Heat
A Form of
Energy
(Thermal
energy)
12.1 Temperature & Thermal Energy
Objectives
• Explain heat using the Kinetic-Molecular
Theory
• Define temperature
• Understand the process of thermal
equilibrium
• Describe and use the Celsius & Kelvin
Temperature Scales
Kinetic-Molecular Theory
It was once common belief that heat was an
invisible substance called “caloric.” It was believed
that it could be transferred between objects. To heat
up an object this caloric had to flow into it. This,
they thought, explained why objects expanded when
heated. (But……could not explain heat coming
from a cold log when it was burned)
The Kinetic-Molecular Theory replaced the
“Caloric Theory” in the 19th century.
It stated that all matter is made up of
atoms/molecules in constant motion. The faster
they move, the hotter an object will be.
Molecules and Motion
• The motion of
molecules produces
heat
• The more motion, the
more heat is generated
Thermal Energy
Thermal Energy (also called Internal
energy) is the energy an object or substance
has due to the kinetic and potential energies
associated with the random motions of all
the particles that make it up.
The hotter something is, the faster
its molecules are moving/vibrating,
the higher its temperature.
Temperature and Heat
• Kinetic energy is the
energy of motion
• Temperature is the
measure of the
average kinetic energy
of an object
Heat Transfer
• The movement of heat
from a warmer object
to a colder one
Forms of heat transfer
• Three forms of heat
transfer:
• Conduction
• Convection
• Radiation
Conduction
• Conduction involves the
transfer of heat through
direct contact
• Heat conductors conduct
heat well, insulators do not
water molecule
iron atom
zoomed in view
Convection
• Takes place in liquids and
gases as molecules move in
currents
• Heat rises and cold settles
to the bottom
Radiation
• Heat is transferred through
space
• Energy from the sun being
transferred to the Earth
Questions
• What are the three types of heat
transfer?
• How is conduction different
from radiation?
What type of heat transfer is
involved?
• Heating a room with a
fireplace
• Egg cooking in a frying
pan
• Roof of a house becoming
hot
What type of heat
transfer?
• Warm air mass bringing
a change in the
weather
• Wire getting hot from
an electric appliance
Question
• How is kinetic energy
related to heat
production?
Thermal Equilibrium
Two bodies are said to be at thermal equilibrium if they
are at the same temperature. If they are at different
temperatures they are not in thermal equilibrium, and
energy is flowing from the hot side to the cold side.
hot
heat
cold
KEhot > KEcold
26 °C
26 °C
KE = KE
No net heat flow
The two purple objects are at the same temp and, therefore
are in thermal equilibrium. There is no net flow of heat
energy here.
Thermometer
• A instrument used to measure
temperature
• Thermometers commonly have
alcohol (with dye) or mercury
• Digital thermometers have replaced
older ones
• A thermometer must be allowed to
come to thermal equilibrium with the
object it is in contact with to
determine its temperature
Temperature Scales
• Fahrenheit: water freezes at 32 °F; boils at 212 °F
• Celsius: water freezes at 0 °C; boils at 100 °C
• Kelvin: water freezes at 273.15 K; boils at 373.15 K
A change of 100 °C corresponds to a change of 180 °F.
This means 1 C° change = 1.8 F° change
Since these scales are linear, and they’re offset by 32 °F,
we get the conversion formula: F = 1.8C + 32
Celsius Scale
• Celsius is the metric
scale for measuring
temperature
• Based on water
freezing at 0ºC and
boiling at 100ºC
Kelvin scale
• The Kelvin scale is a metric
temperature scale measured
in Kelvin units (K) where 0K
is absolute zero (or no
kinetic energy)
• Formula (273+ºC)= Kelvin
One step on the Kelvin scale is the same as one step on
the Celsius scale. These scales are off by 273 K, so:
K = C + 273
Room temperature is around 293 kelvins, which is 20 °C,
or 68 °F.
Absolute zero
• The temperature in
which all molecular
motion stops (0 K)
Questions
• What is the formula for converting a
Celsius temperature to a Kelvin
temperature?
• What is the boiling point of water on
the Kelvin scale?
• What is the freezing point of water on
the Kelvin scale?
