CHAPTER 6: TEMPERATURE AND HEAT
6.1 HEAT AND TEMPERATURE
6.1.1

HEAT
Definition- heat is another name for ‘internal energy’ or energy that is stored inside a
substance. It is a form of energy that moves from one place to another when its
temperatures are different.

Heat is not the same as temperature. Temperature is a measure of how fast molecules are
moving. Heat is the movement energy of all molecules added together.

Heat measured in Joules (J)
6.1.2
TEMPERATURE

Definition – temperature is a measure of how fast molecules are moving.

Temperature is measured in degrees on a temperature scale. The most common
temperature scales are Celcius (Centigrade), Fahrenheit and Kelvin.

On the Celcius scale, the freezing point of water is 00C and its boiling point is 1000C. On the
Fahrenheit scale, these two points are at 320F and 2120F.
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CHAPTER 6: TEMPERATURE AND HEAT
6.2 HEAT TRANSFER

Heat spreads in three ways:
i)
Conduction



Conduction is the spread of heat from atom to atom. The atoms in a hot substance
move fast and hit each other. These collisions pass energy on to nearby atoms, and
those atoms pass their energy on to the next atoms, and so on.
If one end of a metal rod is at a higher temperature, then energy will be
transferred down the rod toward the colder end because the higher speed
particles will collide with the slower ones with a net transfer of energy to the slower
ones.
Example: Heat transfer between two plane surfaces,
ii) Convection



iii)
Convection spreads heat in gases ( and liquids).When they are heated their
molecules speed up, collide more often and spread out further apart. The gas (or
liquid) becomes lighter than the surrounding, cooler gas (or liquid) and floats
upwards.
Convection above a hot surface occurs because hot air expands, becomes less
dense, and rises Hot water is likewise less dense than cold water and rises, causing
convection currents which transport energy.
Examples: Rain process and boiling water
Radiation




Radiation spreads heat by means of invisible infrared (heat) rays.
When electromagnetic waves come in contact with an object, the waves transfer the
heat to that object.
Electromagnetic waves travel through empty space. The sun warms the earth
through the radiation of electromagnetic waves.
Example: sun radiation and home heating lamp
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6.3 HEAT CAPACITY AND SPECIFIC HEAT CAPACITY
6.3.1 HEAT CAPACITY, C

Definition – the amount of heat energy needed to increase the temperature of
an object by 1 oC

Measured in Joule per Kelvin (J/K) or Joule per degrees Celsius (J/°C)

Formula:
𝑯𝒆𝒂𝒕 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒚 =
𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝑶𝒇 𝑯𝒆𝒂𝒕
𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆
𝑪
=
𝑸
𝜽
𝑪
=
𝑸
𝒕𝟐 − 𝒕𝟏
@
𝑯𝒆𝒂𝒕 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒚 = 𝒎𝒂𝒔𝒔 𝒙 𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝑯𝒆𝒂𝒕 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒚
𝑪
=
𝒎 𝒙
𝒄
6.3.2 SPECIFIC HEAT CAPACITY, c

Definition – the quantity of heat energy required to increase its temperature of 1 kg of
the substance by 1 oC.


Measured in (J/kgK) or (J/kg°C)
Formula:
𝑺𝒑𝒆𝒄𝒊𝒇𝒊𝒄 𝑯𝒆𝒂𝒕 𝑪𝒂𝒑𝒂𝒄𝒊𝒕𝒚 =
𝒄
𝒄
=
=
𝑸𝒖𝒂𝒏𝒕𝒊𝒕𝒚 𝒐𝒇 𝑯𝒆𝒂𝒕
𝑴𝒂𝒔𝒔 𝒙 𝑪𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝑻𝒆𝒎𝒑𝒆𝒓𝒂𝒕𝒖𝒓𝒆
𝑸
𝒎𝒙𝜽
𝑸
𝒎 𝒙 ( 𝒕𝟐 − 𝒕𝟏 )
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CHAPTER 6: TEMPERATURE AND HEAT

