1. QUANTITY OF HEAT
Heat capacity
Specific heat capacity
Change of states
Latent heat of fusion
Latent heat of vaporization

2. HEAT CAPACITY (C)
 This is the quantity of heat required to raise the
temperature of a given mass of a material by one
degree Celsius or one Kelvin.
It is denoted by C
heat capacity = heat energy absorbed Q/
temperature
change ∆θ
C = Q/∆θ
The SI unit Jk-1

3. Sample Questions
1. Calculate the quantity of heat required to
raise the temperature of aluminium metal
block with heat capacity of 460 Jk-1 from
150 to 450.
Solution
Quantity of heat = C∆Ѳ
= 460 x 30
= 13 800 J.

4. Specific heat capacity (c)
 When comparing heat capacities of various
substances we talk of specific heat capacities .
Specific in physics refers to unit quantity of a
physical property.
 Also called specific heat.
 Quantity of heat required to raise the
temperature of a unit mass of a substance by one
degree Celsius or Kelvin.
 It is denoted as c.
 Si unit is joule per kilogram kelvin. (J/kg K).

5. FORMULAE
 Specific heat capacity = heat capacity/mass
c= Q/∆θ/m =Formulae
Q/m∆θ
Q = mc∆θ
• Materials with high specific heat capacity e.g.
water require large amount of heat to change
their temperatures while those with less
specific heat capacity requires little heat
energy to change their temperatures e.g..
silver

6. Table of specific heat capacities
J/kg K.
Material Specific Heat Material Specific Heat
Capacity Capacity
Aluminium 900 Lead 130
Brass 380 Mercury 140
Glass(ordinary) 400 Methylated spirit 2400
Ice 670 Sea – water 3900
Alcohol 2100 Water 4200
Iron 460 Zinc 380

7. Sample Question 1
How many joules Question given out
Sample of heat are 1
when a piece of iron of mass 50g and
specific heat capacity 460 J/kg K, cools
from 80 C t0 20 C?
solution
Q = mc∆θ
= 0.05 460 (80 – 20)
= 1380 J.

8. Sample question 2
1. A block of metal of mass 1.5 kg which is
suitably Sample is heated from 30 C to 50
insulated question 2
C in 8 minutes 20 seconds by an electric
heater coil rated 54 watts. Find :
a) the quantity of heat supplied by heater.
b) The heat capacity of the block.
c) Its specific heat capacity.

9. solution
a)Quantity of heat supplied = power x time
Solution Q2
Q = 54 x 500
= 27 000J
b) Heat capacity C = Q/∆θ.
C= 27000/(50 – 30)
= 1 350J/K.
c) specific heat capacity = C/m
c = 1 350/1.5 = 900 Jkg-1 K-1 .

10. Sample question 3
 What is the final temperature of the mixture if 100g of water at 70 0 C
is added to 200g of cold water at 100 C and well stirred?(neglect heat
Sample question 3
absorbed by the container)

11. SOLUTION
Heat lost by hot water = Heat received by cold water
Let the final temperature of the mixture = θ 0 C.
change in temperature of hot water = (70 – θ)
change in temperature of cold water = (θ – 10)
Thus , using the formula mc∆θ , we substitute values in the
equation.
0.1 x 4200 x (70 – θ) = 0.2 x 4200 x (θ – 10)
Dividing both sides by 4200
7 – 0.1 θ = 0.2 θ - 2
θ = 30 0 C.

12. ASSIGNMENT
1. The temperature of piece of copper of mass 250g
Assignment
is raised t0 100 0 C and is then transferred to a
well lagged aluminium can of mass 10.0g
containing 120g of methylated spirit at 10.00 C.
calculate the final steady temperature after the
spirit has been well stirred. Neglect the heat
capacity of the stirrer and any loses from
evaporation and use the table of specific heat
capacities for any data required. (32.70 C)

13. Find the final temperature of water if a heater
source rated 42W heats 50g water from 20 0 C in
five minutes. ( specific heat capacity of water is
4200J/kg/K.
The temperature of 500g of a certain metal is
raised to 100 0 C and is then placed in 200g of
water at 15 0 C. if the final steady temperature
rises to 21 0 C, calculate the specific heat
capacity of the metal. (128J/kg/K.)
Reference physics 5th edition A.F . ABBOTT. Page 198-203.

14. DETERMINATION OF SPECIFIC HEAT CAPACITY
 HeatDETERMINATIONenergy transfer hence the
is a form of OF SPECIFIC HEAT CAPACITY
law of energy transfer applies.
 Heat gain equals heat lost.
 There are various methods of determining
specific heat capacity
a)Mixture method.
b)Electrical method.
c)Mechanical method.
 In this coarse we shall look at the first two.

