INTRODUCTION
When you heat a substance, you are
transferring energy into it by placing it in
contact with surroundings that have a higher
temperature
P-26 SC TEAM
OBJECTIVES

define and distinguish
between various units
of heat

define the mechanical
equivalent of heat
discuss everyday
examples to illustrate
these concepts

WHAT IS HEAT?
Heat is energy that flows from a higher temperature
object to a lower one because of difference in
temperature.
Heat can change the state of substances.
When heated
UNIT OF HEAT
Heat is measured in unit of calorie.
 One calorie is defined as the amount of the
heat energy required to raise the temperature
of 1ºC of 1 gram of water.
 It will take 500 calories to heat 500 grams of
water from 20ºC to 21ºC.
 In SI, heat is measured in unit of joule. One
calorie is equal to 4,184 joule, and it is
frequently rounded up to 4,2 joule.

Unit of HEAT
Heat is measured in Joule (J) or calorie (cal).
1 Joule = 0,24 calorie
1 calorie = 4,186 Joule.
1 kilo calorie = 1 kcal = 4.186 Joule.
If an object has become hotter,
it means that it has gained heat energy.
If an object cools down, it means it has
lost energy
MECHANICAL EQUIVALENT OF HEAT
Joule demonstrated that water
can be heated by doing
(mechanical) work, and
showed that for every 4186 J
of work done, the temperature
of water raise by 1⁰C per kg.
THE THERMAL ENERGY (Q)
Specific heat
capacity (J / kg C)
=
Q
m c T
Mass of
object (kg)
Thermal
energy (J)
The temperature
Change (C)
ΔT = T final - Tinitial
THE AMOUNT OF THERMAL ENERGY (Q)
Greater amount of heat is required to raise the temperature of greater
mass of the material.
The amount of thermal energy is influenced by:
Mass
Type of
object
Temperature
change
The bigger mass the
bigger thermal energy.
The difference object has
difference thermal energy.
The higher temperature change,
the higher thermal energy.


If ΔT is positive, Q is also positive. It means that a
substance has an increase in temperature and gets
the heat energy (receives the heat).
If ΔT is negative, Q is also negative; it means that the
substance has a decrease in temperature and will
lose the heat energy (releases the heat).
EXAMPLE
What is the amount of heat required to raise the
temperature of 2 kg of copper from 30 C to 80 C, given
the specific heat capacity of cooper is 400 J / kg C ?
Solution:
m = 2 kg
T = 80 C – 30 C = 50 C
c = 400 J / kg C
Q = m c T
= 2 kg x 400 J / kg C x 50 C
= 40 000 J
So, the amount of heat is 40 000 J
A silver spoon has a mass
of 32 g and then it is cooled
from 60ºC down to 20ºC.
How much heat is released
by the spoon?
given the specific heat
capacity of silver is 235
J/(kgºC)
EXAMPLE


A silver spoon has a mass of
32 g and then it is cooled
from 60ºC down to 20ºC.
How much heat is released
by the spoon?
Known:
m = 32 g = 0.032 kg
c = 235 J/(kgºC)
TInitial = 60ºC
Tfinal = 20ºC
ΔT = Tf – Ti = -40ºC

Question
Q = ..... ?

Solution: Q = m . c . Δ T
= 0.032 kg. 235 J/(kgºC). -40ºC
= - 301 J
spoon releases the heat of
301 J.
So that the spoon becomes
cooler.
Heat Capacity (C)
Temperature
change by T C
Heat capacity (C) is defined as the
amount of thermal energy required
to raise the temperature of body by
1 K or 1 C.
C= Q
T
Specific Heat Capacity (c)
The amount of thermal energy
needed by 1 kilogram of substance
to increase its temperature up to 1
Kelvin is called specific heat
capacity
m kg
Temperature
changes byT C
c=
Q
m T
SPECIFIC HEAT CAPACITY OF SOME MATERIALS
Material
Aluminum
Brass
Cooper
Ice
Iron
Lead
Specific Heat
capacity
(J / kg C )
920
380
400
2100
460
130
Material
Steam
Mercury
Methylated spirit
Sea water
Water
Zinc
Specific Heat
capacity
(J / kg C )
2010
140
2400
3900
4200
390
Water and other substances that have high specific heat
capacity can absorb more heat energy with less change
of temperature.
LATENT HEAT (L) AND SPECIFIC LATENT HEAT
Latent heat of fusion, (Lf) is heat per unit mass needed to
change a substance from solid to liquid without change in
temperature.
Lf = Q
m
Q = m Lf
Latent heat of vaporization, (LV) is heat per unit mass
needed to change a substance from liquid to solid without
change in temperature.
LV = Q
m
Q = m LV
PHASES OF MATTER

Heat required for phase changes:




Vaporization: liquid  vapour
Melting: solid  liquid
Sublimation: solid  vapour
Heat released by phase changes:



Condensation: vapour  liquid
Fusion: liquid  solid
Deposition: vapour  solid
PHASES OF MATTER
GRAPH OF PHASE CHANGES
Useful data for thermal energy calculation
Specific latent heat of water = 336 000 J/kg
Specific latent heat of steam = 2 260 000 J/kg
Specific heat capacity of water = 4200 J/kg C
Specific heat capacity of ice = 2100 J/kg C
Specific heat capacity of water = 1 cal/gram C
Specific heat capacity of ice = 0,5 cal/gram C
Example:
What is the amount of thermal energy required to change
the state of 1 gram of ice at – 30 C to steam at 120  C?
Solution:
The figure below indicates the experimental results obtained
when energy is gradually added to the ice. Let us examine
each portion of the red curve.
T (C)
120
D
100
C
Steam
Water + steam
B
0
A
Ice +
Water
Water
- 30
Ice
E
Energy added (J)
Part A:
On this portion of the curve, the temperature of the ice changes from
- 30. °C to 0°C. The specific heat of ice is 2100 J / kg C.
Q = m ice c ice T
= (1 x 10 -3 kg) (2100 J/kg C) (0 C – (-30 C))
= 63 J
Part B:
When the temperature of the ice reaches 0.0°C, the ice–water
mixture remains at this temperature - even though energy is being
added - until all the ice melts.
Q = m ice Lf
= (1 x 10-3 kg) (336 000 J/kg)
= 336 J
Part C:
Between 0.0°C and 100.0°C, nothing surprising happens. No phase
change occurs, and so all energy added to the water is used to
increase its temperature. The amount of energy necessary to
increase the temperature from 0.0°C to 100.0°C is:
Q = m water c water T
= (1 x 10-3 kg) ( 4 200 J/kg °C) (100 °C)
= 420 J
Part D:
At 100.0°C, another phase change occurs as the water changes from
water at 100.0°C to steam at 100.0°C.
Q = m water LV
= (1 x 10-3 kg) (2 260 000 J/kg)
= 2 260 J
Part E:
On this portion of the curve, as in parts A and C, no phase change
occurs; thus, all energy added is used to increase the temperature of
the steam. The energy that must be added to raise the temperature of
the steam from 100.0°C to 120.0°C is
Q = m steam c steam T
= (1 x 10-3 kg) (2 010 J/kg °C) (120 – 100 °C)
= (1 x 10-3 kg) (2 010 J/kg °C) (20 °C)
= 40.2 J
The total amount of energy that must be added to change 1 gram of
ice at - 30.0°C to steam at 120 0°C is the sum of the results from all
five parts of the curve.
Q = QA + QB + QC + QD + QE
= 63 J + 336 J + 420 J + 2 260 J + 40.2 J
= 3119.2 J
CALORIMETERS
BLACK’S PRINCIPLE
BLACK’S PRINCIPLE