Announcements
Applied Physics
02-14-08
Temperature & Heat
• Florence Henderson (74)
• One new quiz up today. Get to work
on them NOW!
• We will be burning food in lab on Tuesday. You
may want to wear clothes that can get soot on them.
• Homework Problems: 5.11, 5.18, 5.22, 5.26, 5.29
Questions
Temperature
• A method to determine thermal equilibrium.
• A definition requires looking at matter on an
atomic level.
• Are there any?
• Temperature is a measure of the average KE of
the molecules of a substance.
• Two objects with the same temperature have
molecules with the same average KE.
• KE is transfered between objects through
collisions.
Q01
Each molecule of a diatomic gas has mass m1. The gas is at
temperature T. A second gas is monatomic and each atom
has a mass of m2 = 2 x m1.
A)The diatomic molecules are moving twice as fast as the
monatomic atoms.
B)The diatomic molecules are moving half as fast as the
monatomic atoms.
C)The diatomic molecules are moving slower, but more
than half as fast, than the monatomic atoms.
D)The diatomic molecules are moving faster, but not quite
twice as fast, than the monatomic atoms.
•
•
Anders Celsius (Swedish Astronomer)
•
Modern Scale sets 0 oC as the melting point of water and 100 oC
as the boiling point.
•
Step size on the Celsius scale is larger than the step size on the
Fahrenheit scale.
•
Celsius scale is the standard throughout science and most of the
world.
•
•
•
T(oC) = (5/9)[T(oF) - 32o]
Proposed a version of the celsius temperature scale in 1742.
His version had 0 oC as the boiling point of water and 100 oC as
its freezing point.
Average body temp ~ 37 oC
-40 oF = -40 oC
Other Scales
Scales
•
•
Daniel Gabriel Fahrenheit
•
•
Proposed the Fahrenheit temperature scale (1724).
•
•
Historically 100oF was the temperature of the human body.
Invented alcohol thermometer (1709) and mercury thermometer
(1714) and:
Historically 0oF was the temperature of an equal mixture of salt
and ice.
Modern Scale sets 32 oF as the melting point of water and 212 oF
as the boiling point; normally body temperature ~98.6oF
Thermal Expansion
• As average KE increases, things tend to get larger!
• The expansion is generally linear
• !L = L " !T
• Coefficient of Linear Expansion "
• Units 1/ C
o
o
Expansion Example
Expansion Example
What is the change in length of a 3.00 cm long column of
mercury if its temperature changes from 37.0 oC to 40.0 oC,
assuming it is unconstrained?
What is the change in length of a 3.00 cm long column of
mercury if its temperature changes from 37.0 oC to 40.0 oC,
assuming it is unconstrained?
!L = Lo " !T
!L = Lo " !T
Q02
What is the original length (Lo) of the mercury
column? Express your answer in cm.
Expansion Example
Expansion Example
What is the change in length of a 3.00 cm long column of
mercury if its temperature changes from 37.0 oC to 40.0 oC,
assuming it is unconstrained?
What is the change in length of a 3.00 cm long column of
mercury if its temperature changes from 37.0 oC to 40.0 oC,
assuming it is unconstrained?
!L = Lo " !T
Q03
What is the temperature change in the mercury?
Express your answer in oC.
!L = Lo " !T
Q04
What is the linear expansion coefficient for mercury?
Express your answer in 1/oC.
Expansion Example
Expansion Example
What is the change in length of a 3.00 cm long column of
mercury if its temperature changes from 37.0 oC to 40.0 oC,
assuming it is unconstrained?
What is the change in length of a 3.00 cm long column of
mercury if its temperature changes from 37.0 oC to 40.0 oC,
assuming it is unconstrained?
!L = Lo " !T
Q05
What is the change in the length of the mercury
column? Express your answer in cm.