15. Mixture Method
1. Solids
Experiment 9.2: To determine the specific heat capacity by method of
mixture.
Apparatus :
metal block, beaker ,water, tripod stand, heat source, well lagged
calorimeter, stirrer, thermometer and cardboard.
procedure : the learner to read and follow the procedure on page 264
secondary klb bk 3
2. Liquids
The learner to follow the procedure on page 265 of secondary physics
klb bk 3.

16. Sample Question 1
1. A lagged copper calorimeter of mass 0.75kg contains 0.9kg of water at 20 0 C.
A bolt of mass 0.8kg is transferred from oven at 400 0 C to the calorimeter
and a steady temperature of 50 0 C is reached by water after stirring.
Calculate the specific heat capacity of the material of the bolt.( specific heat
capacity of copper is 400 Jkg-1 K-1 and that of water 4 200Jkg-1 K-1 .
2. A block of iron of mass 1.25 kg at 120 0 C was transferred to an aluminium
calorimeter of mass 0.3kg containing a liquid of mass 0.6kg at 25 0 C. the
block and the calorimeter with its contents eventually reached a common
temperature of 50 0 C. given the specific heat capacity of iron is 450 Jkg-1 k-1
,calculate the specific heat capacity of the liquid.

17. Electrical Method
 Solids
Experiment: to determine the specific heat capacity of
material by electrical method.
The learner to read the procedure for experiment 9.3 on
page 266-268.

18. Sample question 2
1. A metal cylinder of mass 0.5kg is heated electrically .
If the voltmeter reads 15v,the ammeter 3.0v and the
temperature of the block rises from 20 0 C to 85 0 C
in 10 minutes. Calculate the specific heat capacity of
the metal cylinder.
heat supplied = heat gained
vIt = mc∆θ
15 x 3 x 10 x 60= 0.5 x c x 65
c= 831 J/kg/K

19. Sample Question 3
In an experiment to determine specific heat
capacity of water , an electrical heater was
used. If the voltmeter reading was 24 V and
that of ammeter reading was 2.0 A. calculate
the specific heat capacity of water if the
temperature of a mass 1.5kg of water in a
0.4kg copper calorimeter rose by 6 0 C after
13.5 minutes. (specific heat capacity of copper
is 400J/kg / K.

20. CHANGE OF STATE
 Heating leads to a rise in temperature . Sometimes no
observable changes is noted.
 When the ice is about -10 0 C is heated, heat energy is
used in raising its temperature to 0 0 C .
 Heat energy supplied to the ice at 0 0 C is used to
change ice from solid to liquid.
 Heat supplied to ice does not change the temperature
of ice but change its state from solid to liquid.

21. LATENT HEAT
 This is heat involved in change of state of a substance.
It can either be heat loss or heat gain.
 This heat is ‘ latent’ means hidden or concealed
because it does not show its presence by change in
temperature as the extra heat goes into change in state.
 There two types of latent heat:
a) Latent heat of fusion
b) Latent heat of vapourization

22. Latent of fusion
 This is heat required to change the state of a material
from solid to liquid or from liquid to solid without
change in temperature.
 As liquid changes to solid latent heat of fusion is given
out and the amount of heat is absorbed when a solid
changes to liquid.
 The graph below shows temperature vs time(s)

23. Temperature against time.
Temp(0 C)
(+)
Time(s)
(-)

24. Experiment
Experiment : To explore the change of state of
naphthalene using cooling curve.
Apparatus: Naphthalene, test tube, water, thermometer
the learner to copy fig. 9.8 plus the procedure.
Secondary physics klb bk 3 pg 273.

25. The Cooling Curve Of Naphthalene
Temp(0 C)
melting pt -----------------------------------
……………………………………………………

26. EXPLANATION
OP – liquid naphthalene cooling
PQ – liquid naphthalene changes to solid at
constant temperature( latent heat of
fusion is given out)
this point is also known as melting
point.
QR – solid naphthalene cools to room
temperature

27. SPECIFIC LATENT HEAT OF FUSION((Lf )
This is the quantity of heat required to
change the unit mass of a substance from
solid to liquid without change in
temperature.
 Lf = Q/m
Q = mLf
 SI unit of specific latent heat is Jkg-1

