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Chapter 3
Matter and Energy
Tro's "Introductory Chemistry",
Chapter 3
2
Matter
• Matter is anything that occupies space and has mass.
• Even though it appears to be smooth and continuous, matter is actually composed of a lot of tiny little pieces we call atoms and molecules.
Tro's "Introductory Chemistry",
Chapter 3
3
Atoms and Molecules
• Atoms are the tiny particles
that make up all matter.
• In most substances, the
atoms are joined together in
units called molecules.
Tro's "Introductory Chemistry",
Chapter 3
4
Classification of Matter
Matter can be classified in two ways:
1- according to its physical state
2- according to its composition
4
Tro's "Introductory Chemistry",
Chapter 3
5
Classification of Matter According to its
Physical State (STATES OF MATTER)
Matter can be classified as solid, liquid, or gas
SOLID e.g. stone, charcoal,
diamond
LIQUID e.g. water, wine,
vinegar
GAS e.g. oxygen,
carbon dioxide
MATTER
Tro's "Introductory Chemistry",
Chapter 3
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Solids • The particles in a solid are packed
close together and are fixed in position. Due to this close parking solids are
incompressible.
• The inability of the particles to move around results in solids retaining their shape and volume when placed in a new container and prevents the particles from flowing.
• Solids may be further classified as: Crystalline Solids
Amorphous Solids
Tro's "Introductory Chemistry",
Chapter 3
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Crystalline vs Amorphous • Some solids have their particles
arranged in an orderly geometric pattern—we call these crystalline solids.
Salt and diamonds.
• Other solids have particles that do not show a regular geometric pattern over a long range—we call these amorphous solids.
Plastic and glass.
Tro's "Introductory Chemistry",
Chapter 3
8
Liquids
• The particles in a liquid are closely packed, but they have some ability to move around.
• The close packing results in liquids being incompressible.
• The ability of the particles to move allows liquids to take the shape of their container and to flow. However, they don’t have enough freedom to escape and expand to fill the container.
Tro's "Introductory Chemistry",
Chapter 3
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Gases
• In the gas state, the particles have complete
freedom from each other.
• The particles are constantly flying around,
bumping into each other and the container.
• In the gas state, there is a lot of empty space
between the particles.
On average.
Tro's "Introductory Chemistry",
Chapter 3
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Gases, Continued
• Because there is a lot of empty
space, the particles can be
squeezed closer together.
Therefore, gases are
compressible.
• Because the particles are not
held in close contact and are
moving freely, gases expand to
fill and take the shape of their
container, and will flow.
Tro's "Introductory Chemistry",
Chapter 3
11
Summary: Properties of Solids,
Liquids and Gases
Volume Shape Compressible?
Solids fixed fixed NO
Liquids fixed not fixed NO
Gases not fixed
not fixed
YES
Tro's "Introductory Chemistry",
Chapter 3
12
Classification of Matter by Composition
Pure Substance
Constant Composition
Mixture
Variable Composition
Matter
Tro's "Introductory Chemistry",
Chapter 3
13
Classification of Matter as
a pure substance
A pure substance is made of a single type of
particle (i.e., atom or molecule).
The composition of a pure substance does not
change from one sample to another and because
of this, all samples have the same
characteristics (or properties)
Tro's "Introductory Chemistry",
Chapter 3
14
Classification of Matter as a Pure Substance
•Pure substances can further be sub-divided into
two groups:
Elements
Compounds
Elements: Pure substances that cannot be broken down
into simpler substances e.g., copper, helium, sodium
Compounds: Pure substances that are made of two or
more elements in definite proportions e.g., sodium
chloride NaCl, Carbon dioxide CO2
Tro's "Introductory Chemistry",
Chapter 3
15
Classification of Matter as a Mixture
• A mixture is a combination of two or more substances in
which each substance retain their distinct characteristics (or properties). e.g., a mixture of rice and sand
• Mixtures can also be further sub-divided into two groups:
Homogeneous mixture
Heterogeneous mixture
Homogeneous mixture: Has uniform composition through out the
sample e.g., solutions such as tea with sugar, salt solutions
Heterogeneous mixture: Does not have a uniform composition
through out the sample. e.g., sand and water
Note: HOMO means the same while HETERO means different
Tro's "Introductory Chemistry",
Chapter 3
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Example—Classify the Following as
Homogeneous or Heterogeneous
Mixtures
• A cup of coffee
• A mixture of table sugar and black pepper
• A mixture of sugar dissolved in water
• Sand and water
- homogeneous
- homogeneous
- heterogeneous
- heterogeneous
17
Classifying Matter
Tro's "Introductory Chemistry",
Chapter 3
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How Do We Distinguish Matter?
