Chapter #3
Matter and Energy
Matter
•
•
Anything occupying space and having mass.
Matter exists in three states.
 Solid
 Liquid
 Gas
 Plasma
Mixture Separation
Mixtures can be separated based on different
physical properties of the components.
Different Physical Property
Technique
Boiling point
Distillation
State of matter
(solid/liquid/gas)
Filtration
Adherence to a surface
Chromatography
Volatility
Evaporation
Mixture Separation by Boiling Point
Distillation of a Solution Consisting of Salt Dissolved in Water
Mixture Separation by State of Matter
Called Filtration
Separates a liquid from a
solid.
Mixture Separation by Adherence to a surface
This method is called
Chromatography
sand
Law of Conservation of Mass
• Antoine Lavoisier
• “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.
• The total mass of all the reactants is equal to
the total mass of all the products.
Conservation of Mass
• Total amount of matter remains constant in a
chemical reaction.
• 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
Energy
• There are things that do not have mass and
volume.
• These things fall into a category we call energy.
• Energy is anything that has the capacity to do
work.
• Although chemistry is the study of matter, matter
is effected by energy.
– It can cause physical and/or chemical changes in
matter.
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
change its form.
Matter Possesses Energy
• When a piece of matter
possesses energy, it can
give some or all of it to
another object.
– It can do work on the other
object.
• All chemical and physical
changes result in the matter
changing energy.
Kinetic and Potential Energy
• 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.
Forms of Energy
• Electrical
– 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
– Potential energy in the nucleus of atoms.
• Chemical
– Potential energy in the attachment of atoms or because of
their position.
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.
Using Energy
• We use energy to accomplish all kinds of
processes, but according to the Law of
Conservation of Energy we don’t really use it up!
• When we use energy we are changing it from one
form to another.
– For example, converting the chemical energy
in gasoline into mechanical energy to make
your car move.
“Losing” Energy
• If a process was 100% efficient, we could
theoretically get all the energy transformed into
a useful form.
• Unfortunately we cannot get a 100% efficient
process.
• The energy “lost” in the process is energy
transformed into a form we cannot use.
There’s No Such Thing as a Free Ride
• When you drive your car, some of the chemical
potential energy stored in the gasoline is released.
• Most of the energy released in the combustion of
gasoline is transformed into sound or heat energy
that adds energy to the air rather than move your car
down the road.
Units of Energy
• Calorie (cal) is the amount of energy needed to raise
one gram of water by 1 °C.
– kcal = energy needed to raise 1000 g of water 1 °C.
– food calories = kcals.
Energy Conversion Factors
1 calorie (cal)
1 Calorie (Cal)
1 kilowatt-hour (kWh)
=
=
=
4.184 joules (J)
1000 calories (cal)
3.60 x 106 joules (J)
Energy Use
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
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
Unit
joule (J)
Chemical Potential Energy
• The amount of energy stored in a material is its
chemical potential energy.
• The stored energy arises mainly from the
attachments between atoms in the molecules and
the attractive forces between molecules.
• When materials undergo a physical change, the
attractions between molecules change as their
position changes, resulting in a change in the
amount of chemical potential energy.
• When materials undergo a chemical change, the
structures of the molecules change, resulting in a
change in the amount of chemical potential
energy.
Energy Changes in Reactions
• Chemical reactions happen most readily when
energy is released during the reaction.
• Molecules with lots of chemical potential
energy are less stable than ones with less
chemical potential energy.
• Energy will be released when the reactants
have more chemical potential energy than the
products.
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.
An Exothermic Reaction
Surroundings
reaction
Potential energy
Reactants
Amount
of energy
released
Products
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.
An Endothermic Reaction
Surroundings
reaction
Potential energy
Products
Amount
of energy
absorbed
Reactants
Energy and the Temperature of Matter
• The amount the temperature of an object
increases depends on the amount of heat energy
added (q).
– If you double the added heat energy the
temperature will increase twice as much.
• The amount the temperature of an object
increases depending on its mass.
– If you double the mass, it will take twice as
much heat energy to raise the temperature the
same amount.
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.
• It is less for ice and steam.
27
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.
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
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, and how much the
temperature changed.
