Physics Notes HEAT MEASUREMENT S4

SSEKWE ROBERT
HEAT MEASUREMENT
Terminologies used:
i) Heat capacity, C:
This is the amount of heat required to raise the temperature of a body by 1K or 10C.
𝑚𝑜 𝑛 𝑜
𝑝 𝑦
𝑛 𝑛 𝑚𝑝
-1
The SI unit of heat capacity is joules per Kelvin (J/K or JK )
Also, 𝑝 𝑦 𝑚 𝑝 𝑝 𝑦
ii) Specific heat capacity, c:

This is the amount of heat required to raise the temperature of 1kg mass of a substance by 1K or
10C.
𝑚𝑜 𝑛 𝑜
𝑝 𝑝 𝑦
𝑀
The SI unit of specific heat capacity is Joules per kilogram per Kelvin (J/kgK or Jkg-1K-1).
Quantity of heat,
Where – change in temperature.
Specific heat capacity
𝑚 Mass of substance.
Note:
N.B: The specific heat capacity is different for different substances and the table below shows
values of specific heat capacities of some common substances.
Substance Specific heat capacity (Jkg-1K-1)
Water 4200
Ice 2100
Aluminium 900
Copper 400
NOTE:
Water has the highest specific heat capacity of 4200Jkg-1K-1. The high specific heat capacity of
water makes water a very good liquid for cooling machines.
Question:
-1 -1
“The specific heat capacity of water is 4200Jkg K ” What is meant by the statement?
This means that 1kg mass of water requires 4200J of heat to raise its temperature by 1K.
ssekwerobert@gmail.com 0752053997 Page 1

SSEKWE ROBERT
Heat calculations
The following should be noted:
Always mass must be in Kilograms (kg)
In questions with the phrase “the temperature rises by” or “the temperature rose by”, the
temperature value given is the change in temperature
Examples:
o o
1. How much heat is required to raise the temperature of 5kg of iron from 30 C to 40 C if the
-1 -1
specific heat capacity of iron is 440Jkg K ?
𝑚
( )
o
2. When a block of iron of mass 2kg absorbs 19kJ of heat its temperature rises by 10 C. Find
the specific heat capacity of iron.
𝑚
3. How much heat is given out when an iron metal of mass 2 𝑘 and specific heat capacity
−1
460 𝐽 𝑘 𝐾−1 cools from 300℃ to 200℃.
𝑚
( )
4. Calculate the specific heat capacity of gold if 108 J of heat raises the temperature of a 9
mass from 0℃ to 100℃.
𝑚 𝑘
𝑚
( )
−1
5. 5KJ of heat is supplied to a metal whose specific heat capacity is 400 𝐽𝑘 𝐾−1, if the
temperature of the metal rises by 5𝐾. Find the mass of the metal.
𝑚
𝑚
𝑚
5kg

SSEKWE ROBERT
6. 1200J of heat is supplied to 100 of water at 20℃. Calculate the final temperature of water if
its specific heat capacity is 4200𝐽𝑘 −1𝐾−1
𝑚 𝑘
𝑚
( )
CALORIMETRY:
This is the measurement of flow of heat.
The instrument used in calorimetry is called calorimeter.
Calorimeter:
 It is made up of copper.
 It is lagged with an insulator and placed in a jacket with a plastic cover which has two holes
for a thermometer and a stirrer.
METHODS OF MEASURING SPECIFIC HEAT CAPACITY:

There are two common methods namely;
 Electrical method.
 Method of mixtures.
Experiment to determine Specific heat capacity of a solid by electrical method.

SSEKWE ROBERT
 A metal block of mass, m whose S.H.C is to be determined is drilled with two holes, one for
thermometer and the other for heater. Both the heater and thermometer must be in good
contact with the block.
 The initial temperature, θ1 of the block is recorded from the thermometer before closing the
switch.
 The heater is then switched on by closing switch, K until the temperature of block changes to
θ2, in a given time, t.
 The ammeter and voltmeter readings I and V respectively are noted and recorded.
Assuming there are no heat losses,
But
( )
But also power,
( )
Example:
1. 98,000J of electrical heat are needed to raise the temperature of 2kg of a substance from
o o
51 C to 65 C. Calculate the specific heat capacity of a substance.
( )
( )
2. A heater rated 2KW is used for heating the solid of mass 6kg, if its temperature rises from
30℃ to 40℃. In 12s, find the S.H.C of the solid.
( )
( )

SSEKWE ROBERT
Experiment to determine Specific heat capacity of a solid by method of mixtures.
Thermometer Stirrer
Lagging
Copper calorimeter
Water Solid
 A solid of mass, 𝑚 whose specific heat capacity, is required is heated to a temperature, .

 A solid is then transferred quickly to a calorimeter of mass, 𝑚 and specific heat capacity
containing water of mass, 𝑚 both at a temperature, .
 The mixture is well stirred until a steady final maximum temperature, is reached.
Assuming there is no heat loss during the experiment,
( ) ( ) ( )
( ) ( )
( )
 Hence specific heat capacity, of a solid can be calculated

SSEKWE ROBERT
Experiment to determine Specific heat capacity of a liquid by method of mixtures.
Thermometer Stirrer
Lagging
Copper calorimeter
Liquid Solid
 A solid of mass, 𝑚 and specific heat capacity, is heated to a temperature, .

 A solid is then transferred quickly to a calorimeter of mass, 𝑚 and specific heat capacity
containing a liquid of mass, 𝑚 whose specific heat capacity is required both at a
temperature, .
 The mixture is well stirred until a steady final maximum temperature, is reached.
Assuming there is no heat loss during the experiment,
( ) ( ) ( )
( ) ( )
( )
 Hence specific heat capacity, of a liquid can be calculated.
Examples:
1. A piece of metal of mass 0.5kg is heated to 100oC and then placed in 0.4kg of water at 10oC,
if the final temperature of the mixture is 30oC. Calculate the specific heat capacity of the
metal. (Neglect heat absorbed by container with water and S.H.C of water is 4200Jkg-1K-1)
( ) ( )

SSEKWE ROBERT
2. The temperature of a piece of copper of mass 250g is raised to 100℃ and it is then
transferred to a well- lagged aluminum can of mass 10.0g containing 120g of methylated
spirit at 10.0℃. Calculate the final steady temperature after the spirit has been well stirred.
Neglect the heat capacity of the stirrer and any losses from evaporation. (S.H.C of copper,
aluminum and spirit respectively = 400 𝐽 𝑘 −1 𝐾−1, = 900 𝐽 𝑘 −1 𝐾−1, = 2400 𝐽 𝑘 −1 𝐾−1)
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
3. A metal of mass 0.2kg at 100℃ is dropped into 0.08kg of water at 13℃ contained in
calorimeter of mass 0.12kg and S.H.C 400Jkg-1K-1. The final temperature reached is 35℃.
Determine the S.H.C of the metal
( ) ( ) ( )
( ) ( ) ( )

SSEKWE ROBERT
4. A liquid of mass 200g in a calorimeter of heat capacity 500 𝐽kg-1𝐾−1 and mass 1kg is heated
such that its temperature changes from 25℃ to 50℃. Find the S.H.C of the liquid if the heat
supplied was 14,000J.
( ) ( )
( ) ( )
o o o
5. 450g of water at 60 C is to be cooled to 35 C by addition of cold water at 20 C.
Calculate the mass of cold water added. (S.H.C of water is 4200 𝐽𝑘 𝐾 )
( ) ( )
6. Hot water of mass 0.4kg at 100℃ is poured into calorimeter of mass 0.3kg and S.H.C of
400Jkg-1K-1 and contains 0.2kg of a liquid at 10℃. The final temperature of mixture is 40℃
determines the S.H.C of a liquid. (S.H.C of water is 4200 𝐽𝑘 𝐾 )
( ) ( ) ( )
( ) ( ) ( )