!L = Lo " !T
Lo = 3.00 cm
!T = 40.0 oC - 37.0 oC = 3.0 oC
" = 60 x 10-6 1/oC
!L = (3.00 cm)(60 x 10-6 1/oC)(3.0 oC) = 5.4 x 10-4 cm
Density
• D = mass/volume
• Units gm/cm
• Units kg/m
3
3
1000gm
gm
1000gm
1kg
=
= 10−3 3
=
3
3
3
(100cm)
cm
1, 000, 000cm
1m
1
gm
kg
= 10−3 3
3
m
cm
Heat
• Thermal Energy (Internal Energy)
• This is related to Temperature (remember the
relation to average KE).
• Heat will “flow” from a hotter object (higher
temperature) to a cooler object (lower
temperature).
• In two objects that are at the same Temperature,
heat flow between them is equal.
Calculating Heat Flow
• Specific Heat
Q = s m #T
“Q” is heat
“m” is mass
“#T” is the change in the temperature
“s: is the specific heat
There is a table of “s’s” on page 108 (Table 5.3).
Water has a relatively high specific heat.
Water is a good “storer” of heat energy.
Another Example
Example Problems.
A 50 gm substance loses 23 calories of heat energy when the
temperature falls from 35°C to 20°C. What is the specific
heat of the substance?
m = 50 gm
Q = 23 cal
!T = 35°C - 20°C = 15°C
Q = s m !T
23 cal = s (50 gm)(15 °C)
s = (23 cal)/[(50 gm)(15 °C)]
s = 3.07x10-2 cal/(gm °C)
Another Example
A 5.0 gm piece of food is tested in a calorimeter to ascertain how much heat
energy it contains. If the water in the calorimeter has a mass of 150 gm and
the temperature of the water increases by 3°C, how much heat energy was
released by the burning of the food?
A 5.0 gm piece of food is tested in a calorimeter to ascertain how much heat
energy it contains. If the water in the calorimeter has a mass of 150 gm and
the temperature of the water increases by 3°C, how much heat energy was
released by the burning of the food?
The heat from the burning food is used to raise the temperature of the water.
So to determine the energy released by the food, simply determine the amount
of heat that entered the water.
The heat from the burning food is used to raise the temperature of the water.
So to determine the energy released by the food, simply determine the amount
of heat that entered the water.
Q = s m !T
Q06
What is the mass in this equation, for this particular problem.
Express your answer in gm’s.
Q = s m !T
Q07
What is the specific heat for this particular problem.
Express your answer in cal/(gm oC).
Another Example
Another Example
A 5.0 gm piece of food is tested in a calorimeter to ascertain how much heat
energy it contains. If the water in the calorimeter has a mass of 150 gm and
the temperature of the water increases by 3°C, how much heat energy was
released by the burning of the food?
A 5.0 gm piece of food is tested in a calorimeter to ascertain how much heat
energy it contains. If the water in the calorimeter has a mass of 150 gm and
the temperature of the water increases by 3°C, how much heat energy was
released by the burning of the food?
The heat from the burning food is used to raise the temperature of the water.
So to determine the energy released by the food, simply determine the amount
of heat that entered the water.
The heat from the burning food is used to raise the temperature of the water.
So to determine the energy released by the food, simply determine the amount
of heat that entered the water.
Q = s m !T
Q08
What is the temperature change for this particular problem.
Express your answer in oC.
Another Example
A 5.0 gm piece of food is tested in a calorimeter to ascertain how much heat
energy it contains. If the water in the calorimeter has a mass of 150 gm and
the temperature of the water increases by 3°C, how much heat energy was
released by the burning of the food?
The heat from the burning food is used to raise the temperature of the water.
So to determine the energy released by the food, simply determine the amount
of heat that entered the water.
m = 150 gm
s = 1.0 cal/(gm °C)
!T = 3°C
Q = s m !T
Q = [1.0 cal/(gm °C)](150 gm)(3 °C)
Q = (450 cal)
Q = s m !T
Q09
How much heat goes into the water?
Express your answer in cal.