Chapter 4 : Heat
JPN Pahang
Physics Module Form 4Teachers Guide
Chapter 4: Heat
4.1 : UNDERSTANDING THERMAL EQUILIBRIUMBy the end of this
subtopic, you will be able to
Explain thermal equilibrium
Explain how a liquid-in glass thermometer works
1. The net heat will flow from A to B until the temperature of A
is the ( same, zero as the temperature of B. In this situation, the
two bodies are said to have reached thermal equilibrium.2. When
thermal equilibrium is reached, the net rate of heat flow between
the two bodies is (zero, equal)3. There is no net flow of heat
between two objects that are in thermal equilibrium. Two objects in
thermal equilibrium have the same temperature.4. The liquid used in
glass thermometer should (a) Be easily seen
(b) Expand and contract rapidly over a wide range of
temperature
(c) Expand uniformly 5. List the characteristic of mercury
(a) Opaque liquid
(b) Does not stick to the glass
(c) Expands uniformly when heated
(d) Freezing point -390C
(e) Boiling point 3570C6. ( Heat, Temperature ) is a form of
energy. It flows from a hot body to a cold body.
7. The SI unit for ( heat , temperature) is Joule, J.
8. ( Heat , Temperature ) is the degree of hotness of a body
9. The SI unit for (heat , temperature) is Kelvin, K.
10. Lower fixed point (l 0 )/ ice point : the temperature of
pure melting ice/00C11. Upper fixed point( l 100)/steam point: the
temperature of steam from water that is boiling under standard
atmospheric pressure /1000C
Exercise 4.1Section A: Choose the best answer1. The figure shows
two metal blocks. Which the following statement is false?
A. P and Q are in thermal contact
B. P and Q are in thermal equilibrium
C. Energy is transferred from P to Q
D. Energy is transferred from Q to P2. When does the energy go
when a cup of hot tea cools?A. It warms the surroundings
B. It warms the water of the tea
C. It turns into heat energy and disappears.3. Which of the
following temperature corresponds to zero on the Kelvin scale?A.
2730 C
B. 00C
C. -2730 C
D. 1000 C
4. How can the sensitivity of a liquid- in glass thermometer be
increased?
A. Using a liquid which is a better conductor of heat
B. Using a capillary tube with a narrower bore.C. Using a longer
capillary tube
D. Using a thinner-walked bulb
5. Which instrument is most suitable for measuring a rapidly
changing temperature?
A. Alcohol-in glass thermometer
B. Thermocouple
C. Mercury-in-glass thermometer
D. Platinum resistance thermometer6. When shaking hands with
Anwar, Kent Hui niticed that Anwars hand was cold. However, Anwar
felt that Kent Hui hand was warm. Why did Anwar and Kent Hui not
feel the same sensation?
A. Both hands in contact are in thermal equilibrium.
B. Heat is flowing from Kent Huis hand to Anawrs hand
C. Heat is following from Anwars hand to Kent Hui hand.Section
B: Answer all the questions by showing the calculation1. The length
of the mercury column at the ice point and steam point are 5.0 cm
and 40.0cm respectively. When the thermometer is immersed in the
liquid P, the length of the mercury column is 23.0 cm. What is the
temperature of the liquid P?
Temperature, = l l0 x 1000C l100 l0
= 23 5 x 1000C
40 - 5
= 51.430C2. The length of the mercury column at the steam point
and ice point and are 65.0 cm and 5.0cm respectively. When the
thermometer is immersed in the liquid Q, the length of the mercury
column is 27.0 cm. What is the temperature of the liquid Q?
Temperature, = l l0 x 1000C
l100 l0
= 27 5 x 1000C
65 - 5
= 36.670C3. The distance between 00C and 1000C is 28.0 cm. When
the thermometer is put into a beaker of water, the length of
mercury column is 24.5cm above the lower fixed point. What is the
temperature of the water?