SSEKWE ROBERT
NOTE: Since mass is proportional to volume, then mass of liquid is equal to its
volume for a unit substance.
o
7. A copper metal of mass 250g is heated to 145 C and then placed in a copper calorimeter of mass
3 o
250g which contains 250cm of water at 20 C. Calculate the maximum temperature attained by
-1 -1
water [specific heat capacity of water is 4200Jkg K and specific heat capacity of copper is
-1 -1
400Jkg K ]
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
PRECAUTIONS TAKEN WHEN USING METHOD OF MIXTURES

 The solid/specimens should be transferred as quickly as possible to the calorimeter to
avoid heat losses.
 The calorimeter must be insulated.
 Stirring must be done to ensure uniform distribution of heat.
 The calorimeter must be polished inside to avoid heat loss by radiation.
EXERCISE
1) A piece of copper of mass 100g is heated to 100℃ and is then transferred to a well lagged
copper can of mass 50g containing 200g of water at 10℃. Neglecting heat loss, calculate the
final steady temperature of water after it has been well stirred. Take S.H.C of copper and
water to be 400Jkg-1K-1 and 4200Jkg-1K-1 respectively. Ans; [14℃]

SSEKWE ROBERT
2) A heating coil is placed in thermal flask containing 0.6kg of water for 600s. The temperature
of water rises by 25℃ during this time. Water is replaced by 0.4kg of another liquid. And the
same temperature rise occurs in 180s. Calculate the S.H.C of the liquid given that S.H.C of
water is 4200Jkg-1K-1. State any assumption. Ans; [1890Jkg-1K-1]
3) Copper calorimeter of mass 120g contains 100g of paraffin at 15℃. If 45g of aluminum at
100℃ is transferred to the liquid and the final temperature is 27℃. Calculate the S.H.C of
paraffin [S.H.C of aluminum and copper are 1000 Jkg-1K-1 and 400 Jkg-1K-1 respectively].
Ans; [ 2.4 x103Jkg-1K-1
4) A liquid of mass 250g is heated to 80℃ and then quickly transferred to a calorimeter of heat
capacity 380JK-1 containing 400g of water at 30℃. If the maximum temperature recorded is
55℃ and specific heat capacity of water is 4200Jkg-1K-1. Calculate the S.H.C of the liquid.
Ans; [8240Jkg-1K-1]
5) 500g of water is put in a calorimeter of heat capacity 0.38JK-1 and heated to 60℃. It takes
2minute for the water to cool from 60℃ to 55℃. When the water is replaced with 600g of a
certain liquid, it takes 1½ minutes for the liquid to cool from 60℃ to 55℃. Calculate the
S.H.C of the liquid.
Ans; [2624.8kgJ-1K-1]
6) 400g of a liquid at a temperature 70℃ is mixed with another liquid of mass 200g at a
temperature of 25℃. Find the final temperature of the mixture, if the S.H.C of the liquid is
4200 𝐽 𝑘 −1 𝐾−1. Ans; [55℃]
7) 60 kg of hot water at 82℃ was added to 300 kg of cold water at 10℃. Calculate the final
temperature of the mixture (S.H.C of water =4200 𝐽 𝑘 −1 𝐾−1) Ans; [=22℃]
8) Calculate the final steady temperature obtained when 0.8 kg of glycerine at 25℃ is put into a
copper calorimeter of mass 0.5 kg at 0℃ ( S.H.C of copper =400 𝐽 𝑘 −1 𝐾−1, . . 𝑜
𝑙𝑦 𝑛 = 250 𝐽 𝑘 −1𝐾−1). Ans; [12.5℃]
9) A block of metal of mass 0.01 kg at a temperature of 100℃ was dropped in a container of

water at 20℃. The final temperature was 40℃. Calculate the S.H.C of the metal ,if
S.H.C of water 4200 𝐽 𝑘 −1 𝐾−1. Ans; [7000 −
K− ]
10) A copper block of mass 250g is heated to a temperature of 145℃ and then dropped into a
copper calorimeter of mass 250g which contains 2500𝑚3 of water at 20℃. Calculate the final
temperature of water. (S.H.C of copper = 400𝐽𝑘 −1 ℃−1, S.H.C of water = 4200 𝐽 𝑘 −1 ℃−1).
Ans; [30℃]
11) The temperature of heat which raises the temperature of 0.1 kg of water from 25℃ to 60℃ is
used to heat a metal rod of mass 1.7 kg and S.H.C of the rod was 20℃. Calculate the final
temperature of the rod. Ans; [48.8℃]

SSEKWE ROBERT
LATENT HEAT (HIDDEN HEAT)

This is the amount of heat required to change the state of substance without change in
temperature.
There are two types of latent heat namely:
 Latent heat of fusion
 Latent heat of vaporization
KINETIC THEORY EXPLANATION OF LATENT HEAT

Question: Explain why during change of state, the temperature of a substance remains constant
 According to kinetic theory, when a substance is changing state, there is no change in
temperature because all the heat supplied is only used to break the intermolecular forces
between the molecules and increase the molecular spacing of the substance.
Latent heat of fusion:

This is the amount of heat required to change a substance from solid state to liquid state at constant
temperature.
Specific latent heat of fusion ( ):

This is the amount of heat required to change 1kg mass of substance from solid state to liquid
state at constant temperature.
The SI unit of specific latent heat of fusion is
Examples:
1. The specific latent heat of fusion of ice is 340,000 . What do you understand by this
statement?
It means that 1kg of ice needs 34,000J of heat energy to change to a liquid.
2. How much heat is needed to melt 10g of ice at 0oC? [Specific latent heat of fusion of ice =
3.36 x 105Jkg-1]

SSEKWE ROBERT
Experiment to determine specific latent heat of fusion of ice by electrical method.
Switch
A
V
Electrical heater
Pure melting ice
Funnel
Beaker
Collected water from melted ice
 An electrical heater is placed in a funnel.

 Small pieces of ice are packed around the heater.
 The heater is then switched on for a known time, t.
 The ammeter and voltmeter readings I and V respectively are noted and recorded.
 The mass, m of collected water from melted ice is measured and recorded
 Assuming there are no heat losses in the experiment,
 Hence specific latent heat of fusion of ice 𝐿 can be calculated.

SSEKWE ROBERT
Examples;
o
1. A 3kW electrical heater is left for 2 minutes in a container packed with ice at 0 C. If 100g
of ice melted into water, calculate the specific latent of fusion of Ice.
Note:
If the ice is not at its melting point , the heat supplied first increases/raises its temperature
to .
𝑜 ,
o o
2. How much heat is needed to melt 10g of ice at -10 C to water at 0 C? [Specific latent heat of
5 -1
fusion of ice = 3.36 x 10 kg and specific heat capacity of ice = 2100Jkg K-1]
𝑚𝑝 𝑜 𝑜𝑚 𝑜 𝑜𝑚 𝑙
( ( ))

SSEKWE ROBERT
Experiment to determine specific latent heat of fusion of ice by method of mixtures.
Thermometer Stirrer
Water Copper calorometer
Pieces of pure melting ice Lagging
 Hot water of mass, 𝑚 and specific heat capacity, is poured in a calorimeter of mass, 𝑚
and specific heat capacity, .
 The initial temperature of the hot water and calorimeter is then recorded from the
thermometer.
 Small pieces of pure melting ice at 0oC are placed in a calorimeter.
 Stir the mixture gently until all the ice melts.
 The final steady temperature, of the mixture is recorded from the thermometer.
 Weigh the calorimeter and its contents and determine the mass, 𝑚 of melted ice from
( ) ( )
 Assuming there are no heat losses
( ) ( ) ( )
( ) ( )
Latent heat of vaporization:

This is the amount of heat required to change a substance from liquid state to gaseous state at
constant temperature.
Specific latent heat of vaporization, :

This is the amount of heat required to change 1kg mass of a substance from liquid state to
gaseous state constant temperature.
The SI unit of specific latent heat of vaporization is .