• Looking at both flasks, it is had to tell the difference between the two substances
• How can we tell if one flask contain water and the other contains an acid?
• We can only tell by studying their properties
water acid
Tro's "Introductory Chemistry",
Chapter 3
19
Properties of Matter
• Each sample of matter is distinguished by its characteristics.
• The characteristics of a substance are called its properties.
Tro's "Introductory Chemistry",
Chapter 3
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Types of Properties of Matter
• Physical Properties
Properties of matter that can be observed
without changing its composition.
e.g., coke is dark brown
• Chemical Properties
Properties of matter that can be observed only
when matter changes its composition.
e.g., gasoline is a very flammable liquid
Tro's "Introductory Chemistry",
Chapter 3
21
Some Physical Properties Mass Volume Density
Solid Liquid Gas
Melting point Boiling point Volatility
Taste Odor Color
Texture Shape Solubility
Electrical
conductance
Thermal
conductance Magnetism
Malleability Ductility Specific heat
capacity
Tro's "Introductory Chemistry",
Chapter 3
22
Example: Some Physical Properties of Iron
• Iron melts at 1538 °C
• Iron boils at 4428 °C.
• Iron’s density is 7.87 g/cm3.
• Iron conducts electricity, but not as well as most other common metals.
Tro's "Introductory Chemistry",
Chapter 3
23
Some Chemical Properties
Acidity Basicity
Causticity Corrosiveness
Reactivity Stability
Inertness Explosiveness
Flammability Combustibility
Oxidizing ability Reducing ability
Tro's "Introductory Chemistry",
Chapter 3
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Example: Some Chemical Properties of Iron
• Iron is easily oxidized in
moist air to form rust.
• When iron is added to
hydrochloric acid, it produces
a solution of ferric chloride
and hydrogen gas.
• Iron is more reactive than
silver, but less reactive than
magnesium.
Tro's "Introductory Chemistry",
Chapter 3
25
Example—Decide Whether Each of the Observations
About Table Salt Is a Physical or Chemical Property
• Salt is a white, granular solid.
• Salt melts at 801 °C.
• A liquid burns with a blue flame.
• 36 g of salt will dissolve in 100 g of water.
• Salt solutions conduct electricity.
• When a clear, colorless solution of silver nitrate is added to a salt solution, a white solid forms.
• When electricity is passed through molten salt, a gray metal forms at one terminal and a yellow-green gas at the other.
physical
physical
chemical
chemical
physical physical
chemical
Tro's "Introductory Chemistry",
Chapter 3
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Changes in Matter
• Changes that alter the state or appearance of
matter without altering its composition are called
physical changes.
• Changes that alter the composition of matter are
called chemical changes.
Tro's "Introductory Chemistry",
Chapter 3
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Physical Changes in Matter
• Physical Changes—Changes that do not affect composition of matter.
Heating water.
Raises its temperature, but it is still water.
Evaporating butane from a lighter.
Dissolving sugar in water.
Even though the sugar seems to disappear, it can easily be separated back into sugar and water by evaporation.
Tro's "Introductory Chemistry",
Chapter 3
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Changes in Matter, Continued
• Chemical Changes— involve a
change in composition.
A chemical reaction.
