Matter and Energy
An Introduction
Still Another View
What is Matter?
• Matter
– Anything that
occupies space
and has mass
• Atoms
– Tiny particles too
small to see
• Molecules
– Atoms bonded
together- also too
small to see
Classifying Matter
• States of Matter
• Solid
– Fixed Volume
– Rigid Shape
• Liquid
– Fixed volume
– Assume the shape of the
container
• Gas
– Atoms separated by large
distances
– Assume the volume and shape of
the container
Classifying Matter
Based on Composition
Pure substance
• Matter that has a uniform
and unchanging
composition
Mixture
• Two or more substances
physically placed
together in any
proportion; each retains
its characteristic
properties
Classifying Matter
• Mixtures
– Heterogeneous Mixture
• Two or more regions with
different composition
– Homogeneous Mixture
• Uniform composition
• Elements
– Cannot be broken into simpler
substances
• Compounds
– Composed of two or more
elements
Classifying Matter
Classifying Matter
• Matter
• Pure substance
• Element
• Compound
• Mixture
• Heterogeneous
• Homogeneous
Physical and Chemical Properties
• Physical properties
– No change of composition
• Chemical properties
– Change of composition
Physical and Chemical Properties
• Physical properties
–
–
–
–
–
–
Boiling point
Odor
Taste
Color
Melting point
Density
• Chemical properties
– Oxidation  Rust
– Flammability
– Chemical activity
Matter and Energy
Conservation of Mass
• Antoine Lavoisier
• Law of Conservation
of Mass
• “Matter is neither
created nor
destroyed”
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.
58 grams + 208 grams = 176 grams + 90 grams
266 grams
=
266 grams
Energy
• Energy
• Law of Conservation
of Energy
• “Energy is neither
created nor
destroyed”
Energy
• Energy
– Capacity to do work- what does this mean?
• Types of Energy
–
–
–
–
Kinetic Energy – motion
Potential Energy – position
Electrical Energy – flow of electrons
Chemical Energy – chemical changes
• Units of Energy
– Joule (J)
– Calorie (cal)
Conversion of Energy Units
A candy bar contains 225 Cal of nutritional
energy. How many joules does it contain?
Given: 225 Cal
Find: J
Conversion Factors: 1000 calories = 1 Cal
4.184 J = 1 calorie
Solution Map: Cal  cal  J
225 Cal X 1000 cal X 4.184 J = 9.41 X 105 J
1 Cal
1 cal
Heat (q)
Heat is a common form of energy associated with the
energy transfer that results from thermal differences
between objects or between the system and its surroundings
Heat gained q and change in T are positive
Heat lost
q and change in T are negative
“flows” spontaneously
from higher T 
lower T
“flow” ceases at
thermal equilibrium
EOS
Temperature Changes
• Heat Capacity
– Quantity of heat energy required to
change the temperature of a given
amount of substance by a standard
increment.
• Specific Heat Capacity
–

–
–
Amount of substance = exactly 1 gram
T = exactly 1 degree
Units of heat in joules or calories
Can be used to identify materials
Many metals have
low specific heats.
The specific heat of
water is higher than
that of almost any
other substance.
Calorimeter- Student Style
Calculations
• Specific Heat
– Used to quantify relationship between heat
added and the temperature change
• Equation
Heat =
q
Mass X Heat Capacity X Temperature
Change
= m
X
C
X
T
Suppose that you are making a cup of tea. How much
heat energy will be needed to warm 236 grams of
water (about 8 ounces) from 25 0C to 100 0C ?
235 grams of water  m
25 0C  initial temperature  Ti
100.0 0C  final temperature  Tf
Find: Amount of Heat needed  q
Equation: q = m X C X T
Solution Map: C, m, T  q
Given:
Equation: q = m X C X T
Solution Map: C, m, T  q
C = 4.18 J/g 0C
m = 235 grams
T = 100.0 0C – 25 0C = 75 0C
q
= m X C
X T
= 235 g X 4.18 J/g 0C X 75 0C
= 7.4 X 104 J
A chemistry student finds a shiny rock that she suspects is gold.
She weighs the rock on a balance and obtains the mass, 14.3 g.
She then finds that the temperature of the rock rises from 25 0C
to 52 0C upon absorption of 174 J of heat. Find the heat capacity
of the rock and determine if the value is consistent with the heat
capacity of gold.
Given: m = 14.3 g
q = 174 J
T = 52 0C – 25 0C = 27 0C
Find: C
Equation: q = m X C X T
Solution Map: m, q, T  C
q
C
= m X C X T
=
=
q
m X T
174 J
14.3 g X 27 0C
= 0.45
J
g 0C
Conceptualizing an Exothermic Reaction
Surroundings are at 25 °C
25 °C
Typical situation:
some heat is released
to the surroundings,
some heat is retained
by the solution.
T increases
Hypothetical situation: all heat
is instantly released to the
surroundings. Heat = qrxn
32.2 °C
35.4 °C
In an isolated system, all heat
remains in the system. Maximum
temperature rise.
Heat Capacity of Water
• Water
– High Heat Capacity
– Changing water
temperature requires a lot
of heat energy
– Important property for
living organisms
Matter and Energy
Chapter in Review
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Matter
Classification of Matter
Properties and Changes in Matter
Conservation of Matter
Energy
Heat Capacity
Specific Heat Capacity
Matter and Energy
Chemical Skills
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Classify Matter
Physical and Chemical Properties
Chemical and Physical Changes
Conservation of Mass
Conservation of Energy Units
Energy, Temperature Change, and
Specific Heat Capacity Calculations