physical
Changes
Examples of physical changes:
• Paper torn into little pieces
(change of size)
• Copper hammered into thin
sheets (change of shape)
• Water poured into a glass
(change of shape)
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26
Learning Check
Classify each of the following as a
1) change of state 2) change of shape
A.
B.
C.
D.
E.
27
Chopping a log into kindling
Water boiling in a pot
Ice cream melting
Ice forming in a freezer
Cutting dough into strips
Solution
Classify each of the following as a
1) change of state 2) change of shape
A. (2)
B. (1)
C. (1)
D. (1)
E. (2)
28
Chopping a log into kindling
Water boiling in a pot
Ice cream melting
Ice forming in a freezer
Cutting dough into strips
Chemical Properties
Chemical properties describe
the ability of a substance
• To interact with other
substances.
• To change into a new
substance.
Example:
Iron has the ability to form rust
when exposed to oxygen.
29
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Learning Check
Classify each of the following properties
as physical or chemical:
A. Ice melts in the sun
B. Copper is a shiny metal
C. Paper can burn
D. A silver knife can tarnish
E. A magnet removes iron particles
from a mixture
30
Solution
Classify each of the following properties
as physical or chemical:
A. Ice melts in the sun (physical)
B. Copper is a shiny metal (physical)
C. Paper can burn (chemical)
D. A silver knife can tarnish (chemical)
E. A magnet removes iron particles
from a mixture (physical)
31
Chemical Change
In a chemical change, a new
substance forms that has
• A new composition
• New chemical properties
• New physical properties
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32
Some Chemical Changes
Silver tarnishes Shiny metal reacts
to form black, grainy coating.
Wood burns A piece of wood burns
with a bright flame to form ash,
carbon dioxide, water vapor, and
heat.
Iron rusts A shiny nail combines
with oxygen to form orange-red rust.
33
Learning Check
Classify each of the following changes as
physical or chemical
A. Burning a candle.
B. Ice melting on the street.
C. Toasting a marshmallow.
D. Cutting a pizza.
E. Iron rusting in an old car
34
Solution
Classify each of the following changes as
physical or chemical
A. Burning a candle. (chemical)
B. Ice melting on the street. (physical)
C. Toasting a marshmallow. (chemical)
D. Cutting a pizza. (physical)
E. Iron rusting in an old car. (chemical)
35
Chapter 3 Matter and Energy
3.3
36
Temperature
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Temperature
Temperature
• Is a measure of how hot or cold
an object is compared to
another object.
• Indicates that heat flows from
the object with a higher
temperature to the object with a
lower temperature.
• Is measured using a
thermometer.
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37
Temperature Scales
Temperature scales
• Are Fahrenheit,
Celsius, and
Kelvin.
• Have reference
points for the
boiling and
freezing points of
water.
38
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Learning Check
A. What is the temperature of freezing water?
1) 0°F
2) 0°C
3) 0 K
B. What is the temperature of boiling water?
1) 100°F
2) 32°F
3) 373 K
C. How many Celsius units are between the boiling
and freezing points of water?
1) 100
2) 180
3) 273
39
Solution
A. What is the temperature of freezing water?
2) 0°C
B. What is the temperature of boiling water?
3) 373 K
C. How many Celsius units are between the boiling
and freezing points of water?
1) 100
40
Fahrenheit Formula
• On the Fahrenheit scale, there are 180°F
between the freezing and boiling points and
on the Celsius scale, there are 100°C.
180°F =
9°F =
1.8°F
100°C
5°C
1°C
• In the formula for calculating the Fahrenheit
temperature, adding 32 adjusts the zero point
of water from 0°C to 32°F.
41
TF
= 9/5 TC + 32
or TF
= 1.8 TC + 32
Solving for °F Temperature
A person with hypothermia has
a body temperature of 34.8°C.
What is that temperature in °F?
TF = 1.8 TC
+ 32
TF = 1.8 (34.8°C) + 32°
exact tenth's
exact
= 62.6 + 32°
= 94.6°F
tenth’s
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42
Celsius Formula
• TC is obtained by rearranging the equation for TF.
TF
=
1.8TC + 32
• Subtract 32 from both sides.
TF - 32
=
1.8TC ( +32 - 32)
TF - 32
=
1.8TC
• Divide by 1.8 = °F - 32
1.8
TF - 32
1.8
43
=
TC
= 1.8 TC
1.8
Learning Check
The normal temperature of a chickadee is 105.8°F.