Silver combines with sulfur in the air to make
tarnish.
Rusting is iron combining with oxygen to
make iron(III) oxide.
Burning results in butane from a lighter to
be changed into carbon dioxide and water.
Tro's "Introductory Chemistry",
Chapter 3
29
Is it a Physical or Chemical Change?
• A physical change results in a different form of the same substance.
The kinds of molecules don’t change. i.e., composition stays the same .
• A chemical change results in one or more completely new substances.
The new substances have different molecules than the original substances. i.e., composition changes.
Appearance may or may not change.
You will observe different physical properties because the new substances have their own physical properties.
Tro's "Introductory Chemistry",
Chapter 3
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Phase Changes Are
Physical Changes
• Vaporizing = liquid to gas.
• Melting = solid to liquid.
• Subliming = solid to gas.
• Freezing = liquid to solid.
• Condensing = gas to liquid.
• Deposition = gas to solid.
• Changes in the state of matter require heating or cooling the substance.
Tro's "Introductory Chemistry",
Chapter 3
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Example—Classify Each Change as Physical
or Chemical, Continued
• Evaporation of rubbing alcohol.
• Sugar turning black when heated.
• An egg splitting open and spilling out.
Physical change
Chemical change
Physical change
Tro's "Introductory Chemistry",
Chapter 3
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Practice—Classify Each Change as Physical
or Chemical
• Sugar fermenting.
• Bubbles escaping from soda.
• Bubbles that form when hydrogen peroxide is
mixed with blood.
Tro's "Introductory Chemistry",
Chapter 3
33
Separation of Mixtures • Separating mixtures based on different
physical properties of their components.
Decanting Density
Evaporation Volatility
Chromatography Adherence to a surface
Filtration State of matter (solid/liquid/gas)
Distillation Boiling point
Technique Different Physical Property
Decanting Density
Evaporation Volatility
Chromatography Adherence to a surface
Filtration State of matter (solid/liquid/gas)
Distillation Boiling point
Technique Different Physical Property
Tro's "Introductory Chemistry",
Chapter 3
34
Distillation
Tro's "Introductory Chemistry",
Chapter 3
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Filtration
Tro's "Introductory Chemistry",
Chapter 3
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Law of Conservation of Mass
• The law of conservation of mass states: “Matter is neither created nor destroyed in a chemical reaction.”
• The total amount of matter present before a chemical reaction is always the same as the total amount after.
• Example: 58 grams of butane burns in 208 grams of
oxygen to form 176 grams of carbon dioxide and 90 grams of water.
butane + oxygen carbon dioxide + water
58 grams + 208 grams 176 grams + 90 grams
266 grams = 266 grams
Tro's "Introductory Chemistry",
Chapter 3
37
Practice—A Student Places Table Sugar and
Sulfuric Acid into a Beaker and Gets a Total
Mass of 144.0 g. Shortly, a Reaction Starts that
Produces a “Snake” of Carbon Extending from
the Beaker and Steam Is Seen Escaping. If the
Carbon Snake and Beaker at the End Have a
Total Mass of 129.6 g, How Much Steam Was
Produced?
Tro's "Introductory Chemistry",
Chapter 3
38
Energy
• Energy is anything that has the capacity to do
work.
• Unlike matter, energy does not have mass and
does not occupy any space.
• Although chemistry is the study of matter, matter
is effected by energy.
It can cause physical and/or chemical changes in
matter.
Tro's "Introductory Chemistry",
Chapter 3
39
Law of Conservation of Energy
• “Energy can neither be created nor destroyed.”
• The total amount of energy in the universe is constant. There is no process that can increase or decrease that amount.
• However, we can transfer energy from one place in the universe to another, and we can also change energy from one form to another.
Tro's "Introductory Chemistry",
Chapter 3
40
Kinds of Energy
Kinetic and Potential
• Potential energy is energy that is stored. Water flows because gravity pulls it
downstream. However, the dam won’t allow it to
move, so it has to store that energy.
