Chapter 3: Matter and Energy
Classification of Matter
Properties of Matter
Temperature
Energy
Specific Heat
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Matter
Matter
• Is the material that makes up all things.
• Has mass and occupies space.
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Matter• Anything that has mass and takes up space.
Matter
Pure Substance Mixture
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Classification of Matter
• Pure Substances: A form of matter that always has a definite and constant composition.
Properties always the same under a given set of conditions (temperature & pressure)
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A pure substance is classified as
• An element when composed of one type of atom.
• A compound when composed of two or more different elements combined in a definite ratio.
Pure Substances
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Pure Substances: Elements
• Element – Pure substance that can not be broken down into simpler
substances by chemical means.
Copper (Cu)Lead (Pb)Aluminum (Al)
– The most basic form of matter.– Each element can be found on the periodic table.
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Pure Substances: Compounds
• Compounds – A chemical combination of 2 or more different
elements.– A pure substance that can be broken down into
simpler substances by chemical means.
Salt (NaCl)Table sugar (C12H22O11)Water (H2O)Carbon monoxide (CO)NOTE: CO is different from the element Co
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Elements in a Compound
“Table salt” is a compound that contains the elements sodium and chlorine.
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Pure SubstancesElements and Compounds
• Elements and compounds have definite compositions, and each has a set of properties that are unique.
• Which pairs of symbols / formulas represent elements and compounds respectively?
(A)
(B)
(C)
(D)
(E)
P-1
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Mixtures
A mixture is a type of matter that consists of• Two or more substances that are physically
mixed, not chemically combined.• A physical combination of two or more pure
substances in which each substance retains its own chemical identity.
• Two or more substances in different (variable) proportions.
• Substances that can be separated by physical methods.
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MixturesExample of a mixture:
• Pasta and water can be separated by using a strainer.
• Uses a physical method to separate the components.
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Homogeneous Mixtures
In a homogeneous mixture,
• The composition is uniform throughout.
• The different components of the mixture are not visible or discernable, one from the other. Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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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.
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Classification of Matter
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Elements v. Compounds v. Mixtures
• Identify the following as an
Q-1
to
Q-5
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Matter
Matter
• Has characteristics called physical and chemical properties.
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Physical Properties
Physical properties are:
• Characteristics observed or measured without changing the identity of a substance.
• Shape, physical state, odor, boiling and freezing points (Changes of state), density, and color of that substance.
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Physical Properties of Copper
Copper has the followingphysical properties:• Reddish-orange
• Very shiny
• Excellent conductor of heat and electricity
• Solid at 25C• Melting point 1083C• Boiling point 2567 C Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings
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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.
Physical Change
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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 Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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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 neon
tube
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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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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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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.Copyright © 2005 by Pearson Education, Inc.Publishing as Benjamin Cummings
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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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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.
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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.
Learning Check
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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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Temperature Scales
Temperature scales
• Are Fahrenheit, Celsius, and Kelvin.
• Have reference points for the boiling and freezing points of water.
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Learning Check P-2
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• 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.
TF = 9/5 TC + 32
or TF = 1.8 TC + 32
Fahrenheit Formula
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Solving for °F TemperatureA person with hypothermia hasa 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’sCopyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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• 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 TC
1.8 1.8
TF - 32 = TC
1.8
Celsius Formula
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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
Learning Check
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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
Learning Check
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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
Learning Check
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The kelvin temperatureHas 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
• 0 K (absolute zero) is the lowest possible temperature .
0 K = –273 °C
Kelvin Temperature Scale
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Temperatures
Table 3.6
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Learning Check P-8
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Energy
• Makes objects move.
• Makes things stop.
• Is needed to “do work”.
Energy
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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.Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Potential EnergyPotential energy isenergy stored for use ata later time. Examples are• Water behind a dam.• A compressed spring.• Chemical bonds in
gasoline, coal, or food. Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Kinetic EnergyKinetic energy is the energy of matter in motion.
Examples are
• Swimming.
• Water flowing over a dam.
• Working out.
• Burning gasoline.
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Learning Check
Identify the energy as potential or kinetic.
1. Roller blading.
2. A peanut butter and jelly sandwich.
3. Mowing the lawn.
4. Gasoline in the gas tank.
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Forms of EnergyEnergy can take many forms.
• Heat (thermal)
• Mechanical (movement)
• Light
• Electrical
• Chemical
• Nuclear
P-3
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Changes in Forms of Energy
Energy can change from one form to
another.
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Heat is measured in joules or calories.
• 4.184 Joules (J) = 1 calorie (cal)
Exact by definition
• 1 kJ = 1000 J
• 1 kilocalorie (kcal) = 1000 calories (cal)
Units for Measuring Energy or Heat
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Examples of Energy In Joules
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Learning CheckHow many calories are obtained from a pat of butter if it
provides 150 J of energy when metabolized?
Solution →
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SolutionHow many calories are obtained from a pat of butter if
it provides 150 J of energy when metabolized?
Given: 150 J Need: caloriesPlan: J cal
Equality: 1 cal = 4.184 J
Set Up: 150 J x 1 cal = 36 cal 4.184 J
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Specific heat
• Is different for different substances.
• Is the amount of heat (q) that raises the temperature of 1 g of a substance by 1°C.
• In the SI system has units of J/gC.
• In the metric system has units of cal/gC.
Specific Heat
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Examples of Specific Heats
Table 3.7
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Note: Small values indicate good conductor of heat.
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SolutionWhat is the specific heat of a metal if 24.8 g absorbs275 J of energy and the temperature rises from 20.2C to 24.5C?
Given: 24.8 g, 275 J, 20.2C to 24.5C Need: J/gC
Plan: SH = Heat/gΔC where ΔoC = (Tf – Ti)
ΔT = 24.5C – 20.2C = 4.3 C SH Equation: SH = heat (q) (mass)(T)
Set Up: 275 J = 2.6 J/gC
(24.8 g)(4.3C)
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Rearranging the specific heat expression givesthe heat equation.
Heat(q) = g x Δ°C x J = J g°C
The amount of heat lost or gained by a substanceis calculated from the
• Mass of substance (g).
• Temperature change (T or Δ°C ).
• Specific heat of the substance (J/g°C).
Heat Equation
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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 (J) = 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
Using Specific Heat
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How many kilojoules are needed to raise the temperature of 325 g of water from 15.5°C to 77.5°C?
77.5°C – 15.5°C = 62.0°C
heat = g x T x SH
325 g x 62.0°C x 4.184 J x 1 kJ
g °C 1000 J
= 84.3 kJ
Solution
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Calculating MassAluminum 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 in grams
g = heat
ΔT x SH
Set Up: 3250 J
25.0°C x 0.897 J/g oC
= 145 g Al
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Transferring Heat Energy
Heat energy
• Flows from a warmer object to a colder object.
• Provides kinetic energy for the colder object.
• Energy lost by the warmer object is equal to the heat energy gained by the colder object.
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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) gainedCopyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Measuring Heat ChangesA 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)
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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 = 1000 cal1 Cal = 1 kcal1 Cal = 4184 J
1 Cal = 4.184 kJ
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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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Energy Values for Some Foods
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Table 3.9
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Energy Requirements
The amount of energy needed eachday depends on• Age• Gender• Physical activity
Table 3.11
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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) 50 kcal (50 Cal)
2) 80 kcal (80 Cal)
3) 170 kcal (170 Cal)
Learning Check
Solution →