SSEKWE ROBERT
Example:
o
1. How much heat is needed to change 10g of water at 100 C to steam at constant temperature?
6
[Specific latent heat of vaporization of water = 2.3 x 10 kg]
Experiment to determine specific latent heat of fusion of steam by method of mixtures.

Steam
Thermometer Stirrer
Flask
Cold water
Boiling water
Heat
 Cold water of mass,𝑚 and specific heat capacity, is poured in a calorimeter of mass, 𝑚
and specific heat capacity, .
 The initial temperature, of cold water and calorimeter is recorded.
 Steam from pure boiling water at is passed through the cold water in the calorimeter.
 Stir the mixture gently until a steady final temperature, is reached.
 Weigh the calorimeter and its contents to determine the mass, 𝑚 of condensed steam from,
( ) ( )
 Assuming there is no heat loss during the experiment.
( ) ( ) ( )
( ) ( ) ( )
 Hence specific latent heat of vaporization of steam, 𝐿 can be calculated.

SSEKWE ROBERT
Examples:
1. A calorimeter of mass 35.0g and specific heat capacity of 840Jkg-1K-1 contains 143.0g of
water at 7oC. Dry steam at 100oC is passed through the water in the calorimeter until the
temperature of water rises up to 29oC. If the mass of steam which condenses is 5.6g,
calculate
i) The heat gained by water and calorimeter
ii) Specific latent heat of vaporization of water
(S.H.C of water 𝐽𝑘 𝐾
i)
( ) ( )
( ) ( )
ii)
( )
( )
2. The temperature of water of mass 2kg and specific heat capacity of 4200Jkg-1K-1 is raised
from 20oC to 80oC by steam at 100oC. Calculate the mass of steam needed if the specific
latent heat of vaporization of water is 2.3x106Jkg-1K-1. (Neglect heat absorption by container
with water)
( ) ( )
( ) ( ) ( )
NOTE:
If the water is not at its boiling point , the heat supplied first increases/raises its
temperature to so as to be converted to vapour.

SSEKWE ROBERT
o o
3. How much heat is needed to melt 10g of ice at -10 C to steam at 100 C?
5 -1 -1
[Specific latent heat of fusion of ice = 3.36 x 10 kg, specific heat capacity of ice = 2100Jkg K , specific
-1 -1 6 -1
heat capacity of water = 4200Jkg K , specific latent heat of vaporization of water = 2.3 x 10 Jkg ]
( ( )) ( )
4. Find the heat required to change 2 kg of ice at 0℃ into water at 50℃.

(S.L.H of fusion of ice = 3.36𝑥105Jk −1 , . . 𝑜 𝑤 = 4200𝐽𝑘 −1
𝐾−1).
( )
5. An ice making machine removes heat from water at a rate of 20 J −1. How long will it take to
convert 0.5kg of water at 20℃ to ice at 0℃. (S.L.H of fusion of ice = 3.36𝑥105Jk −1 , S.H.C
of water = 4200𝐽 𝑘 −1 𝐾−1).
( )
6. A calorimeter with heat capacity of 80𝐽℃−1 contains 50g of water at 40℃. What mass of ice
at 0℃ needs to be added in order to reduce the temperature to 10℃. Assume no heat is lost
to the surrounding (S.H.C of water = 4200Jk −1℃−1, S.L.H of fusion of ice = 3.4x105Jkg-1)
( ) ( ) ( )
( ) ( )

SSEKWE ROBERT
7. Steam at 100℃ is passed into a copper calorimeter of mass 150g containing 340g of water at
15℃. This is done until the temperature of the calorimeter and its content is 71℃. If the
mass of the calorimeter and its contents is found to be 525g. Calculate the specific latent heat
of vaporization of water. (S.H.C of copper = 400Jkg-1K-1)
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
EXERCISE:
1. Ice at 0℃ is added to 200g of water initially at 70℃ in a vacuum flask. When 50g of ice is
added and has all melted, the temperature of the flask and content is 400℃. When further
80g of ice has been added and has been melted, the temperature of the whole becomes 10℃.
Calculate the S.L.H of fusion of ice neglecting any heat loss of surrounding.
-1
Ans; [3.78x105 Jkg ]
2. Calculate the heat required to melt 200g of ice at 0℃ . (S.L.H of ice= 3.4x105Jkg-1 )
Ans; [ 6.8x104 J]
3. Calculate the heat required to turn 500g of Ice at 0℃ into water at 100℃.
(S.L.H of ice= 3.4x105Jkg-1, S.H.C of water = 4200Jkg-1 )
Ans; [3.8x ]
4. Calculate the heat given out when 600g of steam at 100℃ condenses to water at 20℃ [S.L.H
of steam = 2.26x106 Jkg-1, S.H.C of water = 4200 Jkg-1].
Ans; [1.56x106J]
5. 1kg of vegetables, having a specific heat capacity 2200 Jkg-1 at a temperature 373K are
plugged into a mixture of ice and water at 273K. How much is melted. [S.L.H of fusion of
the ice = 3.3x105 Jkg-1]
Ans; [0.67kg]
6. 0.02kg of ice and 0.10kg water at 0℃ are in a container. Steam at 100℃ is passed in until all
the ice is just melted. How much water is now in the container? (S.L.H of vaporistion of steam
= 2.3x106Jkg-1, S.L.H of fusion of ice = 3.4x105Jkg-1, S.H.C of water = 4.2 x103Jkg-1K-1
Ans; [0.1225kg]

SSEKWE ROBERT
QUESTION; Explain why specific latent heat of vaporization of a substance is always greater
than specific latent heat of fusion the same substance e.g. ( ice, water and steam)
 For Latent heat of fusion (solid to liquid); heat required is small
because it only increases slightly increases the molecular spacing by
breaking the intermolecular forces.
 For latent heat of vaporization (liquid to gas); heat required is large
because it has to increase the molecular spacing by breaking the
intermolecular forces and also has to provide energy that enables
molecules to escape from the surface of the liquid.
NOTE: The phenomenon above explains why a person feels much heat when burnt by steam
than when burnt by water at the same temperature.
EFFECTS OF HEAT ON MATTER:

Evaporating
Melting
Solid Liquid Gas
Freezing Condensing
When a solid is heated it changes to a liquid at its melting point.

Definition:
Melting point is a constant temperature at which a solid changes to a liquid.
When a liquid is cooled it changes to a solid at its freezing point.
Definition:
Freezing point is a constant temperature at which a liquid changes to a solid.
When a liquid is heated it changes to a gas (vapour) at its boiling point.
Definition:
Boiling point is the constant temperature at which a liquid changes into a gas.
When a liquid is cooled it condenses and changes to a liquid.
HEATING AND COOLING CURVES (graphs of temperature against time)

The heating curve when ice below its melting point is heated.
Temperature (oC) F
E
100
B
0 Time

SSEKWE ROBERT
Explanation of the shape of graph.