Questions
• Describe absolute zero.
• What is absolute zero
on the Celsius scale?
-273 0C
Practice Problems
• Practice Problems 1-3 pg 247
• Due Monday 3/31
12.1 Heat and Thermal Energy
Objectives
• Distinguish heat from thermal energy
• Define Specific Heat
• Calculate temperature changes due to heat
transfer
Measuring Heat
• Addition of heat
• Causes an increase in
temperature
• Removal of heat
• Causes a decrease in
temperature
Calories
• Unit for measuring heat
• The amount of heat
needed to raise 1 gram
of water one degree
Celsius
Temperature Changes
• Joule is another unit for
measuring amount heat
(heat is just energy)
• Mass and type of
substance determine the
amount of temperature
change
Units for Measuring Heat Flow
2) The calorie is also related to the Joule, the
SI unit of heat and energy
• named after James Prescott Joule
• 4.184 J = 1 cal
 Heat Capacity - the amount of heat needed
to increase the temperature of an object
exactly 1 oC
• Depends on both the object’s mass and its
chemical composition
29
Specific Heat
• The ability of a substance to
absorb heat energy (specific heat)
• Different substances absorb heat
at different rates
• The greater the mass of the object
the more heat is absorbed
Heat Capacity and Specific Heat
Specific
Heat Capacity (abbreviated
“C”) - the amount of heat it takes to raise
the temperature of 1 gram of the
substance by 1 oC
• often called simply “Specific Heat”
has a HUGE value, when it is
compared to other substances
Water
31
Heat Capacity and Specific Heat
water, C = 4.18 J/(g oC) in Joules,
and C = 1.00 cal/(g oC) in calories.
Thus, for water:
• it takes a long time to heat up, and
• it takes a long time to cool off!
Water is used as a coolant and regulates
Earths temperatures!
For
32
Heat Capacity and Specific Heat
To
calculate, use the formula:
Q = mass (in grams) x T x C
heat is abbreviated as “q”
T = change in temperature
C = Specific Heat
• Units are either: J/(g oC) or cal/(g
oC)
33
Specific Heat Equation
Q = mC T
Q = thermal energy
m = mass
C = specific heat
T = change in temp
Ex: The specific heat of silicon is 703 J / (kg· ºC). How
much energy is needed to raise a 7 kg chunk of silicon 10 ºC?
answer: Q = 7 kg · 703 J
·10 ºC = 49 210 J
kg· ºC
Note that the units do indeed work out to be energy units.
*Use table 12-1 on p. 248 for some specific heats of different substances.
Questions
• How can heat be
measured?
• What is the unit used
to measure heat?
• What is specific heat?
Homework
• Practice Problems 5-8 page 248
• Due Wed. 4/2/14
*Remember T is the same in oC or K
(eg: a change of 10 oC is a change of 10K)
So you can use specific heats in J/kg-K or J/Kg-C
they are the same number
12.1 Calorimetry: Measuring
Specific Heat
Objectives
• Use the law of conservation of energy to
determine temperature changes using
calorimetry
• Use Calorimetry to determine specific heats
What is a calorimeter?
• Device used to measure
changes in thermal energy
• It can be used to measure
heat given off during
chemical reactions
(measure energy required to
burn off a food product)
Calorimetry
A horseshoe at 275 ºC, is dropped into bucket of water and
covered. The bucket and cover are made of an insulating
material. The bucket contains 2.5 L of water originally at 25 ºC.
The 1.9 kg shoe is made of iron, which has a specific heat of 448 J /
(kg·ºC). Let’s find the temp of the horseshoe and water once
equilibrium is reached.
At thermal equilibrium the water and
shoe are at the same temp. The total
thermal energy in the bucket does not
change, but it is redistributed.
continued on next slide
Calorimetry
(cont.)
Let T = the equilibrium temperature.
