SPH3U: Energy, Work, Power
Temperature & Heat
Molecules and Thermal Energy

The kinetic molecular theory of matter
states that matter is made up of particles
(molecules, atoms, ions) that are in constant
random motion. These particles are
continually vibrating, rotating, and colliding
with one another.
Molecules and Thermal Energy

The particles that make up matter have kinetic
energy because of their constant motion. As
particles vibrate, their energy constantly
alternates between kinetic energy and elastic
potential energy. Particles also have potential
energy due to the forces of attraction or
chemical bonds between them.
Molecules and Thermal Energy

Thermal energy is the total kinetic and
potential energy of the particles within an
object. Thermal energy is internal energy –
the energy associated with the movement and
interactions of the particles. An object has
thermal energy in addition to any potential
energy or kinetic energy it may have as a
whole.

For example, a car moving over a bridge has kinetic
energy because it is moving, Gravitational potential
energy because it is high on the bridge, and thermal
energy because of the particles that make up the car.
Temperature

As particles move, they collide, and when they
do, they transfer kinetic energy back and forth
among themselves. In this way, energy can be
transferred from one part of an object to
another, or even from one object to another
object.
Temperature

The temperature of a substance is the
average kinetic energy of the particles. The
greater the average kinetic energy of the
particles, the higher the object’s temperature.

Temperature is measured in either Celsius or
Kelvin. The standard SI unit is the Kelvin (K).
0 C = 273 K
Temperature

Absolute zero is the lowest temperature that is
theoretically possible. It occurs at a
temperature of 0 Kelvin, or -273C. At
absolute zero, molecular motion and energy
would be minimal.
Heat

Heat is the energy transferred from a warmer
object to a cooler one.
Example: When we place hot chocolate in a
mug, the particles in the hot chocolate have a
lot of kinetic energy because they are hot.
They will in turn collide with particles in the
mug, and transfer some of their energy to the
particles in the mug through these collisions.
As a result, over time, the hot chocolate loses
heat and the mug gains heat.
Heat

Heat is always transferred from a hotter object
to a colder object. If two objects are the same
temperature, there will be no heat transfer.
Heat

However, when two contacting surfaces are
rubbed against one another, friction exists as
the particles in the substance to move past
each other and their kinetic energy gets
transferred into thermal energy, causing a rise
in temperature. You can see this when your
rub your hands together – they get warm.
 The conversion of kinetic energy to thermal
energy by friction is why no mechanical
process can be 100% efficient.
Ways to transfer Thermal Energy
Conduction
The transfer of thermal energy through
direct contact of substances. Example:
hot chocolate in a mug
Ways to transfer Thermal Energy
Convection
The transfer of thermal energy through
the bulk movement of particles from
one location to another. This occurs in
liquids and gases. Convection is a
natural stirring of a fluid: hotter, less
dense material flows up while cooler,
denser material flows down.
Eventually all the material is heated to
a uniform temperature. Example:
boiling water.
Ways to transfer Thermal Energy
Radiation
Radiation is the transfer of thermal
energy through fast moving particles
or electromagnetic waves. Example:
heat given off by the sun
Heat Capacity

Materials are affected by thermal energy
differently. Heat capacity (c) is the amount of
thermal energy needed to change the
temperature of an object or system by 1
Kelvin, or 1C.
 You need a different amount of thermal energy
to raise the temperature of water by 1 Kelvin
than you would to heat milk by 1 Kelvin. In
fact, it takes 4190 Joules of energy to raise the
temperature of 1 kg of water by 1 Kelvin,
where as it takes 3930 Joules to raise the
temperature of 1 kg of milk by 1 Kelvin.
Heat Capacity

The amount of thermal energy (Q) you need to
heat an object depends on three things:
 The mass of the object
 How much temperature change you are
trying to get
 The type of material (specific heat
capacity)
Heat Capacity

The following formula is used to relate these
four things:
Q = mcΔT.
(Q = thermal energy, m = mass, c = specific heat
capacity,and ΔT = temperature change)
Example 1

A 515-g granite rock cools from 450C to
100C. The specific heat capacity of
granite is 790. Calculate how much
thermal energy is lost by the rock.
Example 1

A 515-g granite rock cools from 450C to 100C. The specific heat
capacity of granite is 790 J/kgK. Calculate how much thermal energy is
lost by the rock.
ΔT = 350 K
 m = 0.515 kg
 c = 790 J/kgK




Q = mc ΔT
Q = (0.515)(790)(350)
Q = 142397.5 J
Therefore the total
amount of thermal
energy lost by the
rock is approximately
142 000 Joules.
The Principle of Heat Exchange

When two substances at different
temperatures are mixed, the amount of
thermal energy lost by the hotter
substance in cooling is equal to the
amount of thermal energy gained by the
colder substance in warming.
The Principle of Heat Exchange

The thermal energy gained by the cooler
substance equals the thermal energy lost by
the warmer substance.
 If you mix together two substances that are at
initial temperatures T1 and T2, you can use
the principle of heat exchange to predict the
final temperature, Tf.



Q1 + Q2 = 0
m1c1(ΔT1) + m2c2(ΔT2)=0
m1c1(Tf - T1) + m2c2(Tf – T2 )=0
Example 2

An insulated cup containing 255 grams
of water at 21.6C is emptied into
another insulated cup containing 407
grams of water at 63.8C. Determine the
final temperature of the mixture,
assuming that no thermal energy is lost.
Example 2 - Solution
m1 = 0.255 kg
m2 = 0.407 kg
c1 = c2 = c
T1 = 21.6C
T2 = 63.8C
Q1 + Q2 = 0
m1c1(ΔT1) + m2c2(ΔT2)=0
m1c1(Tf - T1) + m2c2(Tf – T2 )=0
m1c(Tf - T1) + m2c2(Tf – T2 )=0
c [m1 (Tf - T1) + m2(Tf – T2 ) ]=0
(0.255) (Tf – 21.6) + (0.407)(Tf – 63.8) =0
0.255Tf – 5.508 + 0.407Tf - 25.966 = 0
0. 662 Tf – 31. 474 = 0
Tf = 47.5 C
Homework – Extra Practice
Attempt the homework questions on the
handout
 Pg. 287 # 1-6
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