• Kinetic energy is energy of motion, or energy that is being transferred from one object to another. When the water flows over the dam,
some of its potential energy is converted to kinetic energy of motion.
Tro's "Introductory Chemistry",
Chapter 3
41
Some Forms of Energy • Electrical Energy
Kinetic energy associated with the flow of electrical charge.
• Heat or Thermal Energy
Kinetic energy associated with molecular motion.
• Light or Radiant Energy
Kinetic energy associated with energy transitions in an atom.
• Nuclear Energy
Potential energy stored in the nucleus of atoms.
• Chemical Energy
Potential energy in compounds.
Tro's "Introductory Chemistry",
Chapter 3
42
Converting Forms of Energy • When water flows over the dam, some of its
potential energy is converted to kinetic energy. Some of the energy is stored in the water because it is
at a higher elevation than the surroundings.
• The movement of the water is kinetic energy.
• Along the way, some of that energy can be used to push a turbine to generate electricity. Electricity is one form of kinetic energy.
• The electricity can then be used in your home. For example, you can use it to heat cake batter you mixed, causing it to change chemically and storing some of the energy in the new molecules that are made.
Tro's "Introductory Chemistry",
Chapter 3
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How do We Measure Energy?
Units of Energy
• The SI unit for energy is the joule (J)
Other units such as calorie, Calorie, KWh are widely used
• calorie (cal) is the amount of energy needed to raise
one gram of water by 1 °C.
1 kcal = energy needed to raise 1000 g of water 1 °C.
1 food calories (Cal) = 1 kcals.
Energy Conversion Factors
1 calorie (cal) = 4.184 joules (J)
1 Calorie (Cal) = 1000 calories (cal)
1 kilowatt-hour (kWh) = 3.60 x 106 joules (J)
Tro's "Introductory Chemistry",
Chapter 3
44
Some Common Uses of Energy Use
Unit
Energy Required
to Raise
Temperature of 1 g
of Water by 1°C
Energy
Required to
Light 100-W
Bulb for 1
Hour
Energy
Used by
Average
U.S. Citizen
in 1 Day
joule (J) 4.18 3.6 x 105 9.0 x 108
calorie (cal) 1.00 8.60 x 104 2.2 x 108
Calorie (Cal) 1.00 x 10-3 86.0 2.2 x 105
kWh 1.1 x 10-6 0.100 2.50 x 102
Example 3.5—Convert 225 Cal to Joules
225 Cal = 9.41 x 105 J Round: 6. Significant figures and
round.
Solution: 5. Follow the solution map to
Solve the problem.
Solution
Map:
4. Write a Solution Map.
1 Cal = 1000 cal
1 cal = 4.184 J
Conversion
Factors:
3. Write down the appropriate
Conversion Factors.
? J Find: 2. Write down the quantity
you want to Find and unit.
225 Cal Given: 1. Write down the Given
quantity and its unit.
Cal
cal 1
J .1844
J 1041.9cal 1
J .1844
Cal 1
cal 1000Cal 252 5
cal J
Cal 1
cal 1000
3 sig figs
3 significant figures
46
Practice 1: The complete combustion of a wooden
match produces 512 cal of heat. How many
kilojoules are produced?
Practice 2: An energy bill indicates that the
customer used 955 KWh. How many calories did
the customer use?
Answer: 8.22 x exp(8) cal.
Answer: 2.14 kJ
Tro's "Introductory Chemistry",
Chapter 3
47
Practice 1: The complete combustion of a wooden match
produces 512 cal of heat. How many kilojoules are
produced?
Answer: 2.14 kJ
48
Practice 2: An energy bill indicates that the
customer used 955 KWh. How many
calories did the customer use?
Answer: 8.22 x exp(8) cal.
Tro's "Introductory Chemistry",
Chapter 3
49
Energy Changes
• Processes that involve energy
changes can be:
Exothermic
Endothermic
Tro's "Introductory Chemistry",
Chapter 3
50
Exothermic Processes
• When a change results in the release of energy it is
called an exothermic process.
