Do Now:
Why is CO2 a gas and H2O a liquid
at room temperature?
What can we do to change H2O to a
gas?
How can we change CO2 to a
liquid?
Heat and the States of Matter
Homework:
Read pages293 – 299
Answer the following questions on
page 299 #4, 6, 7, 9
AND
The even # questions on the handout
Heat
• Def: the total amount
of kinetic energy of the
particles in a sample of
matter
• Units: Joule (J)
• Heat always flows from
objects at higher temps
to objects at lower
temps, until both
objects are at the same
temperature
Temperature
• Def: a measure of the average kinetic energy
of the particles in a sample of matter
(not a form of ENERGY)
Measuring Heat
• Calorimeter is the
device used to measure
the quantity of heat lost
or gained in a reaction
– Endothermic – heat
absorbed
– Exothermic – heat
released
Specific Heat Capacity (c)
Def: the amount of heat needed to raise the
temp of 1 gram of a substance 1 ⁰C
- All substances have different specific heat
capacities depending on their composition
- Water’s specific heat capacity is listed on
Table B
Calculating heat lost or gained
q=mcDt
Where,
q = heat lost or gained (J)
c = specific heat capacity (J/g⁰C)
D t = change in temperature (⁰C)
(final temp - initial temp)
How much heat is required to raise
50.0 g of water from 5.00 ⁰C to 20.0
⁰C?
q=
Q = mcDt
m=
= (50.0 g)(4.18J/g C) (20.0 C – 5.00 C)
c=
= 3138 J
ti =
= 3140 J
tf =
How much heat is required to raise the
temp of 200.0g of water 24.5C?
If 500.0 J of heat are added to 10.0
g of water at 30.0 C. What is the
final temp of the water?
Using a Calorimeter
• While studying heat
exchanges we define
the experiment that is
being performed as the
system, and everything
outside of the
experiment as the
surroundings.
Law of Conservation of Energy
• Energy can neither be
created nor destroyed
in a chemical reaction
but may be converted
from one form to
another.
Heat lost = Heat gained
The picture shows a reaction to
be carried out in a
calorimeter.
The heat released by the
reactants (system) is
absorbed by the
surrounding water.
Measuring the temperature
change of the water will
allow us to calculate the
heat lost during the reaction.
We can calculate the heat loss
using q = mcDt for the
water.
The following is a description of an experiment
used to determine the specific heat of copper.
• In a laboratory
investigation, a 50.0gram sample of copper
is at 100.0°C in a boiling
water bath.
• A Styrofoam cup with a
lid is used as a
calorimeter.
• The cup contains 100.0
grams of distilled water
at 23.2°C.
• The hot copper is
poured into the cup of
water, and the cup is
quickly covered with
the lid.
• A thermometer is
inserted through the lid.
• The coppe rand water
are gently stirred in the
cup.
• The temperature is
checked periodically.
• The highest
temperature noted is
26.3°C.
• In terms of energy flow, explain why the
temperature of the water in the calorimeter
increases.
• Using the information given, complete the
data table provided in your answer booklet.
Calculate the number of joules of heat
gained by the water.
Calculate the specific heat capacity of
the copper.
Homework
Go to Chemthink and complete the
Particulate Nature of Matter
section under the introduction section.