AB: temperature of ice is increasing from A to its melting point
o
BC: ice is changing to water at 0 C
CD: the temperature of water is increasing from 𝑜 𝑜 𝑙 𝑛 𝑝𝑜 𝑛
o
DE: water is changing to steam at 100 C
EF: temperature of steam is increasing
The states of water along different regions are;

AB – solid state (ice)
BC – solid state and liquid state (water + ice)
CD – liquid state (water)
DE – liquid state and gaseous state (water + vapour)
EF – gaseous state (steam or vapour)
The cooling curve when water above its boiling point is cooled
Temperature (00C)
A
100 B C
D E
0 Time
F
Explanation of shape of the graph
AB: temperature of steam is decreasing from to boiling point 100 .
o
BC: steam is changing to water at 100 C
CD: the temperature of water is decreasing from 100 to freezing point
o
DE: water is changing to ice at 0 C
EF: temperature of ice is decreasing
The states of water along different regions are;

EF – solid state (ice)
DE – solid state and liquid state (water + ice)
CD – liquid state (water)
BC – liquid state and gaseous state (water + vapour)
AB – gaseous state (steam or vapour)

SSEKWE ROBERT
Example:
1. 2kg of ice at −5℃ was heated up to steam at 100℃.
i) Sketch a temperature time graph curve for the ice up to steam
ii) Find the heat at each section of the graph drawn. (S.H.C of ice = 2000 J 𝑘 1K 1, S.H.C of
water = 4200 J 𝑘 1K 1, S.L.H. of fusion of ice = 3.36𝑥105Jkg-1, S.L.H. of vaporization of
water = 2.26𝑥106Jkg)
i)
100
-5
ii)
( ( ))
( )

SSEKWE ROBERT
EXERCISE:
Where necessary assume the following
Specific heat capacity of water = 4200Jkg-1K-1
Specific heat capacity of copper = 400 Jkg-1K-1
Specific heat capacity of iron = 450 Jkg-1K-1
Specific heat capacity of aluminium = 880 Jkg-1K-1
Specific heat capacity of ice = 2100 Jkg-1K-1
Specific latent heat of fusion of ice = 336,000 Jkg-1
-1
Specific latent heat of vaporization of water = 2,250,000 Jkg
o o
1. How much heat is required to raise the temperature of 50g of aluminium from -100 C to 120 C?
Ans: 9,680J
o o
2. If 98,000J of heat are needed to raise the temperature of 2kg of a substance from 51 C to 65 C.
What is the specific heat capacity of a substance?
Ans: 3500Jkg-1K-1
3. An electric fire has a power of 1,800W. When used to heat a liquid of 5kg, it takes 6 minutes to
o
raise the temperature by 90 C. What is the specific heat capacity of the liquid?
Ans: 1440 Jkg-1K-1
o o
4. A 30g block of copper is heated from -20 C to 180 C. How much heat does it absorb during heating?
Ans: 2400J
o
5. How much heat energy is needed to melt 0.01kg of ice at 0 C?
Ans: 3360J
o o
6. How much heat energy is needed to change 0.2kg of ice at 0 C into steam at 100 C?
Ans: 601,200J
o
7. An electric heater marked 225,000W keeps water boiling at 100 C. What mass of water
evaporates in a second?
Ans: 0.1kg
o o
8. An electric heater was used to heat 2kg of water from 20 C to 50 C in 25 minutes. If the
voltage across the heater was 24V, what was the current through the heater?
Ans: 7.0A
9. 5kg of ice cubes are removed from the freezing compartment of a refrigerator into a home freezer.
o o
The refrigerator‟s freezing compartment is kept at -40 C the home freezer is kept at -17 C. How
much heat does the freezer‟s cooling system remove from the ice cubes?
Ans: 241,500J
10. What is the heat capacity of 5.5kg of aluminium?
Ans: 4,840JK-1

SSEKWE ROBERT
GAS LAWS
Gas laws describe the behavior of gases when subjected to physical factors such as pressure and
temperature.
These laws express the relationships between pressure (P), volume (V) and temperature (T) of a
fixed mass of a gas.
There are three gas laws namely;
 Boyle’s law.
 Pressure law.
 Charles’ law.
Boyle’s law:
It states that the volume of a fixed mass of a gas is inversely proportional to its pressure
at constant temperature.
𝐾 𝑜𝑛 𝑛 𝑜 𝑝 𝑜𝑝𝑜 𝑜𝑛 𝑙 𝑦
If the volume of gas changes from 𝑜 and its pressure changes from 𝑜 .
𝑛
Where 𝑛 𝑙𝑝
𝑛 𝑙𝑝
𝑛 𝑙 𝑜𝑙 𝑚
Examples:
3
1. The volume of a fixed mass of a gas at constant temperature is 250cm when the pressure is
3
720mmHg. Find the pressure when the volume is increased to 600cm .

SSEKWE ROBERT
-5 3
2. The volume of a fixed mass of a gas at constant temperature is 2.0 x 10 m when the
6 -4 3
pressure is 7.2 x 10 Pa, find the pressure when the volume is increased to 6.0 x 10 m .
3
3. The volume of a fixed mass of a gas at constant temperature is 4cm when the pressure is 6
atmospheres, find the volume when the pressure is increased to 12 atmospheres.
4. The pressure of a fixed mass of a gas is 5 atmospheres when its volume is 𝑚 . Find the
pressure when the volume is
i) halved
) doubled
ii)
( )

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Experiment to verify Boyle’s law

Graduated glass tube
Volume scale
Pressure gauge
Trapped air
Air from foot pump
Oil
 Trap air above oil in the graduated glass tube.

 Read the initial volume, V and pressure, P of trapped air.
 Increase the pressure of trapped air by using a foot pump connected to a pressure gauge as shown
above.
 Allow the air to cool to room temperature.
 Read and record the new values of V and P.
 Increase the pressure again to get different values of V and P.
 Record your results in a suitable table including values of
P V
 A graph of P against V is then plotted.

P
 A graph of P against is also plotted.

P
 From the above graphs it shows that pressure is inversely proportional to volume which is
Boyle’s law.

SSEKWE ROBERT
Pressure law (Gay Lussac law):

It states that the pressure of a fixed mass of a gas is directly proportional to its absolute
temperature at constant volume.
𝑜𝑛 𝑛 𝑜 𝑝 𝑜𝑝𝑜 𝑜𝑛 𝑙 𝑦
If the temperature of gas changes from to and its pressure changes from to .
Where 𝑛 𝑙 𝑚𝑝
𝑛 𝑙 𝑚𝑝
𝑛 𝑙𝑝
𝑛 𝑙𝑝
Definition:
Absolute temperature is the temperature at which the volume of a gas reduces to zero.
Or Absolute temperature the temperature at which the molecules of a gas have the lowest
kinetic energy
N.B: The temperature must always be in kelvins.
Examples:
o
1. The pressure of a fixed mass of a gas at 127 C is 600mmHg. Calculate its pressure at
o
constant volume if the temperature reduces to 27 C.
2. The pressure of a gas is 75𝑁𝑚−2 at −73℃. What is its pressure when a gas is heated up to
127℃.

SSEKWE ROBERT
3. A sealed flask contains a gas at a temperature of 27℃ and a pressure of 90 𝑘 . If the

temperature rises to 127℃. What will be the new pressure?
Experiment to verify Pressure law
Thermometer
Rubber tubing
Pressure gauge
Water
Flask
Dry air
Heat
 The apparatus is set up as shown above.

 The flask containing dry air is placed in a metal can with water such that water is almost to
the top of its neck.
 The can is heated from the bottom while stirring.
 Pressure, P is then recorded for different values of temperatures.
 The results are recorded in a suitable table
P T(0C) T(K)

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 A graph of P against temperature is then plotted.