Temperature, = l l0 x 1000C
l100 l0
= 24.5 0 x 1000C
28 - 0
= 87.50C
4. The distance between 00C and 1000C is 25 cm. When the
thermometer is put into a beaker of water, the length of mercury
column is 16cm above the lower fixed point. What is the temperature
of the water? What is the length of mercury column from the bulb at
temperatures i) 300C
Temperature, = l l0 x 1000C
l100 l0
= 16 0 x 1000C
25 - 0
= 64.00C
Temperature, = l l0 x 1000C
l100 l0
300C = x 0 x 1000C
25 - 0
x = 7.5cmSECTION C: Structured Questions1. Luqman uses an
aluminium can, a drinking straw and some plasticine to make a
simple thermometer as shown in figure below. He pours a liquid with
linear expansion into the can.
(a) Suggest a kind of liquid that expands linearly. (1m)
.(b) He chooses two fixed points of Celsius scale to calibrate
his thermometer. State them (2m)
(c) If the measurement length of the liquid inside the straw at
the temperature of the lower fixed point and the upper fixed point
are 5cm and 16 cm respectively, find the length of the liquid at
82.50C.
(d) Why should he use a drinking straw of small diameter?
(e) What kind of action should he take if he wants to increase
the sensitivity of his thermometer?
2. What do you mean by heat and temperature?
....
4.2 : UNDERSTANDING SPECIFIC HEAT CAPACITY
By the end of this subtopic, you will be able to
Define specific heat capacity State that c = Q/MC
Determine the specific heat capacity of a liquid
Determine the specific heat capacity of a solid
Describe applications of specific heat capacity
Solve problems involving specific heat capacity
1. The heat capacity of a body is the amount of heat that must
be supplied to the body to increase its temperature by 10C.2. The
heat capacity of an object depends on the
(a) .(b) .(c) 3. The specific heat capacity of a substance is
the amount of heat that must be supplied to increase the
temperature by 1 0C for a mass of 1 kg of the substance. Unit Jkg-1
K-14. The heat energy absorbed or given out by an object is given
by Q = mcO.5. High specific heat capacity absorb a large amount of
heat with only a small temperature increase such as plastics.6.
Conversion of energy
7. Applications of Specific Heat Capacity
Explain the meaning of above application of specific heat
capacity:
(a) Water as a coolant in a car engine
(i) Water is a has high specific heat capacity. It is used as a
cooling agent to prevent overheating of the engine .Therefore,
water acts as a heat reservoir as it can absorb a great amount of
heat with small increase in temperature.(b) Household apparatus and
utensils-The base and the body of a cooking pot is made up of
material that has small specific heat capacity so that it can
absorb heat faster and become hot quickly.
-The handle of cooking pot is made up of wood that has large
specific heat capacity so that the handle become not too hot (also
acts as insulator). (c) Sea breeze
(d) Land breeze
Exercise 4.2SECTION A : Choose the best answer
1. The change in the temperature of an object does not depend
on
A. the mass of the object
B. the type of substance the object is made of
C. the shape of the object
D. the quantity of heat received
2. Which of the following defines the specific heat capacity of
a substance correctly?A. The amount of heat energy required to
raise the temperature of 1kg of the substance
B. The amount of heat energy required to raise 1kg of the
substance by 10C.
C. The amount of heat energy required to change 1kg of the
substance from the solid state to the liquid state.
3. Heat energy is supplied at the same rate to 250g of water and
250g of ethanol. The temperature of the ethanol rises faster. This
is because the ethanol..
A. is denser than water
B. is less dense than water
C. has a larger specific heat capacity than water
D. has a smaller specific heat capacity than water4. In the
experiment to determine the specific heat capacity of a metal
block, some oil is poured into the hole containing thermometer. Why
is this done?A. To ensure a better conduction of heat
B. To reduce the consumption of electrical energyC. To ensure
the thermometer is in an upright position.
D. To reduce the friction between the thermometer and the wall
of the block.