Complete both the tutorial and the
question set
Phases of Matter
The Kinetic Molecular
Theory
• Matter is composed of smaller particles
– Physical properties determined by distance between
particles, and forces (IMF) they exert on one another
• Particles are in constant motion
– Average kinetic energy is determined by temperature
• Collisions between particles transfer energy without a
loss of energy
• As substances change from a solid  liquid  gas,
the amount of motion of the particles increases
• As substances change from solid  liquid  gas, the
particles have enough energy to slowly overcome the
forces of attraction between particles
Solids
•
•
•
•
have a definite shape, definite volume (incompressible)
particles of substance are closely packed together
particles vibrate
2 types of solids
– Crystals – particles of solid arranged in a repeating
geometric pattern
ex) NaCl, Diamond
– Amorphous solids – particles are not arranged in
any order
ex) wax, glass
Liquids
• definite volume, no definite shape
(assume the shape of container, relatively
incompressible)
• particles of a substance not held together
tightly
• particles are in constant motion (can slide
past one another)
Gas
No definite shape, no definite volume (expand and
contract, compressible)
particles are widely separated
Gases exert a pressure by striking the walls of a container
(each collision exerts a force on the container)
Measured using a barometer
Air pressure measured torr the air pressure at sea level is
760 torr, 1 atmosphere, or 101.325 kPa
Volume of a gas can change under different conditions of
temp and pressure
 Standard temp and press (STP) have been assigned
OC and 101.3 kPa (1 atm)
Phase Changes
Solid / Liquid
• Freezing – the physical change of state from a liquid
to a solid
- energy is released
liquid  solid + heat energy
• normal freezing point – the temperature at which a
liquid changes to a solid at 101.3 kPa (normal
atmospheric pressure)
• temperature does not change during phase change
Solid/Liquid phase changes
• Melting – the change of state from a solid to a
liquid
- energy is added to solid, giving the particles
sufficient energy to break free from other
particles, no temperature change
solid + heat energy  liquid
• occurs at the same temp as normal freezing point
Heat changes during
melting/freezing
• Heat of Fusion (Hfus) – the amount of heat energy needed to change
a mass of solid into a certain mass of liquid at constant temp
for H2O (s) Hfus = 334 J/g (table B)
• when a mass of H2O (l) freezes, 334 J/g are released
• when a mass of H2O(s) melts, 334 J/g are absorbed
• to find the amount of heat energy lost or gained during melting or
freezing use:
q = mass x Hfus
If 32 g of ice are melted at 0 C, how
much heat has been absorbed by the
ice?
q = m x Hfus
q = 32 g x 334 J/g
q = 10688 J
q = 11000 J
How many grams of ice at 0C can be
melted to water at 0C by absorbing 3347 J
of heat?
q = m x Hfus
q =m
Hfus
3347 J = m
334 J/g
10.0 g
=m
Why does a balloon stay inflated?
(explain in terms of the behavior of the
molecules that fill the balloon)
Liquid/Gas
• Vaporization – change from a liquid to a gas
– Boiling - change of a liquid to bubbles of vapor that
appear throughout
- temp at which a liquid boils is known as normal boiling
point
– Evaporation – vaporization that takes place at the
surface of a liquid
- occurs at any temp
- particles at surface with greater average kinetic energy
overcome the intermolecular forces of attraction that keep
particles in liquid state, and escape into gaseous state
Liquid/Gas
• Condensation – reverse of vaporization
Gas  liquid + heat
Heat calculations for
boiling/condensation
Heat of vaporization – amount of heat required
to change a mass of a liquid to a mass of gas
at constant temp
* for water Hvap = 2260 J/g (Table B)
at 100C
q = mass x Hvap
How many joules of heat are needed to
change 20. g of H2O at 100C to steam at
100C?
q = m x Hvap
q = 20. g x 2260 J/g
q = 45000 J
But wait! There’s more
Vapor Pressure – the pressure exerted by a gas
from the liquid, when confined above the
liquid at a given temperature
• as temp increases, KE increases,
• more particles in gaseous state,
• vapor pressure increases
• Table H
Vapor pressure and boiling points
• ****Boiling point – the temperature at which
the vapor pressure of a liquid is equal to the
atmospheric pressure
• at high altitudes, atmospheric pressure is lower
than at sea level
 the boiling point is lower
ex) Salt Lake City and Denver, the BP of water is
95C
Heating Curve
Cooling curve
Sublimation
Phase change from a solid to a gas
Occurs in substances with weak
intermolecular forces of attractions and
high vapor pressures
Ex) dry ice
CO2(s) + energy → CO2(g)
Deposition is the opposite process
What would the heating curve look
like for a substance that undergoes
sublimation?