Q lost by iron = Q gained by water
or Eiron + Ewater = 0
miron Ciron Tiron + mwater Cwater Twater = 0
miron Ciron (Tf-Ti,iron)+ mwater Cwater (Tf-Ti,water) = 0
We’ve got a simple linear equation in which we
want to solve for Tf. Isolating Tf to one side gives:
miron Ciron Ti,iron + mwater Cwater Ti,water
Tf = ____________________________
miron Ciron+ mwater Cwater
*You can use this for determining Tf of any substance
Calorimetry
Tf =
(cont.)
miron Ciron Ti,iron + mwater Cwater Ti,water
____________________________
miron Ciron+ mwater Cwater
(1.9 kg)(448 J / kg·ºC)(275 ºC) + (2.5 kg)(4186 J / kg·ºC)(25 ºC)
Tf = _____________________________________________________
(1.9 kg)(448 J / kg·ºC) + (2.5 kg)(4186 J / kg·ºC)
Tf = 43.8 0C
We’ve got a simple linear equation in
which we want to solve for Tf. Solving it
gives us Tf = 43.8 ºC. This is the
equilibrium temp--the final temp for both
the shoe and water.
Practice Problems
• Practice Problems 9-12, page 252
• Due Thurs 4/3
12.2 Changes of State
Objectives
• Define heats of fusion and vaporization
• Calculate heat transfers needed to cause
changes of state
States of Matter
• Solid, Liquid and Gas (won’t discuss
plasma’s yet……)
• States change when the bonds between
particles (atoms/molecules) are over come
by the thermal energy of the particles
Heat and Changes of
State (phase change)
• A phase change is a
physical change that
requires a change in heat
energy
• Addition or removal of
HEAT
Latent Heat
The word “latent” comes from a Latin word that means “to lie
hidden.” When a substance changes phases (liquid  solid or
gas  liquid) energy is transferred without a change in
temperature. This “hidden energy” is called latent heat.
For example energy is required to change 0 ºC ice into 0 ºC
water. When frozen, water molecules are in a crystalline
structure, and energy is needed to break this structure. The
energy needed is called the latent heat of fusion.
Additional energy is also needed to change water at 100 ºC to
steam at 100 ºC, and this is called the latent heat of vaporization.
Heat in Changes of State
1. Heat of Fusion (Hf.) = the energy required for
one kilogram of a substance to melt from a
solid to a liquid
Q = mHf. (no temperature change)
Values given in Table 12-2, page 254
2. Heat of Solidification ( - Hf.) = the heat lost
when one kilogram of liquid solidifies (or
freezes) to a solid
Q = - mHf. (no temperature change)
47
Heat in Changes of State
• Heat absorbed by a melting solid is
equal to heat lost when a liquid
solidifies
– Thus, Hfus. = - Hsolid.
• Note Table 12-2, page 254 for the number
values. There is no value listed for the heat
of solidification it is just -Hf
48
Heats of Vaporization and Condensation
• When liquids absorb heat at their boiling
points, they become vapors.
3. Heat of Vaporization (Hv.) = the amount
of heat necessary to vaporize one
kilogram of a given liquid.
Q = mHv. (no temperature change)
• Table 12-2, page 254
49
Heats of Vaporization and Condensation
• Condensation is the opposite of
vaporization.
4. Heat of Condensation (Hcond.) = amount of
heat released when one kilogram of vapor
condenses to a liquid
Q = mHcond. (no temperature change)
• Hcond = - Hv
• . or …..Q = -mHv
50
Summary: Latent Heat Formulas
Q = m Hf or
Q = m Hv
Q = thermal energy
m = mass
H = heat of fusion or vaporization
H is the energy per unit mass needed to change the state
of a substance from solid to liquid or from liquid to gas.
Ex: Hf (the latent heat of fusion) for gold is 6440 J/kg.
Gold melts at 1063 ºC. What is the heat energy needed
to melt 5 grams of solid gold at 1063 0C?
(6440 J/kg) (0.005 kg) = 32 J. The liquid gold will still be
at 1063 ºC.
Latent Heat / Specific Heat Example
Superman vaporizes a 1800 kg ice monster with
his heat ray vision. The ice monster was at
-20 ºC. After being vaporized he is steam at
135 ºC. How much energy did Superman expend?
Substance
Specific Heat (in J / kg·ºC)
ice
2090
liquid water
4186
steam
1970
For water: Hf = 3.33 ·105 J / kg; Hv = 2.26 ·106 J / kg
Q = (1800 kg)(2090 J / kg·ºC)(20 ºC) heating ice to melting pt.
