Chapter #3Matter and Energy
Matter• Anything occupying space and having mass.• Matter exists in three states.
Solid Liquid Gas Plasma
Mixture Separation
EvaporationVolatility
ChromatographyAdherence to a surface
FiltrationState of matter
(solid/liquid/gas)
DistillationBoiling point
TechniqueDifferent Physical Property
Mixtures can be separated based on different physical properties of the components.
Mixture Separation by Boiling PointDistillation of a Solution Consisting of Salt Dissolved in Water
Mixture Separation by State of Matter
Separates a liquid from a solid.
Called Filtration
Mixture Separation by Adherence to a surface
sand
This method is called Chromatography
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 Factors1 calorie (cal) = 4.184 joules (J)1 Calorie (Cal) = 1000 calories (cal)
1 kilowatt-hour (kWh) = 3.60 x 106 joules (J)
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
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
Pot
enti
al e
nerg
y
Reactants
Products
Surroundings
reaction
Amount of energy released
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
Pot
enti
al e
nerg
y
Products
Reactants
Surroundings
reaction
Amount of energy absorbed
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.
27
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.
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 CapacitiesSubstance 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 jg- °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 jg- °C
7.40 g
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 jg- °C
7.40 g 20 °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 jg- °C
7.40 g 20 °C= 620 j
Heat Problems1. 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 Problems1. 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 Problems1. 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 Problems1. 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 Problems1. 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 j3.22 g
Heat Problems1. 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 j3.22 g (57.0 – 20.0) °C
Heat Problems1. 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 j3.22 g (57.0 – 20.0) °C
=3.71j/g- °C
Heat Problems1. 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 Problems1. 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 Problems1. 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 Problems1. 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 103cal
Cal
Heat Problems1. 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 103cal
Cal 88.6 g
Heat Problems1. 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 103cal
Cal 88.6 g (97.0-20.0) °C
Heat Problems1. 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 103cal
Cal 88.6 g (97.0-20.0) °C
1.22 g
Heat Problems1. 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
g-°C 103cal
Cal 88.6 g (97.0-20.0) °C
1.22 g= 5.59 Cal/g
Calories and ExerciseActivity
Energy Expended (Cal/h)by 150 lb Adult
1.0 lb body fat = 3500 Cal
1.0 Pt of ice cream = 6.00 X 102 Cal
Sleeping 80Sitting 100Walking (2.5 mph) 324Cycling (5.5 mph) 330Skiing (downhill) 486Basketball 564Swimming (fast crawl) 636
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 miHr
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 miHr
Hr324 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?
2.5 miHr
Hr324 Cal
6.00 X 102 Calpt
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 miHr
Hr324 Cal
6.00 X 102 Calpt
1.0 pt
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 miHr
Hr324 Cal
6.00 X 102 Calpt
1.0 pt= 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
324 Cal
3500 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
324 Cal
3500 Cal
Lb
1.0 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
324 Cal
3500 Cal
Lb
1.0 Lb= 27 mi
The End