SECTION B: Answer all questions by showing the calculation1. How
much heat energy is required to raise the temperature of a 4kg iron
bar from 320C to 520C? (Specific heat capacity of iron = 452 Jkg-1
0C-1).
Amount of heat energy required, Q = mc
= 4 x 452 x (52-32)
= 36 160 J
2. Calculate the amount of heat required to raise the
temperature of 0.8 kg of copper from 350C to 600C. (Specific heat
capacity of copper = 400 J kg-1 C-1).
Amount of heat required, Q = mc
= 0.8 x 400 x (60-35)
= 8 000 J3. Calculate the amount of heat required to raise the
temperature of 2.5 kg of water from 320C to 820C. (Specific heat
capacity of water = 4200 J kg-1 C-1).
Amount of heat required, Q = mc
= 2.5 x 4200 x (82-32)
= 525, 000 J4. 750g block of a aluminium at 1200C is cooled
until 450C. Find the amount of heat is released. . (Specific heat
capacity of aluminium = 900 J kg-1 C-1).
Amount of heat released, Q = mc
= 0.75 x 900 x (120-45)
= 50 625 J5. 0.2 kg of water at 700C is mixed with 0.6 kg of
water at 300C. Assuming that no heat is lost, find the final
temperature of the mixture. (Specific heat capacity of water = 4200
J kg-1 C-1)
Amount of heat required, Q = Amount of heat released, Q
mc = mc 0.2 x 4200 x ( 70- ) = 0.6 x 4200 x ( - 30)
= 40 0CSECTION C: Structured questions1. In figure below, block
A of mass 5kg at temperature 1000C is in contact with another block
B of mass 2.25kg at temperature 200C.
Assume that there is no energy loss to the surroundings.
(a) Find the final temperature of A and B if they are in thermal
equilibrium. Given the specific heat capacity of A and B are 900
Jkg-1 C-1 and 400 Jkg-1 C-1 respectively.
Amount of heat required, Q = Amount of heat released, Q
mc = mc
5.0x 900 x ( 100- ) = 2.25 x 400 x ( - 20)
= 86.670C(b) Find the energy given by A during the process.
Energy given = mc
= 5 x 900 x (100 86.67)
= 60 000 J
(c) Suggest one method to reduce the energy loss to the
surroundings.
..
4.3 UNDERSTANDING SPECIFIC LATENT HEATBy the of this subtopic,
you will be able to State that transfer of heat during a change of
phase does not cause a change in temperature Define specific latent
heat State that l = Q/m Determine the specific latent heat of
fusion and specific latent heat of vaporisation Solve problem
involving specific latent heat.1. Four main changes of phase.
2. The heat absorbed or the heat released at constant
temperature during a change of
phase is known as latent heat. Q= ml3. Complete the diagrams
below and summarized.(a) Melting
(b) Boiling(c) Solidification(d) Condensation
4. is the heat absorbed by a melting solid. The specific latent
heat of fusion is the quantity of the heat needed to change 1kg of
solid to a liquid at its melting point without any increase in ..
The S.I unit of the specific latent heat of fusion is Jkg-1.
5. ... is heat of vaporisation is heat absorbed during boiling.
The specific latent heat of vaporisation is the quantity of heat
needed to change 1kg of liquid into gas or vapour of its boiling
point without any change in .. The S.I unit is Jkg-1.
6. Explain the application of Specific Latent Heat above:(d)
Cooling of beverage
(e) Preservation of Food
(f) Steaming Food
(g) Killing of Germs and Bacteria
EXERCISE 4.3Section A:1. The graph in figure below shows how the
temperature of some wax changes as it cools from liquid to solid.
Which section of the graph would the wax be a mixture of solid and
liquid?
A. PQ
B. QR
C. RS
D. ST
2. Figure show a joulemeter used for measuring the electrical
energy to melt some ice in an experiment. To find the specific
latent heat of fusion of ice, what must be measured?