Example 1:
An object of mass 3.5 kg has absorbed 4500 J of thermal energy when it is heated from
40°C to 48 °C. Calculate:
a) Specific heat capacity,
b) heat capacity
Solution :
a) Specific heat capacity
m  3.5 kg
Q
m x
4500
c
3 .5 x 8
c  160.714 J / kgC
c
Q  4500 J
t1  40 C
t 2  48 C
  48  40  8 C
b) Heat capacity
Q  4500 J
C
t1  40 C
t 2  48 C
  48  40  8 C
C
Q

4500
8
C  562.5 J / C
OR
m  3.5 kg
c  160.714 J / kgC

Cmxc
C  3.5 x 160.714
C  562.499 J / C
Example 2:
The specific heat capacity of metal X is 1200 J/kg°C. What is the heat capacity of a piece of
metal X of mass 300 g?
m  300 g  kg
 0.3 kg
c  1200 J / kgC
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Cmxc
C  0.3 x 1200
C  360 J / C
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CHAPTER 6: TEMPERATURE AND HEAT

Exercise 1:
What is the heat capacity for a 5 kg water? Assume the specific heat capacity of water is
4200 J/kg°C.

Exercise 2:
If 20.5 J of energy is needed to raise the temperature of 150 g of copper from 25 °C to 60 °C,
find the specific heat capacity of copper?

Exercise 3:
The heat energy required to raise the temperature of 400 g of copper by 5 °C is X Joule. If
the specific heat capacity of copper is 380 J/kg°C, calculate the value of X?
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CHAPTER 6: TEMPERATURE AND HEAT
6.4 QUANTITY OF HEAT, Q


Since heat is a form of energy, the units in which a quantity of heat is measured are
fundamentally units of energy.
Formula:
Where is,
𝑸 = 𝒎𝒙𝒄𝒙 𝜽

Q
m
c
θ
=
=
=
=
quantity of heat (J)
mass (kg)
specific heat capacity (J/kg°C)
change in temperature (°C)
Example 1:
The specific heat capacity of an object of mass 1.5 kg is 0.90 kJ/kgoC. Determine the heat
energy of the object to raise the temperature by 1 oC.
Solution :
m  1.5 kg
c  0.9 kJ / kgC
  1 C
Q?

Heat energy, 𝑄 = m x c x 
= 1.5 x 0.9 x 1
= 1.35 kJ
= 1350 J
Example 2:
How much heat is lost when a 640 g piece of copper cools from 26 oC to 15 oC? (The
specific heat of copper is 385 J/Kg oC)
Solution :
m  640 g  kg
 0.64 kg
c  385 J / kgC
t1  26 C, t 2  15 C
  26  15
Heat lost, Q  m x c x
 0.64 x 385 x 11
 2710.4 J
 11 C
Q?
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
Example 3:
The heat energy required to raise the temperature of 0.85 kg iron by 20 °C is 2500 J. What is
the specific heat capacity of iron?
Solution :
m  0.85 kg
c?
  20 C
Q  2500 J

Heat required , Q  m x c x 
2500  0.85 x c x 20
2500  17 x c
c  147.059 J / kgC
Exercise 1:
Water at 40 °C needs 20 kJ of energy to heat the water until 65 °C. Calculate the mass of
water. (Specific heat capacity of water = 4200 J/kg-1C-1)

Exercise 2:
What would be the final temperature of a 300 g sample of copper with an initial temperature
of 80 °C, after it loses 6000 J? (Specific heat capacity of copper = 390 J/Kg°C )
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CHAPTER 6: TEMPERATURE AND HEAT

Exercise 3:
The initial temperature of a glass block with a mass of 2 kg is 45 oC. If it received 20.5 kJ of
heat energy, find the final temperature of glass block? Assume specific heat capacity for
glass is 670 J kg-1C-1.
6.5 THERMAL EQUILIBRIUM
6.5.1 HEAT ENERGY LOST OR GAINED THROUGH CHANGE IN TEMPERATURE

The terms ‘loss’of heat or ‘gain’ of heat is frequently used in connection with heat energy
transfer from and to a body.