28. TABLE OF VALUES FOR SPECIFIC HEAT
CAPACITIES
MATERIAL SPECIFIC LATENT HEAT OF FUSION
(X 105 JKg-1 )
Copper 4.0
Aluminium 3.9
Water (ice) 3.34
Wax 1.8
Naphthalene 1.5
Solder 0.7
Lead 0.026
mercury 0.013

29. DETERMING SPECIFIC LATENT HEAT OF FUSION OF A
MATERIAL
 There are various methods of determining
specific latent heat of fusion of a material
a) Mixture Method
 Apparatus
 water
 Ice pieces
 Calorimeter
 stirrer

30. PROCEDURE
Find the mass of the calorimeter – M1
 Place water with temperature of about 50 C above
the room temperature into the calorimeter.
 Mass of water + calorimeter – M2
Record temperature of water in calorimeter – θ1
 Add pieces of ice to the calorimeter
 Mass of calorimeter and mixture – M3
Measure the final temp. of the mixture after
stirring – θ2

31. DATA ANALYSIS
Heat lost by warm water + heat lost by
calorimeter = heat gained by ice at 00 C to
water at 00 C. + heat gained by water (00 C to
final temperature)
Let quantity of heat required to melt a unit
mass of ice at 00 C to 00 C to water at 00 C
be Lf.
 (m2-m1)Cw (θ1-θ2)+ m1Cc (θ1-θ2) = (m3-
m2) Lf +(m3-m2)Cw (θ2- 0).

32. ELECTRICAL METHOD
With electrical method the quantity of
heat is calculated as follows.
Heat supplied by the heater = heat gained
by the ice.
Q = mLf = VIt
Lf = VIt /m

33. SAMPLE QUESTION 1
Calculate the quantity of heat required to
melt 4 kg of ice and to raise the
temperature of the water formed to 50 0 C.
take the specific latent heat of ice to be
3.4 x 105 J/kg and the specific heat
capacity of water to be 4200 J/kg K.

34. SAMPLE QUESTION 2
A beaker contains 200 g of water at 15 0 C. 25 g of ice
at 00 C is added to the water which is stirred until the
ice is completely melted.
a) How much heat is needed to melt all the ice.
b) What is the mass of water produced by
melting all the ice.
c) Calculate the lowest temperature of all
mixture, assuming that all the heat to melt the ice is
taken ice is taken from the water and no heat enters
or leaves the system. ( specific latent heat of fusion
of ice 336 000 J / kg.

35. LATENT HEAT OF VAPOURASATION
Heat energy absorbed by a liquid as it changes
its state to vapour without change in
temperature.
OR
 Heat energy given out by a vapour as it
changes its state to liquid without change in
temperature.

36. SPECIFIC LATENT HEAT OF
VAPORIZATION(Lv)
 This the heat required to convert unit
mass of a liquid, at a boiling point, into
vapour without change in temperature.
 The SI unit Jkg-1
Q = m Lv
Lv = Q/m

37. Specific Latent Heat of Vaporization
Material Specific latent heat of vaporization
(x 105 Jkg-1 )
Water 22.6
Benzene 4.0
Petrol 8.5
Alcohol 8.6
Ether 3.5
Turpentine 2.7
Ethanol 8.5

38. SAMPLE QUESTION
1. Dry steam is passed into a well-lagged copper
can of mass 250 g containing 400 g of water
and 50 g of ice at 00 C. The mixture is well
stirred and the steam supply cut off when the
temperature of the can and its content reaches
20 0 C. Neglect heat losses, find the mass of
steam condensed. (specific heat capacities:
water 4200J/kg K; copper 400 J/kg K ;
specific latent heats of fusion of steam 22.6 x
105 J/kg.)

39. SOLUTION
 Using the principle of conservation of energy, we may say
 heat given out stem = heat received by ice , water and
can
 let the mass of steam condensed = m (g)
Heat in joules given out by:
Steam condensing to water at 1000 C = m x 226000
Condensed steam cooling from 1000 C to 20 0 C

40. SAMPLE QUESTION 2
 A jet of dry steam at 1000C is sprayed onto the
surface of 100g of dried ice at 00C placed in a
plastic container of negligible heat capacity.
The temperature of the mixture is 400C when
the total mass of the water in the container
120g. Given that the specific heat capacity of
water is 4200J/ kg/K and latent heat of fusion
of ice is 336KJkg-1; determine the specific
latent heat of vaporization of water.

41. Factors affecting melting and boiling
points
Boiling point
 There are two factors affecting the boiling
point of a liquid.
a) Pressure
b) Impurities
Experiment 9.13: investigates the effects
of increased pressure on boiling points.