A. The time taken for the ice to melt
B. The voltage of the electricity supply
C. The mass of water produced by melting ice
D. The temperature change of the ice.
3. It is possible to cook food much faster with a pressure
cooker as shown above. Why is it easier to cook food using a
pressure cooker?
A. More heat energy can be supplied to the pressure cooker
B. Heat loss from the pressure cooker can be reduced.
C. Boiling point of water in the pressure cooker is raised
D. Food absorbs more heat energy from the high pressure
steam
4. Which of the following is not a characteristics of water that
makes it widely used as a cooling agent?A. Water is readily
available
B. Water does not react with many other substance
C. Water has a large specific heat capacity
D. Water has a large density
5. Figure below shows the experiment set up to determine the
specific latent heat of fusion of ice. A control of the experiment
is set up as shown in Figure (a) with the aim of
A. determining the rate of melting of iceB. ensuring that the
ice does not melt too fast.
C. determining the average value of the specific latent heat of
fusion of ice.
D. determining the mass of ice that melts as a result of heat
from the surroundings
6. Scalding of the skin by boiling water is less serious then by
steam. This is becauseA. the boiling point of water is less than
the temperature of steam
B. the heat of boiling water is quickly lost to the
surroundings
C. steam has a high specific latent heat.
D. Steam has a high specific heat capacity.SECTION B: Answer the
question by showing the calculation
2. 300g of ice at 00C melts. How much energy is required for
this
Q = ml
= 0.3 x 334 000 kJ kg-1
= 100200 JQuestion 2-7 are based on the following information
Specific heat capacity of water = 4 200 J kg-1 C-1 Specific heat
capacity of ice = 2 100 J kg-1 C-1 Specific latent heat of fusion
of ice = 3.34 X 105J kg-1 Specific latent heat of vaporization of
water = 2.26 X 106 J kg-13. An immersion heater rated at 500 W is
fitted into a large block of ice at 00C. How long does it take to
melt 1.5kg of ice?
Q = ml
Pt = 1.5 x 3.34 x 105
500 x t = 501 000
t = 1002 s4. 300 g of water at 400C is mixed with x g of water
at 800C. The final temperature of the mixture is 700C. Find the
value of x.x = 0.9 kg
5. Calculate the amount of heat released when 2 kg of ice at 00C
is changed into water at 00C.668000 J
6. Calculate the amount of heat needed to convert 3 kg of ice at
00C to water at 300C.1.38 x 106 J7. Find the amount of heat needed
to convert 0.5 kg of ice at -150C into steam at 1000C1.523 x 106
J8. Calculate the amount of heat needed to convert 100 g of ice at
00C into steam at 1000C.3.014 x 105 J9. The specific latent heat of
vaporization of water is 2300 kJ kg-. How much heat will be
absorbed when 3.2 kg of water is boiled off at its boiling
point.7360 kJ / 7360 000 J4.4 UNDERSTANDING THE GAS LAWBy the end
of this subtopic; you will be able to : Explain gas pressure,
temperature and volume in terms of the behavior of gas
molecules.
Determine the relationship between
(i) pressure and volume
(ii) volume and temperature
(iii) pressure and temperature
Explain absolute zero and the absolute/Kelvin scale of
temperature
Solve problems involving pressure, temperature and volume of a
fixed mass of gas
1. Complete the table below.
Property of gasExplanation
Volume,V m3
The molecules move freely in random motion and fill up the whole
space in the container.
The volume of the gas is equal to the volume of the
container
Temperature,T K (Kelvin) The molecules are in continuous random
motion and have an average kinetic energy which is proportional to
the temperature.
Pressure,P Pa(Pascal) The molecules are in continuous random
motion. When a molecules collides with the wall of the container
and bounces back, there is a change in momentum and a force is
exerted on the wall The force per unit area is the pressure of
gas
2. The kinetic theory of gas is based on the following
assumptions:(a) The molecules in a gas move freely in random motion
and posses kinetic energy
(b) The force of attraction between the molecules are
negligible.
