In general, if the temperature of a substance increases, so does its volume. This is
thermal expansion.
Ex. p. 360 paragraph 3 about bridge gaps
Questions: Do all substances go through different amounts of expansion for specific
temperatures? YES
Coefficient of volume expansion – thermal characteristics of materials are indicated
by this coefficient.
 Gases have the largest values
 Liquids have much smaller values than gases (liquids are somewhat different due
to the fact that liquids expand when heated and then cooled)
 Solids have the smallest values
II.
Measuring Temperature
a. thermometers – common thermometers are a glass tube with mercury,
color alcohol, or colored mineral spirits
b. Calibrating thermometers requires fixed temperatures.
Reference Points (all at 1 atmosphere)
Ice Point – zero degrees Celsius
Steam Point – one-hundred degrees Celsius
The scale is then found by dividing the distance between the reference marks
into equally spaced units, called degrees.
c. Temperature units depend on the scale used.
Fahrenheit
Celsius
Kelvin
Celsius – Fahrenheit Conversion
TF = 9/5 TC + 32.0
Temp. Fahrenheit = 9/5 * Temp. Celsius + 32.0
Question:
1. What is the purpose of the 32.0 in the equation? DIFFERENCE BEWEEN THE
ICE POINTS ON THE TWO SCALES
2. What is the purpose of the 9/5Tc in the equations? DIFFERENCE IN SIZE OF A
DEGREE IN EACH SCALE
3. What would the conversion be for Celsius to Fahrenheit? PROVE WITH
ALGEBRA
PROOF:
d. Celsius and Fahrenheit all have positive, zero, and negative measurements
which are good for liquids and solids. Gases typically use a different
measurement for temperature.
e. If gases were to be compressed to a zero volume it would equal (-273.15
C). This is designated for the Kelvin Scale at 0.00 K (abbreviated K =
Kelvin and with no degree sign). This is often called absolute zero, which
has never really been reached.
Temperature Scales and Their Uses
Scale
Ice Point
Fahrenheit
32 F
Celsius
0C
Kelvin
273.15 K
Steam Point
212 F
100 C
373.15 K
Celsius – Kelvin Temperature Conversion
T = Tc = 273.15
Kelvin Temp. = Temp. Celsius + 273.15
Reminder: no degree signs written with Kelvin scale and Kelvin scale is rounded to 273
for most problems
Homework: p.363 #1-5
10.2 Defining Heat
Heat – the energy transferred between objects because of a difference in their
temperature
I.
a. Energy is transferred between substances as heat.
b. Heat always moves from an object at higher temp. to an object at lower
temp. (higher gravitational potential energy to a lower gravitational
potential energy)
Question: Visual Strategy p. 365
Read & Discuss p. 366 paragraph 2
II.
The transfer of energy as heat alters an object’s temp.
a. Thermal equilibrium may be understood in terms of energy exchange
between 2 objects at equal temperature. See Fig. 10-9 This makes the net
transfer between 2 objects is zero.
b. Atoms of all objects are in continuous motion so all objects have some
internal energy.
c. Temperature is a measure of the internal energy so all objects have some
temperature.
III.
Heat is the energy transferred from one object to another because of the
difference between them. When there is no temperature difference between a
substance and its surrounding, no net energy is transferred as heat.
a. Energy transfer depends on difference of the temperature of the two
objects.
b. The greater the temperature difference between 2 objects, the greater the
amount of energy that is transferred between them as heat.
c. Read and Discuss p.367 paragraph 1&2
IV.
Heat has the units of energy.
a. Because heat (like work) is energy in transit, all heat units can be
converted to Joules (J) – the SI unit for energy.
b. Read p.368 paragraph 1&2
c. Friction is just one way of increasing a substance’s internal energy. In the
case of a solid, internal energy can be increased by deforming their
structure .
Ex. rubber band stretching or bending of a paper clip Quick Lab p. 368
Friction
V.
VI.
Internal Energy of Nail
Transfer of Energy to Hand to Feel Heat
Total energy is conserved.
Conservation of Energy
ΔPE + ΔKE + ΔU = O
Potential Energy + Kinetic Energy + Internal energy = 0
Mechanical Energy is often conserved. The sum of KE and all forms of PE
ME = KE + ∑PE
Homework: p.370 #1-5
Sample Problem:
M = 11.5 kg
h = 1.3 m
PEi = mgh
KEi= 0
KEf = 0
PEf = 0
g = 9.81 m/s2
ΔPE + ΔKE + ΔU = 0
PEi + KEi + Ui = PEf + KEf + Uf
mgh + 0 + Ui = 0 + 0 + Uf
mgh = Uf – Ui
(11.5)(9.81)(1.3) = ΔU (Uf – Uf)
For each problem write out the long version to fill out:
PEi + KEi + Ui = PEf + KEf + Uf
1. Read the problem and find what it wants you to know
2. Place zeros in what we don’t want to know
3. look at what you have and fill in the formula
10-3 Changes in temperature & phase
I.
Specific Heat
a. each substance has a unique value for the energy requires to change the
temperature of 1 kg of a substance by 1 degree Celsius = Specific Heat
Capacity
- relates mass, temperature change, and energy transferred
Cp = Q / mΔt
Specific heat = energy transferred as heat / mass X change in temp
b. applies to substances that absorb energy from their surrounding and those
that transfer energy to their surroundings
Δt and Q are positive = energy transfer into a substance (energy gained)
Δt and Q are negative = energy transfer from a substance (energy released)
c. calorimeter – was to determine a substance’s specific heat
II.
Latent Heat
a. energy per unit mass that is transferred during a phase change of a
substance
Q = mL
Energy Transferred = mass X latent heat
as heat during phase
change
III.
Phase Change – physical change of a substance from one state to another at a
constant temperature and pressure (Fig. 10-13 p. 376)
a. Phase changes involve PE between particles.
b. PE is associated with electric forces between charges
c. Particles far apart can break and form based on how close together the
molecules are
d. Increase in PE = breaks bonds
Decrease in PE = new bonds form (will increase KE)
e. Phase changes result from a change in PE between particles of a
substance. When energy is added to or removed from a substance
undergoing a phase change, the particles of the substance rearrange
themselves to make up for their change of energy. This occurs without a
change in the average KE of the particles.
f. Energy required to vaporize a substance mostly goes into separating the
molecules.
IV.
Controlling Heat
a. Thermal Conduction – energy is transferred as heat through a material
between 2 points at different temperatures.
b. Convection – displacement of cold by hot matter involves pressure,
conduction, and buoyancy
Ex. Heat over a burner
1. particles collide more causing it to expand
2. density decreases
3. warm air displaced by denser colder air from above
c. Energy is transferred by thermal conduction through particle collisions.