• An exothermic chemical reaction occurs when the
reactants have more chemical potential energy
than the products.
• The excess energy is released into the surrounding
materials, adding energy to them.
Often the surrounding materials get hotter from the
energy released by the reaction.
Tro's "Introductory Chemistry",
Chapter 3
51
An Exothermic Reaction
Pote
nti
al e
ner
gy
Reactants
Products
Surroundings
reaction
Amount
of energy
released
Tro's "Introductory Chemistry",
Chapter 3
52
Endothermic Processes
• When a change requires the absorption of energy
it is called an endothermic process.
• An endothermic chemical reaction occurs when
the products have more chemical potential energy
than the reactants.
• The required energy is absorbed from the
surrounding materials, taking energy from them.
Often the surrounding materials get colder due to the
energy being removed by the reaction.
Tro's "Introductory Chemistry",
Chapter 3
53
An Endothermic Reaction
Pote
nti
al e
ner
gy
Products
Reactants
Surroundings
reaction
Amount
of energy
absorbed
Tro's "Introductory Chemistry",
Chapter 3
54
Temperature Scales
• Fahrenheit scale, °F.
Used in the U.S.
• Celsius scale, °C.
Used in all other countries.
A Celsius degree is 1.8
times larger than a
Fahrenheit degree.
• Kelvin scale, K.
Absolute scale.
Temperature Scales
Celsius Kelvin Fahrenheit -273°C -269°C
-183°C
-38.9°C
0°C
100°C
0 K 4 K
90 K
234.1 K
273 K
373 K
-459 °F -452°F
-297°F
-38°F
32°F
212°F
Absolute
zero
BP helium
Boiling
point
oxygen
Boiling
point
mercury
Melting
point ice
Boiling
point water
0 R 7 R
162 R
421 R
459 R
671 R
Rankine
Room temp 25°C 298 K 75°F 534 R
Tro's "Introductory Chemistry",
Chapter 3
56
The Fahrenheit Temperature Scale
• The Fahrenheit temperature scale was setup
by assigning 0 °F to the freezing point of
concentrated saltwater and 96 °F for normal
body temperature.
• Room temperature is about 72 °F.
Tro's "Introductory Chemistry",
Chapter 3
57
The Celsius Temperature Scale
• Was setup by assigning 0 °C to the freezing
point of distilled water and 100 °C to the
boiling point of distilled water.
Most commonly used in science.
More reproducible standards.
• Room temperature is about 22 °C.
Tro's "Introductory Chemistry",
Chapter 3
58
The Kelvin Temperature Scale
• Both the Celsius and Fahrenheit scales have negative numbers.
• The Kelvin scale avoids negative numbers and therefore the lowest temperature is 0 K.
• 0 K is called absolute zero. It is too cold for matter to exist because all molecular motion would stop.
0 K = -273 ° C = -459 °F.
Absolute zero is a theoretical value obtained by following patterns mathematically.
Tro's "Introductory Chemistry",
Chapter 3
59
• Celsius to Kelvin
K = ° C + 273
• Fahrenheit to Celsius
1.8
32-F C
Converting from one temperature
scale to another
Example 3.7—Convert –25 °C to Kelvins
248 K Round: 6. Significant figures and
round.
Solution: 5. Follow the solution map to
Solve the problem.
Solution
Map:
4. Write a Solution Map.
Equation: 3. Write down the appropriate
Equations.
K ? Find: 2. Write down the quantity
you want to Find and unit.
-25 °C Given: 1. Write down the Given
quantity and its unit.
° C K
273C K
K 248273C) 25(K
K = ° C + 273
Example 3.8—Convert 55° F to Celsius
Units and magnitude are correct.
Check: 7. Check.
12.778 °C = 13 °C Round: 6. Significant figures and
round.
Solution: 5. Follow the solution map to
Solve the problem.