(c) The collisions of the molecules with each other and with the
walls of the container are elastic collisions
4.4.1 Boyles Law
1. Boyles law states that for a fixed mass of gas, the pressure
of the gas is inversely proportional to its volume when the
temperature is kept constant.2. Boyles law can be shown graphically
as in Figure above
3. The volume of an air bubble at the base of a sea of 50 m deep
is 250cm3. If the atmospheric pressure is 10m of water, find the
volume of the air bubble when it reaches the surface of the
sea.
4.4.2 Charless Law
1. Charles law states that for a fixed mass of gas, the volume
of the gas is directly proportional to its absolute temperature
when its pressure is kept constant.2. The temperature -2730C is the
lowest possible temperature and is known as the absolute zero of
temperature.3. Fill the table below.TemperatureCelsius scale
(0C)Kelvin Scale(K)
Absolute zero-2730
Ice point0273
Steam point100373
Unknown point( + 273 )
4. Complete the diagram below. V/m34.4.3 Pressures Law
1. The pressure law states that for a fixed mass of gas, the
pressure of the gas is directly proportional to its absolute
temperature when its volume is kept constant.EXERSICE 4.4Gas
Law
1. A mixture of air and petrol vapour is injected into the
cylinder of a car engine when the cylinder volume is 100 cm3. Its
pressure is then 1.0 atm. The valve closes and the mixture is
compressed to 20 cm3. Find the pressure now.P2 = 5.0 atm2. The
volume of an air bubble at the base of a sea of 50 in deep is 200
cm3. If the atmospheric pressure is 10 in of water, find the volume
of the air bubble when it reaches the surface of the sea.
V2 = 1 200 cm33. The volume of an air bubble is 5 mm3 when it is
at a depth of h m below the water surface. Given that its volume is
15 mm3 when it is at a depth of 2 m, find the value of h.
(Atmospheric pressure = 10 m of water)
h = 26 m4. An air bubble has a volume of V cm3 when it is
released at a depth of 45m from the water surface. Find its volume
(V) when it reaches the water surface. (Atmospheric pressure = 10 m
of water)
V2 = 5.5 V15. A gas of volume 20m3 at 37 oC is heated until its
temperature becomes 87oC at constant pressure. What is the increase
in volume?
V2= 23.2 m36. The air pressure in a container at 33 oC is 1.4 X
105 N m2. The container is heated until the temperature is 550C.
What is the final air pressure if the volume of the container is
fixed?
P2 = 1.5 x 105 Nm-27. The volume of a gas is 1 cm3 at 15 oC. The
gas is heated at fixed pressure until the volume becomes triple the
initial volume. Calculate the final temperature of the gas.
T2 = 864 K (2 = 591 oC8. An enclosed container contains a fixed
mass of gas at 250C and at the atmospheric pressure. The container
is heated and temperature of the gas increases to 980C. Find the
new pressure of the gas if the volume of the container is
constant.(Atmospheric pressure = 1.0 X 105N m2)P2 = 1.25 x 105 Nm-2
/ Pa
9. The pressure of a gas decreases from 1.2 x 105 Pa to 9 x 105
Pa at 400C. If the volume of the gas is constant, find the initial
temperature of the gas.
T1 = 41.73 K
(1 = -231.27 oCPART A: CHAPTER 41. A 5kg iron sphere of
temperature 500C is put in contact with a 1kg copper sphere of
temperature 273K and they are put inside an insulated box. Which of
the following statements is correct when they reach thermal
equilibrium?
D. A iron sphere will have a temperature of 273K
E. The copper sphere will have a temperature of 500C.
F. Both the sphere have the same temperature.
G. The temperature of the iron sphere will be lower than
500C
2. In the process to transfer heat from one object to another
object, which of the following processes does not involve a
transfer to material?
A. Convection
B. Vaporisation
C. Radiation
D. Evaporation
3. When we use a microwave oven to heat up some food in a lunch
box, we should open the lid slightly. Which of the following
explanations is correct?A. To allow microwave to go inside the
lunch box
B. To allow the water vapors to go out, otherwise the box will
explode
C. To allow microwave to reflect more times inside the lunch
box
D. To allow microwave to penetrate deeper into the lunch
box.