Practice—Calculate the Amount of Heat Released
When 7.40 g of Water Cools from 49° to 29 °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j
g- °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j 7.40 g
g- °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j 7.40 g 20 °C
g- °C
Practice—Calculate the Amount of Heat Released When
7.40 g of Water Cools from 49° to 29 °C
First use the specific heat of water 4.184 j/g-C and cross
out all units except the heat unit, the joule (j), using our
four step process
4.184 j 7.40 g 20 °C
g- °C
= 620 j
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity?
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity? j/g-°C
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity? j/g-°C
Now we will organize the information in the problem
to give the required units.
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity? j/g-°C
Now we will organize the information in the problem
to give the required units.
442 j
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity? j/g-°C
Now we will organize the information in the problem
to give the required units.
442 j
3.22 g
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity? j/g-°C
Now we will organize the information in the problem
to give the required units.
442 j
3.22 g (57.0 – 20.0) °C
Heat Problems
1. Calculate the specific heat capacity of a 3.22 g
sample of metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
What are the units for specific heat
capacity? j/g-°C
Now we will organize the information in the problem
to give the required units.
442 j
3.22 g (57.0 – 20.0) °C
=3.71j/g- °C
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
1.0 cal
g-°C
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
1.0 cal Cal
g-°C 103cal
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
1.0 cal Cal
g-°C 103cal
88.6 g
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
1.0 cal Cal
g-°C 103cal
88.6 g (97.0-20.0) °C
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories of a potato chip if a 1.22 g
sample of chip causes 88.6 g of water to rise from
20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
1.0 cal Cal
g-°C 103cal
88.6 g (97.0-20.0) °C
1.22 g
Heat Problems
1. Calculate the specific heat of a 3.22 g sample of
metal, if it absorbs 442 j of heat and its
temperature rises from 20.0°C to 57.0°C.
2. Find the food Calories, per gram, of a potato chip if
a 1.22 g sample of chip causes 88.6 g of water to
rise from 20.0°C to 97.0°C.
Using the specific heat capacity of water, we will
organize the units to give Calories.
1.00 cal Cal
g-°C
103cal
88.6 g (97.0-20.0) °C
= 5.59 Cal/g
1.22 g
Calories and Exercise
Activity
Sleeping
Sitting
Walking (2.5 mph)
Cycling (5.5 mph)
Skiing (downhill)
Basketball
Swimming (fast crawl)
Energy Expended (Cal/h)
by 150 lb Adult
80
100
324
330
486
564
636
1.0 lb body fat = 3500 Cal
1.0 Pt of ice cream = 6.00 X 102 Cal
Sample Problem
How far must one walk at 2.5 mi/hr to burn off the
Calories gained by consuming 1.0 pt of ice cream?
Sample Problem
How far must one walk at 2.5 mi/hr to burn off the
Calories gained by consuming 1.0 pt of ice cream?
2.5 mi
Hr
Sample Problem
How far must one walk at 2.5 mi/hr to burn off the
Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
324 Cal
Hr
Sample Problem
How far must one walk at 2.5 mi/hr to burn off the
Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
6.00 X 102 Cal
pt
324 Cal
Hr
Sample Problem
How far must one walk at 2.5 mi/hr to burn off the
Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
6.00 X 102 Cal 1.0 pt
pt
324 Cal
Hr
Sample Problem
How far must one walk at 2.5 mi/hr to burn off the
Calories gained by consuming 1.0 pt of ice cream?
2.5 mi Hr
6.00 X 102 Cal 1.0 pt
pt
324 Cal
Hr
= 4.6 mi.
Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb
of body fat?
Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb
of body fat?
2.5 mi
hr
Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb
of body fat?
2.5 mi hr
hr
324 Cal
Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb
of body fat?
2.5 mi hr
hr
3500 Cal
324 Cal Lb
Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb
of body fat?
2.5 mi hr
hr
3500 Cal 1.0 Lb
324 Cal Lb
Sample Problem
How far must one walk at 2.5 mi/hr to burn off 1.0 Lb
of body fat?
2.5 mi hr
hr
3500 Cal 1.0 Lb
324 Cal Lb
= 27 mi
The End