P
P
-273 0 T(oC) 0 T(K)
 From the above graphs it shows that pressure is directly proportional to the temperature which is
pressure law.
o
NOTE: The temperature -273 C (0K) is called absolute zero temperature.
Charles’ law:
It states that the volume of a fixed mass of a gas is directly proportional to its absolute
temperature at constant pressure.
𝑜𝑛 𝑛 𝑜 𝑝 𝑜𝑝𝑜 𝑜𝑛 𝑙 𝑦
If the temperature of gas changes from to and its volume changes from to .
Where 𝑛 𝑙 𝑚𝑝
𝑛 𝑙 𝑚𝑝
Example:
o 3
1. The volume of a fixed mass of a gas at 127 C is 300cm . Calculate its volume at constant
o
pressure if the temperature reduces to 27 C.
( ) ( )

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o -4 3
2. The volume of a fixed mass of a gas at 17 C is 5.0 x 10 m . Calculate its temperature at
-4 3
constant pressure if the volume reduces to 2.0 x 10 m
o 3
3. The volume of a fixed mass of a gas at 27 C is 400cm . Calculate its volume at constant
o
pressure if the temperature reduces to -123 C.
4. The temperature of a fixed mass of a gas is 27℃. If the volume is halved, find its new
temperature.

SSEKWE ROBERT
Experiment to verify Charles’ law:
Rubber band
 Trap dry air using the index of concentrated sulphuric acid in a capillary tube.
 Tie the tube on the metre rule using a rubber band.
 Place the tied tube in a metal can containing water.
 Heat the water slowly while stirring gently.
 Read and record the length, L of the trapped air column and the temperature, T from the
thermometer.
 Repeat procedures to obtain other values of L for different temperature values.
 Record the results in a suitable table.
L(cm) T(0C) V
But L is proportional to volume, V so V=L

 Plot a graph of volume V against temperature T.
V V
0 T(K)
-273 0 T(oC)
 From the above graphs it shows that volume is directly proportional to the absolute
temperature which is Charles’s law.

SSEKWE ROBERT
Equation of state for an ideal gas:

This is sometimes referred to as ideal gas equation. It combines the three gas laws.
Combining the three gas laws, we get;
Therefore if the volume of the gas changes from 𝑜 , its pressure changes from 𝑜 and
its temperature from 𝑜
NOTE:
At standard temperature and pressure (s.t.p)
 Standard absolute temperature
 Standard pressure
Examples:
3 o
1. In an experiment 500cm of a gas was collected at a temperature of 97 C and a pressure of
6 6
3.7 x 10 Pa. Find the volume of the gas if the pressure changes to 6.0 x 10 Pa at a temperature
o
of 27 C.
2. A bicycle pump contains 𝑚 of air at and a pressure of 1 atmosphere. Find the

pressure when it is compressed to 𝑚 and its temperature rises to

SSEKWE ROBERT
3 o
3. In an experiment 58cm of a gas was collected at a temperature of 17 C and a pressure of
4
8.0 x 10 Pa. Find the volume the gas at s.t.p.
4. 240 𝑚3 of oxygen gas was collected when a temperature is 20℃ at a pressure of 50cmHg.
Calculate its volume at s.t.p.
5. The volume of hydrogen at 273℃ is 10 𝑚3 at a pressure of 152 cmHg. What is its volume at
s.t.p.

SSEKWE ROBERT
Kinetic theory of Gas laws:

Recall; Kinetic theory of matter states that matter is made up of small particles called
molecules that are in a continuous random motion and possess energy.
Question: Explain what causes gas pressure.

Increase in temperature, increases the speed of molecules and hence with the walls of the
container thus creating pressure.
N.B: This explains why the pressure of a car tyre increases on a hot day.
Boyle’s law:
For a fixed mass of a gas;
When the volume of the fixed mass of a gas is reduced at constant temperature the speed
of the gas molecules increases hence the rate of collision with the walls of the container
increases thus the pressure of the gas increases.
However increasing the volume of a gas reduces the pressure since the speed of the
molecules of the gas reduces hence reducing on the rate of collision with the walls of the
container.
Charles’ law:
When the temperature of the fixed mass of a gas is increased at constant pressure, the
speed of the molecules of a gas increases and the rate of collision with the walls of a container
increases thus increasing the volume of gas.
Pressure law:
When the temperature of the fixed mass of a gas is increased at constant volume, the
speed of the molecules of a gas increases and the rate of collision with the walls of a container
increases hence the pressure of the gas increases.
VAPOURS
Vapour is the gaseous state of a substance below its critical temperature.
Critical temperature is the minimum temperature above which the gas cannot be changed
back to a liquid.
There are two types of vapours namely;

 Saturated vapour
 Unsaturated vapour

SSEKWE ROBERT
SATURATED VAPOUR
This is the vapour that is in dynamic equilibrium with its own liquid.
i.e. 𝑜 𝑝𝑜 𝑜𝑛 𝑜 𝑜𝑛 𝑛 𝑜𝑛
Saturated vapour pressure;

This is the pressure exerted by a vapour that is in dynamic equilibrium with its own liquid.
Explanation of occurrence of saturated vapour pressure (s.v.p) using kinetic theory
Evaporation
Condensation
Consider a liquid enclosed in a container with a piston.
When a liquid in a closed container is heated, some of the liquid molecules get enough
kinetic energy and break the intermolecular forces and escape from the surface of the liquid and
occupy the space just above it and become vapour molecules. This process is called evaporation.
These vapour molecules collide with the walls of the container hence creating vapour
pressure.
When these molecules bounce off from the walls of the container, they strike the liquid
surface and re-enter the liquid. This process is called condensation.
A state of dynamic equilibrium is attained i.e. (rate of evaporation = rate of
condensation) and this point, vapour is said to be saturated and exerting saturated vapour
pressure.
Definition: Vapour pressure is the pressure exerted on the walls of the container by the vapour
molecules.
NOTE:
Gas laws only apply to a fixed/constant mass of a gas.
Therefore, saturated vapours do not obey ideal gas laws because there masses change due to
condensation or evaporation as the conditions change.
It should be noted that saturated vapor occurs for a very short time and at a constant temperature
(boiling point).

SSEKWE ROBERT
UNSATURATED VAPOUR
This is the vapour that is not in dynamic equilibrium with its own liquid.
i.e. 𝑜 𝑝𝑜 𝑜𝑛 𝑜 𝑜𝑛 𝑛 𝑜𝑛
Unsaturated vapour pressure:

This is the pressure exerted by a vapour that is not in dynamic equilibrium with its own liquid.
Differences between saturated vapour and unsaturated vapour
Saturated vapour Unsaturated vapour

 It doesn’t obey gas laws.  It obeys gas laws.
 It is the vapour in dynamic equilibrium  It is the vapour that is not in dynamic
with its own liquid. equilibrium with its own liquid.
 Exists at a fixed temperature.  Exists at any temperature.
Other terms;
Super saturated vapour:
This is the vapour whose rate of evaporation is greater than the rate of evaporation.
Ideal gas:
This is a gas whose intermolecular forces are negligible.
Real gas:
This is a gas whose intermolecular forces are not negligible.
Dew point:
This is the temperature at which atmospheric air is saturated with water vapour.
OR
This is the temperature at which water vapour condenses to liquid water (dew)
Note;
Fog or cloudy film forms on windscreens of cars because the dew point of water vapour
has been exceeded.