4. Water is generally used to put out fire. Which of the
following explanation is not correct?A. Water has a high specific
heat capacity
B. Steam can cut off the supply of oxygen
C. Water is easily available
D. Water can react with some material
5. Given that the heat capacity of a certain sample is 5000
J0C-1. Which of the following is correct?A. The mass of this sample
is 1kg.
B. The energy needed to increase the temperature of 1 kg of this
sample is 5000 J.
C. The energy needed to increase the temperature of 0.5kg of
this sample is 2500J.
D. The temperature of this sample will increase 10C when 5 000 J
energy is absorbed by this sample.6. Which of the following
statement is correct?A. The total mass of the object is kept
constant when fusion occurs.
B. The internal energy of the object is increased when
condensation occurs
C. Energy is absorbed when condensation occurs.
D. Energy is absorbed when vaporization occurs.
7. Water molecules change their states between the liquid and
gaseous statesA. only when water vapour is saturated
B. at all times because evaporation and condensation occur any
time
C. only when the vapour molecules produce a pressure as the same
as the atmospheric pressure
D. only when the water is boiling
8. Based on the kinetic theory of gas which one of the following
does not explain the behaviour of gas molecules in a container?
A. Gas molecules move randomly
B. Gas molecules collide elastically with the walls of the
container
C. Gas molecules move faster as temperature increases
D. Gas molecules collide inelastically with each other
9. A cylinder which contains gas is compressed at constant
temperature of the gas increase becauseA. the average speed of gas
molecules increases
B. the number of gas molecules increases
C. the average distance between the gas molecules increases
D. the rate of collision between the gas molecules and the walls
increases
10. A plastic bag is filled with air. It is immersed in the
boiling water as shown in diagram below.
Which of the following statements is false?A. The volume of the
plastic bag increases.
B. The pressure of air molecules increasesC. The air molecules
in the bag move faster
D. The repulsive force of boiling water slows down the movement
of air molecule
PART B; 1. A research student wishes to carry out an
investigation on the temperature change of the substance in the
temperature range -500C to 500C. The instrument used to measure the
temperature is a liquid in glass thermometer.
ThermometerABCD
LiquidMercuryMercuryAlcoholAlcohol
Freezing point of liquid (0C)-39-39-112-112
Boiling point of liquid (0C)360360360360
Diameter of capillary tubeLargeSmallLargeSmall
Cross section
Table 1(a) (i) State the principle used in a liquid- in glass
thermometer.(1m)
........................................................................................................................................(ii)
Briefly explain the principle stated in (a)(i) (3m)
.
.
.
(b) Table 1 shows the characteristic of 4 types of thermometer:
A,B C and D. On the basis of the information given in Table 1,
explain the characteristics of, and suggest a suitable thermometer
for the experiment.(5 m) ..(c) The length of the mercury column in
uncalibrated thermometer is 6.0cm and 18.5 cm at 00C and 1000C.
respectively. When the thermometer is placed in a liquid, the
length of the mercury column is 14.0cm
(i) Calculate the temperature of the liquid
The temperature of the liquid = 8.0 x 100
12.5
= 64 0C
(ii) State two thermometric properties which can be used to
calibrate a thermometer. (6m)
2. A metal block P of mass 500 g is heated is boiling water at a
temperature of 1000C. Block P is then transferred into the water at
a temperature of 300C in a polystyrene cup. The mass of water in
the polystyrene cup is 250 g. After 2 minutes, the water
temperature rises to 420C.Assuming that the heat absorbed by the
polystyrene cup and heat loss to the surroundings are
negligible.{Specific heat capacity of water 4 200 j kg-1 C-1)
Calculate
(a) the quantity of heat gained by water the polystyrene cup
Q = mc
= 0.250 x 4200 x (42-30)
= 12 600J
(b) the rate of heat supplied to the water
Rate of heat supplied to the water = 12 600J
120s
= 105 Js-1(c) the specific heat capacity of the metal block
P
Heat supplied by metal block P = heat gained by water
0.500 x c x(100 -42) = 12 600J
c = 434 J kg-1 C-13. A student performs an experiment to
investigate the energy change in a system. He prepares a cardboard
tube 50.0 cm long closed by a stopper at one end. Lead shot of mass
500 g is placed in the tube and the other end of the tube is also
closed by a stopper. The height of the lead shot in the tube is 5.0
cm as shown in Figure 3.1. The student then holds both ends of the
tube and inverts it 100 times (Figure 3.2).