42. EFFECT OF PRESSURE ON BOILING POINT
 Increase in pressure increases the boiling point of a
liquid.
 Application of this concept is the pressure cooker. It
has tight fitting lid which prevents free escape of
steam thus making the pressure inside to build up.
Increase in the boiling point to high temperatures
enables food to cook faster.
 Decrease in pressure lowers the boiling point of a
liquid .

43. Effects of impurities on boiling point
Experiment 9.15: investigates the effects
impurities on boiling points
The boiling point for of the salt solution is
higher than that of the distilled water.
The presence of impurities in liquid raises
its boiling points.

44. Melting Point
There are two factors that affect the
melting point of a substance.
a) Pressure
b) Impurities

45. PRESSURE
 EXPERIMENT: To investigate the effect of
pressure on melting point.
 Apparatus : Block of ice, thin copper wire,
two heavy weight, wooded support.
Procedure:
 Attach two heavy weigh to the ends of a thin
copper wire.
 Pass the string over a large block of ice, as
shown in the figure.

46. OBSERVATION
 The wire cuts through the ice block, but
leaves it as one piece.
 This process is known as regelation.

47. Explanation (regelation)
 Weight exerts pressure on the ice beneath; this
pressure makes it melts at a temperature lower
than its melting point.
 The water formed loses its latent heat of fusion to
the wire which hence solidifies again as it is no
longer under pressure.
 The latent heat losed by the water is conducted to
the wire which melts the ice below it.
 This process continues until the wire cuts
through leaving the block.

48. CONCLUSION
 Application of pressure on ice lowers the melting
point.
 NB:
 If the wire with lower thermal conductivity is
used, it will cut through slowly.
 Poor conductors of heat e.g. cotton will not cut
through the block at all because it does not
conduct heat.

49. APPLICATION OF THE EFFECTS OF PRESSURE
ON MELTING POINT OF ICE.
Ice skating
 Weight of the skater exerts pressure on the ice
below causing melting at a lower
temperatures.
 The high pressure reduces melting hence
melting them forming a thin film of water
over which skater slides.

50. Joining ice cubes under pressure
 By pressing ice cubes hard under
pressure, the melting points between
points of contact of the ice is lowered;
water recondenses and the two cubes are
joined together.

51. EFFECTS OF IMPURITIES
Application of impurities lowers the
melting point the melting point of a
substance.
Salt is spread on roads during winter to
prevent freezing of roads.

52. EVAPORATION
 Evaporation occurs on the surface of the
liquid where molecules escape to air.
 Molecules at the surface have higher kinetic
energy than those ones below hence they
break from their attractive forces of the
neighbouring molecules.
 Evaporation takes place at all temperatures.

53. Effects of Evaporation
The following are some effects of
evaporation
 Methylated spirit feels cold on the back of
your hand than water.
 This is because hands feel cold as the spirit
evaporates from the skin. Evaporating methylated
spirit extracts latent heat of vaporisation from the
skin making it feel cold.

54.  A beaker placed on the film of water on a
wooden block as shown on the fig. 9.19 pg
286 klb bk 3.
 The beaker stucks on the to the wooden block after
the air is blown through the ether using a foot
pump.
 This is because blowing increases the rate of
evaporation of ether forming a layer of ice
between beaker and wood. This shows that
evaporation causes cooling.
NB: bubbling increases surface area of ether
exposed to air.

55. Factors affecting rate of evaporation
Temperature
 Increase in temperature increases kinetic
energy of molecules on the surface; these
molecules move faster hence many of them
escape ; enhancing evaporation.
 NB: increase in temperature increases the
rate of evaporation.

56. 2. Surface Area
 Increasing surface area of the liquid exposes
more liquid molecules hence faster molecules
escapes to the environment.
 Large surface area also clears the way for
more molecules to enter the space.

57. 3. Drought
 Passing air over the liquid sweeps away
escaping vapour molecules e.g. clothes
dry faster on a windy day, people take
hot beverages by blowing over it.

58. 4. Humidity
 Humidity is the concentration of water
vapour in the atmosphere .
 High humidity lowers the rate at which
molecules enter the space hence lowers the
rate of evaporation.
 This is why clothes take longer time to dry on
a humid day than on dry one.

59. Comparison Between Evaporation And
Boiling
Evaporation Boiling
Takes place at all Takes place at a fixed
temperatures temperature
Takes place on the Takes place throughout
surface of the liquid the liquid
No bubbles are formed Bubbles of steam are
formed
Decreasing atmospheric Decreasing atmospheric
pressure increases the pressure lowers the
rate of evaporation boiling point