SSEKWE ROBERT
EVAPORATION:
This is the process by which a liquid changes into gas (vapour).
OR
This is the escape of molecules of a liquid from its surface.
Evaporation takes place only at the surface of the liquid.
It takes place at all temperatures but it is greatest when the liquid is at its boiling point.
Explanation of evaporation according to kinetic theory:

(How evaporation causes cooling)
According to kinetic theory, molecules of a liquid are in a state of continuous random motion
and their speed depend on the temperature of the liquid.
Faster moving molecules with the most kinetic energy reach the liquid surface and weaken
the intermolecular forces of attraction and then escape from the surface of liquid causing
evaporation.
The slow moving molecules with the lowest kinetic energy remain in the liquid thus cooling
the liquid.
Recall: Temperature decreases with decrease in speed of molecules. Since some molecules
have low speeds, so they are cold.
Applications of cooling as a result of evaporation:

 Panting of dogs.
 Making of ice by evaporation of a volatile liquid.
 Refrigerators.
Experiment to make ice by evaporation of a volatile liquid:

Air
Glass tube
Metal can
Volatile liquid (ether)
Water
Wooden block
Procedures:
 Place a metal can filled with ether (volatile liquid) on a film of water on top of a wooden
block.
 Blow air through the glass tube.
Observation:
 It is observed that the water under the can turns into ice i.e. it freezes.

SSEKWE ROBERT
Explanation:
Ether will evaporate when it gets necessary heat from water and blowing in air increases
the rate of evaporation.
Since water is supplying heat to ether, it loses heat thus its temperature decreases hence
water freezes to ice.
Definition:
Volatile liquid is a liquid with a low boiling point.
Factors that affect the rate of evaporation:

Rate of evaporation indicates the number of molecules that escape from liquid surface per
second.
The following factors affect the rate of evaporation;
(i) Surface area:
Increasing the surface area increases the rate of evaporation because a large surface exposes
many energetic molecules to escape while small surface exposes fewer molecules to escape.
• This explains why a plate cools porridge faster than a cup since the plate is wider than the cup.
(ii) Temperature:
Increasing temperature increases the rate of evaporation and decreasing the temperature decreases
the rate of evaporation. At high temperature, more molecules will move faster to escape from the
liquid surface but at low temperature fewer molecules move faster to escape from the liquid
surface.
(iii) Wind (air currents):
The rate of evaporation increases if there is too much wind/air blowing because wind blows
away molecules which have already escaped from the liquid so they can’t return back to the
liquid.
This explains why a person can cool porridge while blowing air through it.
(iv) Pressure:
Reducing pressure of air above the liquid surface (atmospheric pressure) increases the rate of
evaporation since low pressure is exerted on the liquid surface.
(v) Intermolecular forces of a liquid:

The stronger the intermolecular forces, the slower the rate of evaporation since molecules will
need much heat to break these forces.

SSEKWE ROBERT
REFRIGERATOR:
This is a cooling device which transfers heat from objects in it to the surrounding.
It is used in preservation of;
 Food in homes and supermarkets
 Blood in hospitals
 Medicines in hospitals and pharmaceuticals.
In a refrigerator, heat is taken in at one point and given out at another point by a volatile liquid or
refrigerant.
HOW A REFRIGERATOR WORKS
Vapour
Volatile liquid
Evaporator pipes
Cooling fins
Condenser tube
(Heat exchanger)
Copper tube
Vapour
Compressor pump
Mode of operation:
 The copper tube contains a volatile liquid which enters the evaporator pipes in the freezer.
 The volatile liquid gets latent heat from the refrigerator contents thus evaporating to vapour.
 This causes cooling of the contents since they lose heat.
 The vapour formed is compressed into the condenser tube and turns into a liquid thus giving
out latent heat.
 The heat given out is lost to the surrounding through the cooling fins by convection and
radiation.
 The liquid returns to the freezer and the process continues.

SSEKWE ROBERT
FUNCTIONS OF THE PARTS:

(a) Compressor pump:
This removes the vapour formed in the freezer and forces the vapour to the condenser tube.
(b) Condenser tube (Heat exchanger):
This where vapour is turned into a liquid giving out latent heat of vaporization to the
surrounding air.
(c) Cooling fins:
These are painted black so that they can give out heat to the surrounding air.
Black colours are good emitters of heat.
(d) Evaporator pipe:
This absorbs heat from the refrigerator contents and gives it to the volatile liquid so as to
evaporate.
BOILING:
Definition:
Boiling is a process which occurs when atmospheric pressure is equal to saturated vapour
pressure.
OR
Boiling is a process by which a liquid changes to vapour at its boiling point.
Boiling occurs at a fixed temperature called boiling point and it takes throughout the liquid.
Boiling involves formation of bubbles.
Differences between boiling and evaporation
Boiling Evaporation
• It occurs at a fixed temperature. • It occurs at any temperature.
• It takes place throughout the liquid. • It takes place at the liquid.
• Doesn’t cause cooling. • Causes cooling.
• Involves formation of bubbles. • Doesn’t involve formation of bubbles.
FACTORS THAT AFFECT BOILING POINT OF A LIQUID

(i) Pressure:
The lower the atmospheric pressure, the lower the boiling point (temperature needed to boil a
liquid). But if pressure is increased, the boiling point also increases.
This is because if the atmospheric pressure is decreased, then the liquid will boil more easily
since it will take less time for its saturated vapour pressure to equal to atmospheric pressure.
This explains why;
 Cooking takes longer at higher altitudes.
 In a pressure cooker, food cooks more quickly.
 During cooking we cover our saucepans.

SSEKWE ROBERT
(ii) Impurities:
Addition of impurities like salt raises the boiling point of a liquid.
Salts in a water will cause water molecules to be more attracted to the salts thus a higher
temperature is required to break the forces of attraction between water molecules thus
increasing the boiling point of water.
This explains local salt “kisula” is added to beans so as they boil easily.
QUESTION:
Why cooking takes a lot of time to boil at high altitudes.
This is because at high altitude, the atmospheric pressure is low therefore, the boiling point of
water is also low. This causes water to boil faster before food is properly cooked.
Hence it takes a lot of time for saturated vapour pressure to equal to the atmospheric pressure.
PRESSURE COOKER:
Pressure cookers are useful in places where the atmospheric pressure is low e.g. at the top of a
mountain because they raise the boiling point of a liquid thus reducing time for cooking.
How a pressure cooker works.

 A pressure cooker has a lid (cover) that prevents steam from escaping.
 As water inside is heated, steam accumulates thus an increase in steam pressure causing the
boiling point of water to rise above 1000C.
Thus food boils quickly thereby saving time and fuel.
FREEZING POINT AND MELTING POINT

Freezing is the process by which a liquid changes to a solid. Freezing occurs at constant temperature
called freezing point.
Melting is the process by which a solid changes to a liquid. Melting occurs at a constant temperature.
Freezing point is a constant temperature at which a substance changes from liquid state to solid
state.
Melting point is a constant temperature at which a substance changes from solid state to liquid
state.

SSEKWE ROBERT
Factors that affect freezing and melting points of a substance.

(i) Pressure:
Increase in pressure lowers the melting/freezing point of a substance and vice versa
(ii) Impurities:
Addition of impurities lowers the melting/freezing points of a substance and vice versa.
• This explains why ice melts why ice melts quickly when salt is sprinkled on it.
Effect of pressure on melting point of ice.
When pressure is increased on the ice by the copper wire;

 The copper wire passes through the ice block since increased pressure by the copper
wire lowers the melting point of ice. So it melts easily at a low temperature.
HEAT ENGINE
A heat engine is a device used to convert heat energy to kinetic energy (mechanical energy).
Why engines are always less than efficient.

 Because some of the energy is lost in overcoming friction since it has moving parts. This is
friction is reduced by lubricating engine parts.
 Some heat is lost to surrounding due to conduction.
 Some energy is wasted in lifting useless loads like pistons.
PETROL ENGINE
It is also called the four stroke cycle engine.

