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```Part One
Heat and
Temperature
What does it mean to have a
temperature of 0 C?
What is
temperature?
Is temperature the same
thing as heat?
Temperature is a measure
of how “hot” or “cold”
something is.
Temperature is measured in
arbitrary units, like
Fahrenheit or Celsius.
Temperature is
proportional to the average
kinetic energy of the
molecules of the substance.
Heat is the thermal energy
transferred from a hot object
to a cold object.
Heat is measured in energy
units -- Joules or calories.
Heat has the symbol q and
is calculated using …
q = mcDT
Quantity
of heat
specific heat
capacity
q = mcDT
mass
temperature
change
Quantity
of heat
specific heat
capacity
q = mcDT
The specific heat
capacity of water
is 4.18 J/gC
How much heat is needed to raise the
temperature of 25.6 grams of water
from 20.0 C to 50.0 C?
q = m c DT
q = (25.6g)(4.18J/gC)(30.0C)
q = 3210 J
What is the final temperature of 27.0
grams of liquid water, initially at 0C,
after it absorbs 700.0 J of energy?
q = m c DT
Hint: start by
solving for DT.
q
DT = m c
Part Two
Phase Changes
We now know that heat is
either absorbed or released
during a phase change.
A process that
absorbs heat is called
endothermic.
A process that gives off
heat is called
exothermic.
Melting (fusion)
Endothermic: Vaporization
Heat is absorbed. Sublimation
Freezing
Exothermic: Condensation
Heat is released. Deposition
Heat is absorbed by the ice.
Ice
And melts.
Heat is absorbed by the ice.
One gram of ice at 0C
absorbs 334 J as it melts
to form water at 0C.
… making liquid water
Heat is
released by
the water as
it freezes.
Ice
water
334 joules is released
when one gram of
water freezes at 0C.
Ice absorbs 334 J per gram
as it melts at 0C
Ice
Water releases 334 J per gram
as it freezes at 0C
Steam
releases
2260 J/g
as it
condenses
at 100 C
Water
absorbs
2260 J/g
as it boils
at 100 C
Hotplate
The heat gained or lost in
phase changes can be
calculated using …
q = mHf
q = mHv
Heat of
fusion
(melting)
Heat of
vaporization
The values for water are …
Hf = 334 J/g
Hv=2260 J/g
Heat of
fusion
(melting)
Heat of
vaporization
How much heat is absorbed
by 150.0 g of ice as it melts
at 0C?
q = m Hf
q = (150.0 g)(334 J/g)
q = 50,100 J or 50.1 kJ
How much heat is released
by 20.0 grams of steam as it
condenses at 100C?
q = m Hv
q = (20.0 g)(2260 J/g)
q = 45,200 J or 45.2 kJ
Part Three
Phase Diagrams
Which phase is in each region?
Pressure
Hint: What happens to ice as temperature increases?
2
1
3
Temperature
The phase
diagram
has three
distinct
regions.
Pressure
The point where all three phases
exist in equilibrium is called the
triple point.
S
L
G
Temperature
At a pressure of 1 atm, most
substances go through all three
phases, as the temperature increases,
S
L
1 atm
G
Temp.
Solids melt to
form liquids,
which
vaporize to
form gases.
S
L
1 atm
G
MP BP
Temp.
Notice the
melting point
and boiling
point.
But the phase diagram for CO2 is a
little different.
Notice that the
triple point is
L
above 1 atm.
5 atm
S
1 atm
G
Temperature
At 1 atm CO2 goes directly from
solid to vapor as the temperature
increases.
S
L
The sublimation
point is –78.5 C
1 atm
G
Temperature
Pressure
What phase change is occurring?
S
L
Melting (fusion)
G
Temperature
Pressure
What phase change is occurring?
S
L
Vaporization
G
Temperature
Pressure
What phase change is occurring?
S
L
Condensation
G
Temperature
Pressure
What phase change is occurring?
S
L
Sublimation
G
Temperature
Pressure
What phase change is occurring?
S
L
G
Temperature
Liquefying a
gas by
increasing
the pressure.
Part Four
Heating and Cooling
Curves
Look at the different regions of
the heating curve for water.
Temp
100
Water
Ice and
water
0
Water and
steam
Steam
Phase changes?
Ice
Time
Part Five
Calorimetry and
Specific Heat
Capacity
Calorimetry is a collection of
laboratory procedures used to
investigate the transfer of heat.
In calorimetry experiments, one
might be looking for a final
temperature or a specific heat
capacity.
What is the law of conservation of
energy?
Energy is neither created nor
destroyed, only changed in form.
The law of conservation of energy
suggests that the heat lost by the
hot object as it cools is equal to the
heat gained by the cool water as it
warms up.
Investigate:
To put it mathematically:
qlost = -qgained
Heat lost by
the hot object
Heat gained by
the cold water
=
And since q = mcDT then
mocDTo = -mwcDTw
Investigate:
The convention for DT is final temperature
minus initial temperature or Tfinal – Tinitial
mhcDTo = -mccDTw
becomes
moc(Tf -Ti) = -mwc(Tf -Ti)
Use your algebra skills, to solve
for Tf , the final temperature.
Specific heat capacity …
• …varies from one substance to
another.
• …a measure of how much heat
something can “hold”.
• …the amount of heat needed to
raise one gram of a substance
by one Celsius degree.
Specific heat capacity lab suggestions:
1. Heat a metal to a known temp.
2. Transfer the metal to a known
quantity of water at a known
temperature.
3. Measure the equilibrium
temperature.
4. Use qlost = -qgained to compute the
specific heat of the metal.
Get the initial
temperature of
the metal.
metal
The temperature
of boiling water.
hotplate
Get initial
temp of
water in
calorimeter
cup.
Transfer the metal to
the calorimeter.
Continue stirring.
Data: Mass of metal
Initial temp of metal
Mass of water
Initial temp of water
Final temp
of water
and metal
qlost = -qgained
mmcmDTm = -mwcwDTw
-mwcwDTw
cm =
mmDTm
Mass of metal
Initial T of metal
Mass of water in calorimeter
Initial T of water
Final T of water and metal
40.0 g
98.0 C
60.0 g
20.0 C
22.9 C
Calculate the specific
heat capacity of the
metal.
Table of selected
specific heats.
What is the
unknown
metal?
Substance
c in
J/g K
Aluminum
Bismuth
Copper
Brass
Gold
Silver
Tin
Zinc
Mercury
Ethanol
Water
Ice
0.900
0.123
0.386
0.380
0.126
0.128
0.233
0.225
0.387
0.140
2.400
4.186
2.050
```