(a) State the energy change each time the tube is inverted.
..
..
(b) What is the average distance taken by the lead shot each
time the tube is inverted? 45.0 cm(c) Calculate the time taken by
the lead shot to fall from the top to the bottom of the tube.
S = ut + at2 0.45 = 0 + (10)t2 t = 0.3s(d) After inverting the
tube 100 times, the temperature of the lead shot is found to have
increased by 30C.
i. Calculate the work done on the lead shot.
Work done = (100) mgh
= 100 x 0.500 x 10 x 0.45
= 225 J
ii. Calculate the specific heat capacity of lead.
mc = 225 J
c = 225
(0.500 x 3)
= 150 J kg-1 oC-1
iii. State the assumption used in your calculation in (d)ii.
...
.
PART C: EXPERIMENT 1. Before travelling on a long journey,
Luqman measured the air pressure the tyre of his car as shown in
Figure (a) He found that the air pressure of the tyre was 200 kPa.
After the journey, Luqman measured again the air pressure of the
tyre as shown in Figure (b) He found that the air pressure had
increase to 245 kPa. Luqman also found that the tyre was hotter
after the journey although the size of the tyre did not change.
Using the information provided by Luqman and his observations on
air pressure in the tyre of his car:
Choose suitable apparatus such as pressure gauge, a
round-bottomed flask and any other apparatus that may he necessary.
In your description, state clearly the following:
i. Aim of the experiment,
ii. Variables in the experiment,
iii. List of apparatus and materials,
iv. Arrangement of the apparatus,
v. The procedure of the experiment including the method of
controlling the manipulated variable and the method of measuring
the responding variable,vi. The way you would tabulate the
data,
vii. The way you would analyse the data. [10 marks]
InferenceAt constant volume, the air pressure depends on the
temperature
HypothesisAt constant volume, the air pressure increase as the
temperature increases
AimTo investigate the relationship between the air pressure and
the temperature at constant volume.
VariableConstant variable : Air temperatureManipulate variable :
Air pressureResponding variable : Volume of air
Material and ApparatusRound-bottom flask, rubber tube, Bourdon
gauge, beaker, stirrer, thermometer, wire gauze, tripod stand and
Bunsen burner.
Arrangement of apparatus
Procedure The apparatus is set up as shown in the diagram above.
The beaker is filled with ice-cold water until the flask is
completely immersed. The water is stirred and the initial
temperature reading taken. The pressure reading from the bourdon
gauge is also taken.
The water is heated and constant stirred. When the water
temperature increases by 100C, the Bunsen burner is removed and the
stirring of water is continued. The temperature and pressure
readings of the trapped air are recorded in the table
The above procedure is repeated until the water temperature
almost reaches boiling point.
Tabulation of Data
Analysis of Data
CHAPTER 4: HEAT
Thermal equilibrium :Keseimbangan terma
Faster. rate of energy transfer
Hot
object
Cold
object
Slower rate of energy transfer
Equivalent to
Equivalent to
No net heat transfer
A
B
l0: length of mercury at ice point
l100: length of mercury at steam point
l: length of mercury at point
Temperature, =
l - l0
l100 - l0
x 1000C
Alkohol
Lower fixed point = freezing point of water.