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OPERATION OF A FOUR STROKE CYCLE ENGINE (PETROL ENGINE)
Intake (inlet) stroke:

 Piston moves down causing a vacuum.
 Inlet valve opens and the air –fuel mixture is forced into the cylinder from carburetor.
 Exhaust valve closes.
Compression stroke:
 Both valves close.
 The piston moves up compressing the air-fuel mixture.
 The fuel is ignited by a spark plug.
Power stroke:
 A spark jumps across the points of a spark plug and explodes the air-fuel mixture.
 Piston is forced to move down.
Exhaust stroke:
 The outlet valve opens pushing the exhaust gases out of the cylinder.
NOTE:
 The operation of a diesel engine is the same as that of a petrol engine.
 The diesel engine use diesel as a fuel yet petrol engines use petrol as a fuel.
Differences between diesel and petrol engines.
Diesel engine Petrol engine

 Uses diesel as a fuel.  Uses petrol as a fuel.
 No spark plug used.  Spark plug is used.
 Has a fuel injector instead of  Has a carburetor instead of fuel
carburetor. injector.
 Produces a lot of smoke.  Produce less smoke.
 Uses less fuel.  Uses a lot of fuel.
 They are expensive.  They are cheap.
 They are heavy  They are lighter.

SSEKWE ROBERT
EXERCISE:
o
1. 200litres of a gas at 0 C are kept under a pressure of 150kPa. If the temperature is raised to
o
273 C, its pressure is raised to 400kPa. Calculate its volume.
Ans: 150 litres
o -3 3
2. The density of argon gas at 27 C is 0.27kgm . A volume of 50m of argon gas is kept under
o o
constant pressure at 27 C. What will be the density of argon if its temperature is raised to 51 C?
Ans: 0.25kg
3
3. The volume of a fixed mass of a gas at constant temperature is 150cm when the pressure is
76cmHg. Calculate the volume when the pressure is 38cmHg.
Ans: 300c
3 o
4. The volume of a fixed mass of a gas at constant pressure is 400cm at a temperature of 27 C.
o
Calculate the volume when the temperature is raised to 78 C.
Ans: 468c
5. The pressure of a fixed mass of a gas at constant volume is 600mmHg at a temperature of
o o
127 C. Calculate the pressure when the temperature falls to 27 C.
Ans: 450mmHg
o
6. Air in a 2.5litre vessel at 127 C exerts a pressure of 3 atmospheres. Calculate the pressure
o
that the same mass air would exert if contained in a 4litre vessel at -73 C
Ans: 0.9375atmospheres.
7. State differences between boiling and evaporation
8. Distinguish between saturated vapour and un-saturated vapour
9. What are the factors that affect the rate of evaporation of a liquid and how
10. Use the kinetic theory to explain effect of increasing temperature of the gas at constant pressure
EXAMINATION QUESTIONS:
1. a) Define specific heat capacity
b) 0.05kg of water at 80oC is mixed with 0.06kg of water at 10oC contained in a vessel of
heat capacity 28Jkg-1. What is the final temperature of the mixture?
Ans: 4000C
c) i) Define specific latent heat of fusion
ii) Describe a simple method to determine the specific latent heat of fusion of ice
c) When 0.005kg of ice at 0oC is added to 0.02kg of warm water at 30oC the final
temperature attained is 8oC. Find the specific latent heat of fusion of ice.
Ans: 336,000Jkg-1
2. a) i) State Boyle‟s law

ii) With the aid of a labeled diagram describe the experiment to show the relationship
between the volume and the temperature of a fixed mass of a gas at atmospheric
pressure.
b) A cylinder with a movable piston contains 0.1m3 of air at a temperature of 27oC.
Calculate the volume of the gas if it is cooled to -33oC at constant pressure.
Ans: 0.08

SSEKWE ROBERT
c) Define the term specific heat capacity

d) A copper block of mass 200g is heated to a temperature of 145oC and then dropped into
a well lagged copper calorimeter of mass 250g which contains 300cm3 of water at 25oC
i) Calculate the maximum temperature attained by the water
ii) Sketch a graph to show the variation of temperature of water with time
Ans: i) 31.70C
3. a) i) Define heat capacity of a substance

ii) Describe an experiment to determine specific heat capacity of a substance by
method of mixtures
iii) State the precautions necessary for accuracy during the experiment above
b) A well lagged copper calorimeter of mass 85g contains 80g of water at 60oC. Dry ice at
0oC is added to the calorimeter and after stirring the mixture attains a steady temperature
of 20 oC. Find the mass of ice added
Ans: 35.20C
c) i) Describe an expiration to show that evaporation produces cooling.
ii) Explain why evaporation produces cooling.
iii) State one application of cooling by evaporation.
4. a) i) What is a saturated vapour

ii) Explain why the boiling point of a liquid depends on altitude.
b) i) Define specific heat capacity.
ii) Describe an experiment to determine the specific heat capacity of a solid
c) A copper block of mass 250g is heated to a temperature of 145°C and then
transferred to a cooper calorimeter of mass 250gwhich contains 250cm3of water at
20°C
i) Calculate the maximum temperature attained by water
ii) Sketch the graph to show the variation of temperature with time
Ans: i) 300C
d) i) What is meant by the term temperature
ii) Give two physical properties which change with temperature
5. a) Define the following terms as used in heat

i) Specific heat capacity
ii) Latent heat of vaporization
b) Describe an experiment to determine the specific heat capacity of a liquid
c) Steam from boiling water is bubbled through 1.5kg of water at 20oC. After this process,
the mass of water was found to be 1.54 kg. What is the new temperature of water?
Ans: 35.990C
d) State four ways in which heat losses can be minimized in a calorimetry experiment
6. a) Define specific latent heat of fusion

b) Describe an experiment to determine the specific latent heat of fusion of ice
c) A copper block of mass 300g is heated to a temperature of 245oC and then dropped into
a well lagged copper calorimeter of mass 350g containing 400g of water at 35oC.

SSEKWE ROBERT
Calculate the maximum temperature attained by the water.

Ans: 43.30C
d) i) What is meant by absolute zero temperature
ii) A sealed flask contains gas at a temperature of 27oC and a pressure of
900Pa. if the temperature rises to 127oC. What will be the new pressure?
Ans: ii) 1200Pa
7. a) i) Define temperature.
ii) The fundamental interval of a mercury-in-glass thermometer is 192mm. Find the
temperature in degrees Celsius when mercury thread is 67.2mm long
b) With the aid of a labeled diagram describe the experiment to show the relationship
between the volume and the pressure a fixed mass of a gas at constant temperature.
c) A copper block of mass 150g is heated to a temperature of 95oC and then dropped into
a well lagged copper calorimeter of mass 200g containing 250g of water at 15oC.
Calculate the maximum temperature attained by the water.
Ans: 19.00C
d) State any two differences between boiling and evaporation
8. a) Define specific latent heat of vaporization

b) A calorimeter of mass 35g and specific heat capacity 840Jkg-1K-1 contains 143g of
water at 7oC. Dry steam at 100oC is bubbled through water in the calorimeter until the
temperature of the water rises to 29oC. If the mass of steam which condenses is 5.6g,
i) Calculate heat gained by the water and calorimeter
ii) Obtain an expression for the heat lost by the steam in condensing at 100oC and in
cooling to 29oC.
iii) Find the specific latent heat of vaporization of water
Ans: i) 13860J ii) 0.0056Lv + 1669.92 iii) 2,176,800
c) Explain in terms of molecules what is meant by a saturated vapour?
d) Describe briefly one application of vaporization
9. a) i) Describe the fixed points of a Celsius scale of temperature

ii) Give two advantages of mercury over alcohol as thermometric liquid
iii) Convert -200oC to Kelvin
Ans iii) 730C
b) Use the kinetic theory to explain the following
i) Cooling by evaporation
ii) Why the temperature of a gas contained in a cylinder increases when it is compressed
c) Explain briefly the transfer of thermal energy by conduction in metals
d) A battery of e.m.f 12V and internal resistance 1Ω is connected for 3minutes across a
heating coil of resistance 11Ω immersed in a liquid of mass 0.2kg and specific heat
capacity 2.0 x 103Jkg-1K-1. Find the rise in temperature of the liquid. State clearly any
assumptions made.
Ans: 4.950C
10. a) With the aid of a labeled diagram, describe the experiment to show the relationship
between temperature and pressure a fixed mass of a gas at constant volume.