Solution
Map:
4. Write a Solution Map.
Equation: 3. Write down the appropriate
Equations.
° C ? Find: 2. Write down the quantity
you want to Find and unit.
55 °F Given: 1. Write down the Given
quantity and its unit. units place
and 2 sig figs
units place and 2 sig figs
° F ° C
1.8
32-F C
1.8
32-F C
C 778.12
1.8
32-F 55 C
Example 3.9—Convert 310. K to Fahrenheit
Units and magnitude are correct.
Check: 7. Check.
98.6 °F = 99 °F Round: 6. Significant figures and
round.
Solution: 5. Follow the solution map to
Solve the problem.
Solution
Map:
4. Write a Solution Map.
Equation: 3. Write down the appropriate
Equations.
°F ? Find: 2. Write down the quantity
you want to Find and unit.
310 K Given: 1. Write down the Given
quantity and its unit. units place
and 3 sig figs
units place and 2 sig figs
1.8
32-F C
32C1.8 F
F 6.9832C 37.81 F
K = °C + 273
°F °C K
°C = K - 273
C 37273310 C
Tro's "Introductory Chemistry",
Chapter 3
63
Practice 1- A sick child has a body
temperature of 40.00 °C. What is the child’s
temperature in Kelvins (K) ?
Practice 2- During one summer day in
Gettysburg, a record temperature of 92.1 ° F
was reported. What is the temperature in K ?
Tro's "Introductory Chemistry",
Chapter 3
64
Practice 1- A sick child has a body
temperature of 40.00 °C. What is the child’s
temperature in Kelvins (K) ?
Tro's "Introductory Chemistry",
Chapter 3
65
Practice 2- During one summer day in
Gettysburg, a record temperature of 92.1 ° F
was reported. What is the temperature in K ?
Tro's "Introductory Chemistry",
Chapter 3
66
Energy and the Temperature of Matter
• The increase in temperature of an object
depends on the amount of heat energy
added (q) and the mass of the object.
If you double the added heat energy the
temperature will increase twice as much.
If you double the mass, it will take twice as
much heat energy to raise the temperature the
same amount.
67
Heat Capacity • Heat capacity is the amount of heat a substance
must absorb to raise its temperature by 1 °C.
cal/°C or J/°C.
Metals have low heat capacities; insulators have
high heat capacities.
• Specific heat = heat capacity of 1 gram of the
substance.
cal/g°C or J/g°C.
Water’s specific heat = 4.184 J/g°C for liquid.
Or 1.000 cal/g°C.
Tro's "Introductory Chemistry",
Chapter 3
68
Specific Heat Capacity • Specific heat is the amount of energy required to raise
the temperature of one gram of a substance by 1 °C.
• The larger a material’s specific heat is, the more energy it takes to raise its temperature a given amount.
• Like density, specific heat is a property of the type of matter.
It doesn’t matter how much material you have.
It can be used to identify the type of matter.
• Water’s high specific heat is the reason it is such a good cooling agent.
It absorbs a lot of heat for a relatively small mass.
Tro's "Introductory Chemistry",
Chapter 3
69
Specific Heat Capacities Substance Specific Heat
J/g°C
Aluminum 0.903
Carbon (dia) 0.508
Carbon (gra) 0.708
Copper 0.385
Gold 0.128
Iron 0.449
Lead 0.128
Silver 0.235
Ethanol 2.42
Water (l) 4.184
Water (s) 2.03
Water (g) 2.02
Tro's "Introductory Chemistry",
Chapter 3
70
Heat Gain or Loss by an Object
• The amount of heat energy gained or lost by an object
depends on 3 factors:
how much material there is
what the material is
how much the temperature changed
Amount of Heat = Mass x Specific Heat Capacity x Temperature Change
q = m x C x DT
J 557.4
C25.0-29.9 0.372g 2.5 Cg
J
q
q
Example 3.10—Calculate Amount of Heat Needed to Raise Temperature of 2.5 g Ga from 25.0 to 29.9 °C
Units and magnitude are
correct. Check: 7. Check.