This is expressed mathematically by the statement that the quantity of heat given out (‘lost’)
or received (‘gained’) is equal to the product of the mass of the substance, the specific heat
capacity of the substance, and its change in temperature.

Formula:
𝑸𝒍𝒐𝒔𝒕
(𝒎 𝒙 𝒄 𝒙 𝜽)𝒍𝒐𝒔𝒕
mc (
temperature )
=
𝑸𝒈𝒂𝒊𝒏𝒆𝒅
= (𝒎 𝒙 𝒄 𝒙 𝜽)𝒈𝒂𝒊𝒏𝒆𝒅
=
mc (
temperature )
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CHAPTER 6: TEMPERATURE AND HEAT

Example 1:
200 g of hot water at 80 °C is mixed with 100 g of cold water at 20 °C. What is the final
temperature of the water?
Mixture
Cold water
Hot water
m = 200 g  kg
= 0.2 kg
+
m = 100 g
= 0.2 kg
=
c = 4200 J/kg°C
c = 4200 J/kg°C
t1 = 80 °C
t2 = ?
θ = 80 – t2
t1 = 20 °C
t2 = ?
θ = t2 –20
Heat lost by hot water
(m x c θ)hot water
0.2 x 4200 x (80 − t 2 )
840 x (80 − t 2 )
67200 − 840t2
67200 + 8400
75600
75600
1260
t2
=
=
t2 = ?
Heat gained by cold water
(m x c x θ) cold water
=
=
=
=
=
=
=
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0.1 x 4200 x (t 2 − 20)
420 x (t 2 − 20)
420t2 − 8400
4200t2 + 840t2
1260t2
t2
𝟔𝟎°𝐂
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CHAPTER 6: TEMPERATURE AND HEAT

Example 2:
A piece of copper having a mass of 1.5 kg is heated to a temperature of 150oC and is then
dropped into a Thermos flask containing 0.85 kg of water at an initial temperature of 25oC.
What will be the final temperature of water? Assume specific heat capacity of copper is 0.378
kJ/kgoC and no heat is absorbed by Thermos flask.
Solution
copper
mixture
water
m = 0.85 kg
m = 1.5 kg
c = 0.378 kJ/kg°C
= 1378 J/kg°C
+
=
c = 4200 J/kg°C
t1 = 25 °C
t2 = ?
θ = t2 - 25
t1 = 150 °C
t2 = ?
θ = 150 – t2
Heat lost by copper
m x c x   copper
1.5 x 1378 x 150  t 2 
2067 x 150  t 2 
=
t2 = ?
Heat gained by water
=
m x c x  water


310050  2067 t 2
=
310050  89250
=
399300
=
0.85 x 4200 x t 2  25
3570 x t 2  25
3570 t 2  89250
3570 t 2  2067 t 2
5637t 2
399300
5637
=
t2
=
70.836 C
=
=
t2
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CHAPTER 6: TEMPERATURE AND HEAT

Example 3:
An 18 kg piece of metal which has a specific heat capacity of 500 J/kg°C is immersed in
oil. The initial temperature of metal is 150 °C and the initial temperature of oil is 18 °C.
Calculate the mass of oil if the final temperature of the mixture is 40 °C. Assume specific
heat capacity of oil is 1580 J/kgoC.
metal
m  18 kg
c  500 J / kgC
+
oil
m?
c  1580 J / kgC
t1  150 C
t 2  40 C
  150  40
t1  18 C
t 2  40 C
  40  18
 110 C
 22 C
mixture
t2  40C
Heat lost by metal
m x c x   metal
=
=
Heat gained by oil
18 x 500 x 110
=
m x c x   oil
990000
=
m x 1580 x 22
m x 34760
990000
34760
=
m
=
28.481 kg
m

=
Exercise 1:
A 130 g sample of zinc at 75 °C is placed in 105 g of water at 20 °C and the final
temperature of the mixture is 35 °C. What is the specific heat capacity of zinc?(specific
heat capacity of water = 4200 J/kg°C )
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CHAPTER 6: TEMPERATURE AND HEAT