What is that temperature on the Celsius scale?
1) 73.8 °C
2) 58.8 °C
3) 41.0 °C
44
Solution
3) 41.0 °C
TC
=
=
=
45
(TF - 32°)
1.8
(105.8° - 32°)
1.8
73.8° = 41.0°C
1.8
Learning Check
A pepperoni pizza is baked at 455°F. What
temperature is needed on the Celsius scale?
1) 423°C
2) 235°C
3) 221°C
46
Solution
A pepperoni pizza is baked at 455°F. What
temperature is needed on the Celsius scale?
2) 235°C
TF - 32°
1.8
= TC
(455° - 32°) = 235°C
1.8
47
Learning Check
On a cold winter day, the temperature is –15°C.
What is that temperature in °F?
1) 19 °F
2) 59°F
3) 5°F
48
Solution
3) 5°F
TF = 1.8 TC + 32
TF = 1.8(–15°C) + 32°
= – 27° + 32°
= 5°F
Note: Be sure to use the change sign key on
your calculator to enter the minus – sign.
1.8 x 15 +/ – = –27
49
Kelvin Temperature Scale
The kelvin temperature
Has 100 units between freezing and boiling points.
100 K = 100°C or
1 K = 1 °C
• Adds 273 to the Celsius temperature.
TK
= TC + 273
• Of 0 K (absolute zero) is the lowest possible
temperature .
0K
= –273 °C
50
Temperatures
Table 3.6
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51
Learning Check
What is normal body temperature of 37°C in kelvins?
1) 236 K
2) 310. K
3) 342 K
52
Solution
What is normal body temperature of 37°C in kelvins?
2) 310. K
TK =
=
=
53
TC + 273
37°C + 273
310. K
Chapter 3 Matter and Energy
3.4
Energy
54
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Pearson Education, Inc.
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Cummings
Energy
Energy
• Makes objects move.
• Makes things stop.
• Is needed to “do work”.
55
Work
Work is done when
• You climb.
• You lift a bag of
groceries.
• You ride a bicycle.
• You breathe.
• Your heart pumps
blood.
• Water goes over a
dam.
56
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Cummings
Potential Energy
Potential energy is
energy stored for use at
a later time.
Examples are
• Water behind a dam.
• A compressed
spring.
• Chemical bonds in
gasoline, coal, or
food.
57
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Education, Inc.
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Cummings
Kinetic Energy
Kinetic energy is the
energy of matter in
motion.
Examples are
• Swimming.
• Water flowing over a
dam.
• Working out.
• Burning gasoline.
58
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Learning Check
Identify the energy as 1) potential or 2) kinetic.
A.
B.
C.
D.
59
Roller blading.
A peanut butter and jelly sandwich.
Mowing the lawn.
Gasoline in the gas tank.
Solution
Identify the energy as 1) potential or 2) kinetic
A. Roller blading. (2 kinetic)
B. A peanut butter and jelly sandwich.
(1 potential)
C. Mowing the lawn. (2 kinetic)
D. Gasoline in the gas tank. (1 potential)
60
Forms of Energy
Energy can take many forms.
• Heat
• Mechanical
• Light
• Electrical
• Chemical
• Nuclear
61
Changes in Forms of Energy
Energy can change from one form to
another.
62
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Units for Measuring Energy
or Heat
Heat is measured in joules or calories.
• 4.184 Joules (J) = 1 calorie (cal) (exact)
• 1 kJ = 1000 J
• 1 kilocalorie (kcal) = 1000 calories (cal)
63
Examples of Energy In
Joules
64
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Learning Check
How many calories are obtained from a pat of
butter if it provides 150 J of energy when
metabolized?
65
Solution
How many calories are obtained from a pat of
butter if it provides 150 J of energy when
metabolized?
Given:
Plan:
Equality:
Set Up:
66
150 J
Need: calories
J
cal
1 cal = 4.184 J
150 J x 1 cal = 36 cal
4.184 J
Chapter 3 Matter and Energy
3.5
Specific Heat
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67
Specific Heat
Specific heat
• Is different for different substances.
• Is the amount of heat that raises the temperature
of 1 g of a substance by 1°C.
• In the SI system has units of J/gC.
• In the metric system has units of cal/gC.
68
Examples of Specific Heats
Table 3.7
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69
Learning Check
What is the specific heat of a metal if 24.8 g
absorbs 275 J of energy and the temperature
rises from 20.2C to 24.5C?