4.557 J = 4.6 J Round: 6. Significant figures and
round.
Solution: 5. Follow the solution map to
Solve the problem.
Solution
Map:
4. Write a Solution Map.
Equation: 3. Write down the appropriate
Equations.
q, J Find: 2. Write down the quantity
you want to Find and unit.
m = 2.5 g, T1 = 25.0 °C,
T2= 29.9 °C, C = 0.372 J/g°C
Given: 1. Write down the Given
quantity and its unit.
2 significant figures
m, C, DT q
TCmq D
TCmq D
Example—Calculate the Amount of Heat Released
When 7.40 g of Water Cools from 49° to 29 °C
q = m ∙ Cs ∙ DT
Cs = 4.18 J/gC (Table 3.4)
The unit and sign are correct.
T1 = 49 °C, T2 = 29 °C, m = 7.40 g
q, J
Check: • Check.
Solution: • Follow the
concept
plan to
solve the
problem.
Solution Map:
Relationships:
• Strategize
Given:
Find:
• Sort
Information
TC mq s Δ
J 106.2J 64.816
C 02- 4.18g 7.40
Δ
2
Cg
J
TCmq s
C 02-
C94 - C 29
12
D
D
T
TTT
Cs m, DT q
Tro's "Introductory Chemistry",
Chapter 3
73
Practice 1- If you hold gallium in your hand, it melts
from body heat. How much heat must 3.5 g of gallium
absorb from your hand to raise its temperature from 25.0
°C to 30.5 °C? The heat capacity of gallium is 0.372
J/g°C.
Practice 2- A bottle containing 24.1 g of an alcohol was
removed from a refrigerator and placed on top of a
counter in a kitchen. The amount of heat absorbed by the
alcohol was determined to be 1.17 kJ. Given that the
specific heat of the alcohol is 2.42 J/g°C
1-Calculate the change in temperature
2-If the initial temperature of the alcohol was 5 °C, find
the final temperature
Tro's "Introductory Chemistry",
Chapter 3
74
Practice 1- If you hold gallium in your hand, it melts
from body heat. How much heat must 3.5 g of gallium
absorb from your hand to raise its temperature from 25.0
°C to 30.5 °C? The heat capacity of gallium is 0.372
J/g°C.
Tro's "Introductory Chemistry",
Chapter 3
75
Practice 2- A bottle containing 24.1 g of an alcohol was
removed from a refrigerator and placed on top of a
counter in a kitchen. The amount of heat absorbed by the
alcohol was determined to be 1.17 kJ. Given that the
specific heat of the alcohol is 2.42 J/g°C
1-Calculate the change in temperature
2-If the initial temperature of the alcohol was 5 °C, find
the final temperature
Tro's "Introductory Chemistry",
Chapter 3
76
Practice 3-
25.3 g of iron was heated to 53.2 °C and let to cool
slowly. If the amount of heat lost by iron during the
cooling process is 353.5 J, find the final temperature of
the iron. The specific heat of iron is 0.499 J/g°C.
Tro's "Introductory Chemistry",
Chapter 3
77
Recommended Study Problems Chapter 3
NB: Study problems are used to check the student’s understanding
of the lecture material. Students are EXPECTED TO BE ABLE
TO SOLVE ALL THE SUGGESTED STUDY PROBLEMS.
If you encounter any problems, please talk to your professor or seek
help at the HACC-Gettysburg learning center.
Questions from text book Chapter 3, p 81
1, 3, 5, 15, 17, 19, 21, 23, 27, 28, 29, 31, 35-37, 39, 41, 44, 45, 47,
51, 52, 55, 56, 57, 60, 63-65, 67, 71, 73, 75, 77, 79, 81, 83, 87, 93,
97, 99, 107, 109, 110
ANSWERS
-The answers to the odd-numbered study problems are found at
the back of your textbook
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