Download Chapter 9 Heat

Chapter 9
Heat
Pages 297 – 320
Coach Kelsoe
Chapter 9
Heat
Section One
Temperature and Thermal Equilibrium
Pages 298 - 304
Section One Objectives
Relate temperature to the kinetic energy
of atoms and molecules
 Describe the changes in the
temperatures of two objects reaching
thermal equilibrium
 Identify the various temperature scales,
and convert from one scale to another

Section One Vocabulary
Temperature – a measure of the average
kinetic energy of the particles in a substance
 Internal energy – the energy of a substance
due to both the random motions of its particles
and to the potential energy that results from
the distances and alignments between the
particles
 Thermal equilibrium – the state in which two
bodies in physical contact with each other
have identical temperatures

Section One Formulas

Celsius – Fahrenheit Temperature Conversion
Fahrenheit temperature = (9/5 x Celsius temperature) + 32.0

Celsius – Kelvin Temperature Conversion
T = TC + 273.15
Temperature & Thermal Equilibrium

The temperature of a substance is proportional
to the average kinetic energy of particles in the
substance.
 A substance’s temperature increases as a
direct result of added energy being distributed
among the particles of the substance.
 The energies associated with atomic motion
are referred to as internal energy, which is
proportional to the substance’s temperature.
Temperature & Thermal Equilibrium

Thermal equilibrium is the basis for measuring
temperature with thermometers.
 By placing a thermometer in contact with an
object and waiting until the column of liquid in
the thermometer stops rising or falling, you
can find the temperature of the object.
 The reason is that the thermometer is in
thermal equilibrium with the object.
 Remember that thermal equilibrium is the state
in which two bodies in physical contact with
each other have identical temperatures.
Practice Problem

What is the equivalent Celsius temperature of
50.0°F?
 Given:
TF = 50.0°F
 Unknown:
TC = ?
 Use the Celsius-Fahrenheit equation to
convert Fahrenheit into Celsius.
TF = 9/5TC + 32.0
TC = 5/9 (TF – 32.0)
TC = 5/9 (50.0 – 32.0)°C = 10.0°C
Practice Problem
What is the equivalent Kelvin
temperature of 50.0°F?
 Use the Celsius-Kelvin equation to
convert Celsius into Kelvin.
T = TC + 273.15
T = (10.0 + 273.15)K = 283.2 K

Chapter 9
Heat
Section Two
Defining Heat
Pages 305 - 312
Section Two Objectives
Explain heat as the energy transferred
between substances that are at different
temperatures
 Relate heat and temperature change on
the macroscopic level to particle motion
on the microscopic level
 Apply the principle of energy
conservation to calculate changes in
potential, kinetic, and internal energy

Section Two Vocabulary

Heat – the energy transferred between
objects because of a difference in their
temperatures
Section Two Formula
 Conservation of Energy
PE + KE + U = 0
Defining Heat
 Heat is the energy transferred between objects because
of a difference in their temperatures.
 The word heat is sometimes used to refer to the
process by which energy is transferred between objects
because of a difference in their temperatures.
Heat
 Heat- the energy transferred between
objects because of a difference in their
temperatures
 Energy is transferred between
substances as heat.
 The symbol for heat is Q and is
measured in joules (J).
Internal Energy
 When objects collide inelastically, not all of their
initial kinetic energy remains as kinetic energy after
the collision. Some of the energy is absorbed as
internal energy by the objects.
 Ex. If you pull out a nail after hammering it into wood,
it feels warm. The nail encounters friction as it is
pulled out; the energy required to overcome the
friction is transformed into internal energy. The
increased internal energy raises the nail’s temperature.
 The symbol for a change in internal energy is ΔU.
Conservation of Energy
 ΔPE + ΔKE + ΔU = 0 (change in potential energy +
change in kinetic energy + change in internal energy =
0)
Specific Heat Capacity
 Defined as the energy required to change the
temperature of 1 kg of that substance by 1° C
 Every substance has a unique specific heat capacity.
 c(p) = Q/mΔT (specific heat capacity = energy
transferred as heat / mass × change in temperature)
Calorimetry
 Calorimetry is used to determine specific heat capacity.
 To measure the specific heat capacity of a substance you
must measure the mass, temperature change, and energy
transferred as heat.
 You can find the mass and ΔT directly, but to find the
measurement of heat you have to measure the heat
transferred between the object and a known quantity of
water.
 cp,wmwΔTw = -cp,xmxΔTx (Specific heat capacity of water ×
mass of water × change of temperature of water = - Specific
heat capacity of substance × mass of substance × change of
temperature of substance)
Latent Heat
 When substances melt, freeze, boil, or condense, the
energy added or removed changes the internal energy
without changing the substance’s temperature. These
changes are called phase changes.
 Latent heat is energy transferred during phase
changes.
Practice Problem
 A 61.4 kg roller skater on level ground brakes from 20.5
m/s to 0 m/s. What is the total change in the internal
energy of the system?
 Given:
m=61.4 kg
vi=20.5 m/s
vf=0 m/s
 Find potential and kinetic energy using the givens.
Practice Problem
 There is no potential energy since the roller skater is in
motion.
 To find kinetic energy use the formula:
KE = ½mv2
 Kinetic energy = 12,870.98 J
 PE + KE + U = 0
0 +12,870.98J + U = 0
U = -12,870.98J
Quantities
U change in internal energy
Q heat
W work
eff efficiency
Units
J joules
J joules
J joules
(unit less)






Sign Convention for heat, q
Heat is transferred into the system ? q > 0
Heat is transferred out of the system ? q < 0
Sign Convention for work, w
Work is done upon the system by the surroundings ?
w<0
Work is done by the system on the surroundings ?
w>0

The principle of energy conservation that takes
into account a system’s internal energy as well
as work and heat is called the first law of
thermodynamics.

A total of 135J of work is done on a gaseous
refrigerant as it undergoes compression. If the
internal energy of the gas increases by 114J
during the process, what is the total amount of
energy transferred as heat? Has energy been
added to or removed from the refrigerant as
heat?


Given: W = -135J
Unknown: Q = ?
U = 114J
(since the work is done on the gas, W is a negative value)
First rearrange the equation U=Q-W to Q=
U+W
 Now plug in the values…
Q=114J+(-135J)
Q= -21J (sign for Q is negative, which indicates energy is

transferred as heat from the refrigerant)



The requirement that a heat engine give up
some energy at a lower temperature in order to
do work does not follow the first law.
Therefore this requirement is the basis of what
is called the second law of thermodynamics.
The second law states that no machine can
transfer all of its absorbed energy as work.
When you shuffle a deck of cards, it is highly
improbable that the cards would end up
separated by suit and in numerical sequence.
Such a highly ordered arrangement can
be formed in only a few ways, but there
are more than 8 x 1067 ways to arrange
52 cards.
A measure of the randomness or disorder of a
system
The greater the entropy of
a system is, the greater a
systems disorder!!!!
Once a system has
reached a state of
greatest disorder, it will
tend to remain in that
state and have
maximum entropy.
A small decrease in entropy =
more order or less disorder
A large increase in entropy =
less order or more disorder
DID YOU KNOW?
Entropy decreases in many systems on Earth. For example,
atoms and molecules become incorporated into complex and
orderly biological structures such as cells and tissues. These
appear to be spontaneous because we think of the Earth
itself as a closed system. So much energy comes from the
sun that the disorder in chemical and biological systems is
reduced, while the total entropy of the Earth, sun, and
intervening space increases.
The entropy of
the universe
increases in all
natural
processes.