SSEKWE ROBERT
b) A gas of volume 1000cm3 at a pressure of 4.0 x105 Pa and temperature of 17oC is heated
to 89.5oC at constant pressure. Find the new volume of the gas.
Ans: 1250c
c) A balloon is filled with 50cm3 of hydrogen and tied to the ground. The balloon alone and
the container it carries have a mass of 2kg. If the densities of hydrogen and air are
9.0 x 10-2kgm-3 and 1.29kgm-3 respectively, how much load can the balloon lift when
released
11. a) What is meant by conduction

b) Draw a labeled diagram of a thermos flask and explain how it is able to keep a liquid
cold for a long time
c) With the aid of a diagram, describe how you would determine the upper fixed point of
un -calibrated thermometer
d) Explain the following observations;
i) A bare cement floor feels colder than a carpeted one
ii) A beam with a notch that is used for constructing a bridge lasts longer when the
notch is on its top surface than when the notch is on its lower surface
12. a) Define the following terms

i) Specific heat capacity
ii) Specific latent heat of fusion
Temperature (oC) A
80 B C
25 D E
F
0 time
b) The figure above shows a cooling curve of a liquid whose boiling point is 80oC and
freezing point is 25oC.
i) Give the states over regions AB, BC, DE and EF
ii) What is happening over region BC?
iii) Use the kinetic theory to explain the differences in states over regions AB and EF
c) An iron rod of mass 0.8kg is pushed into an insulator solid substance through a
distance of 2.3m against frictional force of 400N. The temperature of iron rises by
2.5oC. Calculate the specific heat capacity of iron
Ans: 460Jkg-1K-1
d) i) Explain why when water in a saucepan is heated, the level first falls and then rises after
some time
ii) The length of mercury thread of un-calibrated thermometer is 10cm when the bulb is
in pure melting ice and rises to 20cm in steam. What is the reading of the
thermometer when the mercury thread is 18cm?
Ans: ii) 800C

SSEKWE ROBERT
13. a) Define specific latent heat of vaporization

b) Describe an experiment to determine specific latent heat of vaporization of steam
c) A copper calorimeter of heat capacity 60Jkg-1 contains 0.5kg of water at 20oC. Dry
steam at 1000C is passed into the water in the calorimeter until the temperature of the
water and the container reaches50oC. Calculate the mass of steam condensed
Ans: 4.61kg
d) i) What is meant by saturated vapour pressure
ii) Explain what may happen when one is to cook food from a very high altitude
14. a) i) Define latent heat of fusion.
ii) Describe with the aid of a labeled diagram, an experiment to show the effect of
increase in pressure on the melting point of ice
b) If the melting point of lead is 327oC, find the amount of heat required to melt 200g of
lead initially at 27oC given that specific latent heat of fusion of lead is 2.5 x 106Jkg-1 and
specific heat capacity of lead is 660Jkg-1K-1.
Ans: 539,600J
c) What is meant by the terms?
i) Temperature
ii) Heat
d) State two physical properties which change with temperature.
15. a) Describe an experiment to determine the specific latent heat of fusion of ice
b) i) 2 kg of ice initially at -10oC is heated until it changes to steam at 100oC
ii) Sketch the graph to show how temperature changes with time
iii) Calculate the energy required at each end of the graph
Ans: 42,000J 672,000J 840,000J 4,500,000J
16. a) Differentiate between conduction and convection

b) Describe an experiment which can be performed to show convection in a liquid
c) i) Draw a labeled diagram of a vacuum flask
ii) Explain how a vacuum flask minimizes heat losses
d) Why is a car radiator made of fins and painted black
17. a) State the kinetic theory of matter

b) i) State the law of volume and temperature
ii) The volume of a fixed mass of a gas at a given pressure is 1.5m3 at 300K. at what
temperature will the volume of the gas be at the same pressure
c) Describe an experiment to determine the fixed points of a thermometer
d) i) Mention any three reasons for not using water as a thermometric liquid
iii) When a Celsius thermometer is in a boiling liquid, the mercury thread rises above the
lower fixed point by 19.5cm. Find the temperature of boiling liquid if the
fundamental interval is 25cm.
Ans: 780C
18. a) What is meant by latent heat of vaporation

b) With the aid of a labeled diagram describe how a refrigerator works

SSEKWE ROBERT
c) The cooling system of a refrigerator extracts 0.7kW of heat. How long will it take to
convert 500g of water at 20oC into ice?
Ans: 300s
d) Explain how evaporation takes place
19. a) What is meant by conduction

b) Draw a labeled diagram of a vacuum flask and explain how it is able to keep a liquid hot
a long time
c) With the aid of a labeled diagram describe an experiment to determine the upper fixed
points of an un-calibrated thermometer
d) Explain the following observation a bare cement floor feels colder than a carpeted one
20. a) i) Define latent of fusion

ii) Describe with the aid of a labeled diagram, an experiment to show the effect of
increase in pressure on the melting point of ice
b) What is meant by the terms?
i) Temperature
ii) Heat
c) The fundamental interval of mercury in glass is 192mm. find the temperature in degrees
Celsius when the mercury thread is 67.2mm below the upper fixed point.
Ans: 650C
d) State two physical properties which change with temperature

Physics Notes HEAT MEASUREMENT S4

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SSEKWE ROBERT

ii) Specific heat capacity, c:

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METHODS OF MEASURING SPECIFIC HEAT CAPACITY:

Experiment to determine Specific heat capacity of a solid by electrical method.

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Experiment to determine Specific heat capacity of a solid by method of mixtures.

 A solid of mass, 𝑚 whose specific heat capacity, is required is heated to a temperature, .

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Experiment to determine Specific heat capacity of a liquid by method of mixtures.

 A solid of mass, 𝑚 and specific heat capacity, is heated to a temperature, .

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PRECAUTIONS TAKEN WHEN USING METHOD OF MIXTURES

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9) A block of metal of mass 0.01 kg at a temperature of 100℃ was dropped in a container of

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LATENT HEAT (HIDDEN HEAT)

KINETIC THEORY EXPLANATION OF LATENT HEAT

Latent heat of fusion:

Specific latent heat of fusion ( ):

The SI unit of specific latent heat of fusion is

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Experiment to determine specific latent heat of fusion of ice by electrical method.

Pure melting ice

Collected water from melted ice

 An electrical heater is placed in a funnel.

 Hence specific latent heat of fusion of ice 𝐿 can be calculated.

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Experiment to determine specific latent heat of fusion of ice by method of mixtures.

Water Copper calorometer

Pieces of pure melting ice Lagging

Latent heat of vaporization:

Specific latent heat of vaporization, :

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Experiment to determine specific latent heat of fusion of steam by method of mixtures.

 Hence specific latent heat of vaporization of steam, 𝐿 can be calculated.

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4. Find the heat required to change 2 kg of ice at 0℃ into water at 50℃.

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EFFECTS OF HEAT ON MATTER:

When a solid is heated it changes to a liquid at its melting point.

HEATING AND COOLING CURVES (graphs of temperature against time)

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Explanation of the shape of graph.

The states of water along different regions are;

The states of water along different regions are;

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Experiment to verify Boyle’s law

Air from foot pump