70
Solution
What is the specific heat of a metal if 24.8 g absorbs
275 J of energy and the temperature rises from
20.2C to 24.5C?
Given: 24.8 g, 275 J, 20.2C to 24.5C Need: J/gC
Plan: SH = Heat/gC
ΔT = 24.5C – 20.2C = 4.3 C
SH Equation: SH = heat (q)
(mass)(T)
Set Up:
275 J
= 2.6 J/gC
(24.8 g)(4.3C)
71
Heat Equation
Rearranging the specific heat expression gives
the heat equation.
Heat(q) = g x °C x J = J
g°C
The amount of heat lost or gained by a substance
is calculated from the
• Mass of substance (g).
• Temperature change (T).
• Specific heat of the substance (J/g°C).
72
Using Specific Heat
A layer of copper on a pan has a mass of 135 g.
How much heat in joules will raise the temperature
of the copper from 26°C to 328°C if the specific heat
of copper is 0.385 J/g°C?
The temperature change is 328°C - 26°C = 302°C.
heat (cal) =
g
x
T
x SH(Cu)
135 g x 302°C x 0.385 J
g°C
= 15 700 J or 1.57 x 104 J
73
Learning Check
How many kilojoules are needed to raise the
temperature of 325 g of water from 15.0°C to
77.0°C?
1) 20.4 kJ
2) 77.7 kJ
3) 84.3 kJ
74
Solution
How many kilojoules are needed to raise the
temperature of 325 g of water from 15.5°C to
77.5°C?
3) 84.3 kJ
77.0°C – 15.0°C = 62.0°C
325 g x 62.0°C x 4.184 J x 1 kJ
g °C
1000 J
= 84.3 kJ
75
Calculating Mass
Aluminum is used to make kitchen utensils. What
is the mass of an aluminum spatula if 3.25 kJ of
heat raise its temperature from 20.0°C to 45.0°C.
SHAl = 0. 897 J/g°C?
Given: 3.25 kJ (3250 J), 20.0°C to 45.0°C
ΔT = 25.0°C
Plan: Solve heat equation for mass
m = heat
ΔT x SH
Set Up:
3250 J/g°C
=
145 g Al
25.0°C x 0.897 J
76
Transferring Heat Energy
Heat energy
• Flows from a warmer object to a colder object.
• Provides kinetic energy for the colder object.
• Lost by the warmer object is equal to the heat
energy gained by the colder object.
77
Calorimeters and Heat Transfer
A calorimeter
• Is used to measure heat transfer.
• Can be made with a coffee cup,
water, and a thermometer.
• Indicates the heat lost by a
sample and gained by water.
Heat lost (-q) = Heat (q) gained
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78
Measuring Heat Changes
A 50.0-g sample of tin is heated to 99.8°C and
dropped into 50.0 g water at 15.6°C. If the
final temperature is 19.8°C, what is the specific
heat of tin?
Heat gain (q) by water
= 50.0 g x 4.2°C x 4.184 J/g°C = 880 J
Heat loss (-q) by tin = -880 J
SH tin =
-880 J
= 0.22 J/g°C
(50.0 g)(-80.0°C)
79
Energy and Nutrition
On food labels, energy is shown as the
nutritional Calorie, written with a capital C. In
countries other than the U.S., energy is
shown in kilojoules (kJ).
1 Cal =
1 Cal =
1 Cal =
1 Cal =
80
1000 cal
1 kcal
4184 J
4.184 kJ
Caloric Food Values
The caloric or energy
values for 1 g of a food
is given in
• kJ or
• kcal (Cal)
Table 3.8
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81
Energy Values for Some
Foods
Table 3.9
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82
Energy Requirements
The amount of
energy needed each
day depends on
• Age
• Sex
• Physical activity
Table 3.11
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83
Learning Check
A cup of whole milk contains 12 g carbohydrate,
9.0 g fat, and 9.0 g protein. How many kcal
(Cal) does a cup of milk contain?
1) 48 kcal (48 Cal)
2) 81 kcal (81 Cal)
3) 165 kcal (165 Cal)
84
Solution
3)
85
165 kcal
12 g carbohydrate x 4 kcal/g
=
48 kcal
9.0 g fat
x 9 kcal/g
=
81 kcal
9.0 g protein
x 4 kcal/g
=
36 kcal
Total kcal
=
=
165 kcal
165 Cal