Upper fixed point = boiling point of water
100 = 82.5
16-5 x 5
100x 500 = 907.5
x = 14.08cm
To increases the sensitivity of the thermometer
Use a copper can instead of the aluminum can because it is a
better thermal
conductor
Heat is the energy that transfers from one object to another
object because of a temperature difference between them.
Temperature is a measure of degree of hotness of a body.
Heat capacity
Muatan haba
Specific heat capacity
Muatan haba tentu
Temperature of the body
Mass of the body
Type of material
Specific heat capacity , c =
Q__
m
Electrical energy
Heat energy
Pt = mc
Heater
Power = P
Electrical energy
Potential energy
Kinetic energy
Object falls from
A high position
Moving object stopped due to friction
Power = P
Heat energy
mgh= mc
Heat energy
mv2= mc
Small value of c
Big value of c
Two object of equal mass
Equal rate of heat supplied
Faster increase in temperature
Slower increase in temperature
A
B
1000C
200C
5kg
2.25kg
Put them in a sealed polystyrene box// wrap them with thick
tissue paper // thick cloth//felt cloth
Solid
Solidification
Latent heat released
Boiling
Latent heat absorbed
Condensation
Latent heat released
Liquid
Gas
Temperature
Time
.
Temperature
Time
.
Temperature
Time
.
Temperature
Time
.
Latent heat of fusion
temperature
water
ice
Latent heat absorbed
( melting)
heat lost
( freezing)
Latent heat of vaporisation
temperature
water
gas
Latent heat absorbed
( boiling)
heat lost
( condensation)
When ice melts, its large latent heat is absorbed from
surroundings. This property makes ice a suitable substance for use
as a coolant to maintain other substance at a low temperature.
Beverage can be cooled by adding in several cubes of ice. When the
ice melts a large amount of heat (latent heat) is absorbed and this
lowers the temperature of the drink.
The freshness of foodstuff such as fish and meat can be maintain
by placing them in contact with ice. With its large latent heat,
ice is able to absorb a large quantity of heat from the foodstuff
as its melts. Thus food can be kept at a low temperature for an
extended period of time.
Food is cooked faster if steamed. When food is steamed, the
condensed water vapour releases a quantity of latent heat and heat
capacity. This heat flows to the food. This is more efficient than
boiling the food.
Steam that releases a large quantity of heat is used in the
autoclave to kill germs and bacteria on surgery equipment in
hospitals.
Small volume molecules hit wall more often, greater pressure
P 1
V
That is PV = constant
Or P1V1 = P2V2
Relationship between pressure and volume
0
P
1/V
(b) P directly proportional to 1/V
0
P
V
(a) P inversely proportional to V
P1V1 = P2V2
60m (250 x 10-6)m3 = 10m x V2
1.5 x 10-3 m3 = V2
PI=50m + 10m
V1=250cm3
P2= 10m
V T
that is V = constant
T
Relationship between volume and temperature
Lower temperature
Higher temperature, faster molecules, larger volume to keep the
pressure constant
/0C
100
-273
P T
That is P = constant
T
Higher temperature molecules move faster, greater pressure
Relationship between pressure and temperature
Principle of thermal equilibrium
A system is in a state of thermal equilibrium if the net rate of
heat flow between the component of the system is zero. This means
that the component of the system are at the same temperature
Alkohol freezing point is less than -50C, boiling point higher
than 50C.Thus the alcohol will not boil.
Capillary tube has small diameter will produce a large change in
the length thus making the change clearly visible.
Small diameter increases sensitivity of the thermometer
Change of volume of gas with temperature
Change of electrical resistance with temperature
Figure 2
Figure 3.1
Figure 3.2
Gravitational potential energy kinetic energy heat energy
No heat loss to the surroundings/All the gravitational potential
energy is converted into heat energy
Figure (a)
Figure (b)
State one suitable inference that can be made. [1 mark]
State appropriate hypothesis for an investigation. [1 mark]
Design an experiment to investigate the hypothesis stated in
(b).
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