Exercise 2:
A 320 g of water at 15 oC is poured into a Thermos flask containing 600 g of water at 75 oC.
Assuming no heat loss to the surroundings, what is the final temperature of the mixture?
[specific capacity of water = 4200 J/kg oC ]

Exercise 3:
A copper sphere of mass 5 kg is heated for a several minutes before being immersed into 2
kg of water at 30 oC. The final temperature of the water and the copper sphere is 50 oC.
Calculate the temperature of the copper before it was dropped into the water. ( specific
capacity of copper = 400 J/kg oC and specific capacity of water = 4200 J/kg oC)
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CHAPTER 6: TEMPERATURE AND HEAT
6.6.2 MORE THAN 2 SUBSTANCES

Formula :
Qlost by object
( m c θ ) lost

= Q gained by object
= ( m c θ )gained
+ Q gained by object
+ ( m c θ )gained
Example 1:
A metal spoon of mass 16 g is heated in boiling water and subsequently placed in a 70 g
copper calorimeter at a temperature of 28 oC. The calorimeter contains 30 g of water. Upon
stirring, the temperature of the mixture is 39 oC. Determine the specific heat capacity of metal
spoon. Specific heat of water and copper are 4200 J/kg oC and 400 J/kg oC respectively.
Solution:
calorimeter
Wooden stick
&
Water
m = 16 g  kg
= 0.016 kg
m = 70 g  kg
= 0.07 kg
m = 30 g  kg
= 0.03 kg
c=?
c = 400 J/kg oC
c = 4200 J/kog oC
t2 = 39 oC
t1 = 100 oC
t1 = 28 oC
t1 = 28 oC
t2 = 39 oC
t2 = 39 oC
t2 = 39 oC
θ = 100 – 39
θ = 39 – 28
θ = 39 - 28
= 11 oC
= 11 oC
= 61 oC
The mixture
=
+
Qlost by object
=
( m c θ ) stick
0.016 x 61 x c
0.976 c
Q gained by object
= [ ( m c θ )calorimeter ]
+
[ ( m c θ )water ]
= [ 0.07 x 400 x 11 ]
+ [ 0.03 x 4200 x 11]
= 308 + 1386
1694
0.976
c
=
c
= 1736 J/kg°C
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
Example 2:
A 0.5 kg container is filled with 0.4 kg water and its temperature is 15°C. If 0.3 kg water at
60°C is added into the container, what is the final temperature of the mixture? Assume
Specific Heat Capacity of water and container are 4200 J/kg°C and 390 J/kg°C respectively.
Solution:
Container
Water 1
&
Water 2
m = 0.5 kg
m = 0.4 kg
m = 0.3 kg
c = 390 J/kg oC
c = 4200 J/kg oC
c = 4200 J/kog oC
+
t2 = ? oC
t1 = 15 oC
t1 = 15 oC
t1 = 60 oC
t2 = ? oC
t2 = ? oC
t2 = ? oC
θ = t2 – 15
θ = t2 – 15
θ = 60 – t2
Q lost by the object
m x c x   Water 2
0.3 x 4200 x 60  t 2 
The mixture
=
Q gain by the object
=
=
m x c x   container   m x c   Water1
=
0.5 x 390 x t 2  15
195 t 2  15
1260 x 60  t 2  =

195 t 2  2925
 0.4 x 4200 x t 2  15
1680 t 2  15

75600  1260 t 2
=
75600  2925  25200
=
195 t 2  1680 t 2  1260 t 2
103725
=
3135 t 2
103725
3135
=
t2
t2
=
1680 t 2  25200
33.086 C
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CHAPTER 6: TEMPERATURE AND HEAT

Exercise 1:
A 0.1 kg block of iron which is 100 °C is transferred immediately into a cuprum calorimeter
contain of 0.12 kg of water at 14 °C. Given that the mass and the specific heat capacity of
cuprum calorimeter is 0.05 kg and 400 Jkg-1C-1 respectively. If the temperature of
calorimeter and water have increased to 21 °C, determine the specific heat capacity of
iron? ( Specific heat capacity of the water is 4200 J kg-1C-1)
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CHAPTER 6: TEMPERATURE AND HEAT