+ (1800 kg)(3.33 ·105 J / kg) ice to water, const. temp of 0 ºC
+(1800 kg)(4186 J / kg·ºC)(100 ºC) heating water to boiling pt.
+ (1800 kg)(2.26 ·106 J / kg) water to steam, const. temp of 100 ºC
+ (1800 kg)(1970 J / kg·ºC)(35 ºC) heating steam to 135 ºC
= 5.62 ·109 J total energy expended by Superman
Practice Problems
• Practice Problems 13-15
• Due Monday 4/7
Laws of Thermodynamics
(examples upcoming)
• Zeroth Law: If object A is in thermal equilibrium with object B,
and if object B is in thermal equilibrium with object C, then objects
A and C are also in equilibrium. This is sort of a “transitive
property of heat.”
• First Law: Energy is always conserved. It can change forms:
kinetic, potential, internal etc., but the total energy is a constant.
Another way to say it is that the change in thermal energy of a
system is equal to the sum of the work done on it and the amount
of heat energy transferred to it.
• Second Law: During any natural process the total amount of
entropy in the universe always increases. Entropy can be defined
informally as a measure of the randomness or disorder in a system.
Heat flows naturally from a hot to cooler surroundings as a
consequence of the second law.
Change in Entropy Equation
Because most systems are many up of so many particles, calculating
entropy via probabilities would be very difficult. Fortunately, we are
normally concerned only with changes in entropy. If we have a
system in which energy is not changing forms, the change in entropy
is defined as:
Q
S =
T
S = change in entropy
Q = change in internal energy (heat flow)
T = absolute temperature
The 2nd Law of Thermodynamics says that during any process:
Suniverse = Ssystem + Ssurroundings  0
Change in Entropy Example
A glass rod is heated and then blown by a
glassblower. When it is at 185°C it is brought
outside to cool. 3200 J of heat are transferred
from the glass to the air, which is at 18°C.
Find the change in entropy of the universe:
Suniverse = Ssystem + Ssurroundings
= Sglass + Sair
Qglass
Qair
+
= T
Tair
glass
-3200 J 3200 J
=
+
458 K 291 K
= -7 J/K + 11 J/K = +4 J/K
Entropy & Fluids
Suppose a beaker of very hot water is poured into an aquarium of cool
water. Conservation of energy would not be violated if all the hot
water remained right at the spot where it was poured. But the 2nd Law
demands that the thermal energy eventually become evenly
distributed. The cool water has molecules moving at a wide range of
speeds (red = fast; blue = slow). Since the water is cool, there are
more blues than reds. The hot water poured in has mostly red. The
aquarium has less disorder (entropy) when all the fast molecules are in
one spot than when they are mixed in. With time a much more likely
situation exists, with a much higher entropy.
continued
time
Heat Engines
A heat engine takes advantage of temp differences to produce useful
work. The amount of work done depends on the size of the reservoirs,
engine efficiency, and the temp difference (TH - TC). QH is the heat that
flows from the hot region; QC is the heat flowing into the cold region.
W is the useful work done by engine. The smaller QC is, the more
efficient the engine is. The engine on the right satisfies the 1st Law but
violates the 2nd Law, i.e., 100% efficiency is unattainable.
Hot Reservoir, TH
QH
Hot Reservoir, TH
QH
W
W
Engine
Engine
QC
QC = 0
Cold Reservoir, TC
Real engine. QH = QC + W
Cold Reservoir, TC
Impossible engine. QH = W
Refrigerators
A refrigerator forces heat from a cold region into a warmer one. It
takes work to do this, otherwise the 2nd Law would be violated. Can
a fridge be left open in the summer to provide a make shift air conditioner? Nope, since all heat pumped out of the fridge is pumped back
into the kitchen. Since QH > QC because of the work done, leaving
the refrigerator open would actually make your house hotter!
Hot Reservoir, TH
QH
Hot Reservoir, TH
QH
W
W=0
Engine
Engine
QC
QC
Cold Reservoir, TC
Real fridge. QC + W = QH
Cold Reservoir, TC
Impossible fridge. QC = QH