Exercise 2:
A piece of metal which mass is 0.25 kg is heated to 100°C. Then, the metal is dipped into a
calorimeter filled with water at 30°C. If the final temperature of the mixture are 75°C,
calculate the specific heat capacity of the metal. Given:
mass of water = = 0.4 kg
mass of calorimeter = 0.3 kg
Specific Heat Capacity of water = 4200 J/kg°C
Specific Heat Capacity of calorimeter = 400 J/kg°C
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CHAPTER 6: TEMPERATURE AND HEAT

Exercise 3:
A piece of metal bar which mass is 200 g is heated until its temperature reached 100oC and
subsequently placed in a 500 g calorimeter filled with 300 g of water at 28 oC. Assuming
specific heat of water, calorimeter and metal as 4200 J/kg°C, 400 J/kg°C and 390 J/kg°C
respectively, find the final temperature of the mixture?
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CHAPTER 6: TEMPERATURE AND HEAT
SELF ASSESSMENT
Question 1
Define temperature and heat.
Question 2
Give the definition and formula for heat capacity and specific heat capacity.
Question 3
Explain three methods of Heat Transfer.
Question 4
How much heat must be supplied to a 1.0 kg iron pan to raise its temperature from 20 oC to
100 oC. Given the specific heat capacity of iron is 450 J/kg oC.
Question 5
The heat energy required to raise the temperature of 400 g of copper by 5 oC is 760 J. What is
the specific heat capacity of copper?
Question 6
A 0.25 kg sample of water with an initial temperature of 98.8 oC loses 7500 J of heat. What is
the final temperature of the water? ( Specific heat capacity of water = 4200 J/kg oC)
Question 7
A polystyrene cup is filled with 200 g of water at a temperature of 50 oC. A 10 g of lead block
at 100 oC is placed in the water. What is the temperature of the water when thermal
equilibrium is reached?
Specific heat capacity of water = 4200 Jkg -1 oC-1
Specific heat capacity of lead = 130 Jkg-1 oC-1
Question 8
In an experiment, 50 g of water at 0 oC is mixed with 50 g water at 20 oC. The final
temperature of the mixture is ……………?
Question 9
A container which mass is 0.4 kg is filled with 0.2 kg cold water and their temperature are
10 oC. If 0.3 kg of hot water at 80 oC is added into the container, what is the final temperature
of the mixture? Given the specific heat capacity of the container and water are 390 J/kg oC and
4200 J/kg oC respectively.
Question 10
A block of metal which mass is 0.5 kg is put into a 0.4 kg calorimeter filled with 0.3 kg water.
The initial temperature of the metal is 90 oC and the initial temperature of the calorimeter and
water are 10 oC. If the final temperature of the mixture are 60 oC, determine the specific heat
capacity of the metal?
Specific heat capacity of calorimeter = 400 J/kg oC
Specific heat capacity of water = 4200 J/kg oC
Question 11
The quantity of heat released from 1 kg of liquid is 1.68 x 105 J when its temperature changes by
40 °C. Calculate specific heat capacity of the liquid.
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CHAPTER 6: TEMPERATURE AND HEAT
Question 12
The water in the ice maker of arefrigerator has a mass of 0.4 kg and temperature of water of 22
C. What is the temperature of the water after 33600 J of heat has been removed from it?
Question 13
500 g of liquid A at the temperature of 70 °C is mixed with the liquid B with mass, M at the
temperature 15 °C. If the final temperature of the mixture is 45 °C, ( specific heat capacity of
liquid A = 850 J/kg°C and specific heat capacity of liquid B = 350 J/kg°C)
i) calculate the heat released by liquid A
ii) determine the mass, m of liquid B
Question 14
Calculate the heat quantity needed to raise the temperature of 8 kg water by 40°C. Specific heat
capacity of water is 4.2 kJ/kg°C.
Question 15
Calculate the specific heat capacity of a metal if 880 kJ of heat quantity is needed to raise 2.5 kg
of the metal by 250 °C.
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