chapter 4 heat teacher's guide
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JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
4.1 : UNDERSTANDING THERMAL EQUILIBRIUM (keseimbangan terma)
By the end of this subtopic, you will be able to
1. The net heat will flow from A to B until the temperature of A is the same ,
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
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) Not stick to the glass wall of the capillary tube
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 3570C
6. Heat is a form of energy. It flows from a hot body to a cold body.
CHAPTER 4: HEAT
Faster rate of energy transfer
Hot object
Cold object
Slower rate of energy transfer
Equivalent to Equivalent to
No net heat transfer
10 2
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
7. The SI unit for heat is Joule, J.
8. Heat is measured by joulemeter
9. Temperature is measured by thermometer
10. Temperature is the degree of hotness of a body
11. The SI unit for temperature is Kelvin, K.
12. Lower fixed point (l 0 )/ ice point : the temperature of pure melting ice/00C
13. Upper fixed point( l 100)/steam point: the temperature of steam from water that is boiling
under standard atmospheric pressure /1000C
Exercise 4.1
Section A: Choose the best answer
1. The figure shows two metal blocks. Which the following statement is false?
l0 : length of mercury at ice pointl100 : length of mercury at steam pointlθ : length of mercury at θ point
Temperature, θ = lθ - l0
l100 - l0x 1000C
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
A. P and Q are in thermal contactB. P and Q are in thermal
equilibriumC. Energy is transferred from P to QD. Energy is transferred from Q to P
2. When does the energy go when a cup of hot tea cools?A. It warms the surroundingsB. It warms the water of the teaC. It turns into heat energy and
disappears.
3. Which of the following temperature corresponds to zero on the Kelvin scale?A. 2730 CB. 00CC. -2730
4. How can the sensitivity of a liquid- in –glass thermometer be increased?A. Using a liquid which is a
better conductor of heatB. Using a capillary tube with a
narrower bore.C. Using a longer capillary tubeD. Using a thinner-walked bulb
5. Which instrument is most suitable for measuring a rapidly changing temperature?A. Alcohol-in –glass
thermometerB. ThermocoupleC. Mercury-in-glass
thermometerD. Platinum resistance
thermometer
6. Which of the following thermometers is suitable for measuring a patient’s temperature?
A. Resistance thermometerB. Mercury thermometerC. Alcohol thermometerD. Thermocouple
7. Which of the following thermometer is suitable for measuring temperature which fluctuate?
A. Resistance thermometerB. Mercury thermometerC. Alcohol thermometerD. Thermocouple
8. To produce a scale for a thermometer, the two fixed points required are
A. ice point and triple pointB. yield point and boiling pointC. ice point and steam point
9. In order to make a mercury thermometer more sensitive, we need to
A. decrease the volume of the mercury bulb
B. decrease the diameter of the capillary tube
C. leave the capillary tube open to the air
10. When shaking hands with Anwar, Kent Hui niticed that Anwar’s 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 Hui’s hand to Anawr’s hand
C. Heat is following from Anwar’s hand to Kent Hui hand
Section B: Answer all the questions by showing the calculation
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
1. 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, θ = lθ – l0 x 1000C
l100 – l0
θ = 23 – 5 x 1000C
40 - 5
θ = 51.420C
2. 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, θ = lθ – l0 x 1000C
l100 – l0
θ = 27 – 5 x 1000C
65 - 5
θ = 36.670C
3. 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, θ = lθ – l0 x 1000C
l100 – l0
θ = 24.5 – 0 x 1000C
28 - 0
θ = 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(freeze) fixed point. What
is the temperature of the water? What is the length of mercury column from the bulb at
temperatures 300C
Temperature, θ = lθ – l0 x 1000C
l100 – l0
65cm l1005cm l0
27 cm (Q) lθ
0 cm l0
24.5 cm (Q) lθ
28 cm l100
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
θ = 16 – 0 x 1000C
25 - 0
θ = 64.00C
Temperature, θ = lθ – l0 x 1000C
l100 – l0
300C = x – 0 x 1000C
25 - 0
x = 7.5cm
SECTION C: Structured Questions
1. 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.
Temperature, 82.5 = x – 5 x 1000C
16 – 5
Alkohol
100 = 82.5 16-5 x – 5 100x – 500 = 907.5
x = 14.08cm
Lower fixed point / freezing point of water.
Upper fixed point / boiling point of water
0 cm l0
16 cm lθ
25 cm l100
0 cm l0
(30 oc θ )X cm lθ
25 cm l100
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(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
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/object 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) ………………………………………………………………………………………
Heat capacityMuatan haba
Specific heat capacityMuatan haba tentu
Temperature of the body
Mass of the body
Type of material
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.
Q = m c
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
3. The specific heat capacity of a substance is the amount of heat that must be
supplied to increase the temperature by 1 0C or 1 K for a mass of 1 kg of the
substance.Unit Jkg-1 K-1
4. The heat energy absorbed or given out by an object is given by Q = mcθ.
5. (a)High specific heat capacity absorb a large amount of heat with only a small
temperature increase such as plastics, wood,
(b) Low specific heat capacity absorb a small amount of heat with only high temperature
increase such as aluminium, copper, iron_____________
6. Conversion of energy
7. Applications of Specific Heat Capacity
Specific heat capacity , c = Q__
mθ
Electrical energy Heat energyPt = mcθ
Heater
Power = P
Electrical energy
Potential energy
Kinetic energy
Object falls from
A high position
Moving object stopped
due to friction
Power = P
Heat energymgh= mcθ
Heat energy½ mv2= mcθ
Small value of cCopper (390 J/Kg/K)
Big value of cplastic (3500 J/Kg/K)
Two object of equal mass
Faster increase in temperature
Slower increase in temperature
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
Explain the meaning of above application of specific heat capacity:
(a) Water as a coolant in a car engine
.
(b) Household apparatus and utensils
(c) Sea breeze
Equal rate of heat supplied
Water - high specific capacity (2100
J/kg/K). 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 before it boils
A metal has a low specific heat capacity
and its temperature increases easily when
heated.This is because only a little amount
of heat is needed.
The handles of pots and pans are usually
made of materials of high specific heat
capacity or poor heat conductor
The land is heated to a higher temperature
than the sea because water has a higher
specific heat capacity. This causes the air
above the land to be hotter than the air
above the sea. The hot air above the land
flows up and the cool air from the sea flows
towards the land. The movement of air
cause wind to blow from the sea, therefore a
sea breeze is produced
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(d) Land breeze
Exercise 4.2
SECTION A : Choose the best answer
1. The change in the temperature of an object does not depend onA. the mass of the objectB. the type of substance the object is
made ofC. the shape of the objectD. 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 waterB. is less dense than waterC. has a larger specific heat capacity
than waterD. has a smaller specific heat capacity
than water
4. On a clear calm night, the temperature of surface of the lands falls more rapidly than the temperature of the nearby sea. This is becauseA. land is a solidB. land has a lower specific heat
capacityC. land is stationaryD. land is at a higher level than the sea
5. Steel , which has a low heat capacity, is suitable as a material for making the following with the exception ofA. electrical ironB. wokC. handle of a kettle
At night the air above the land and the sea
release heat to the atmosphere. The
temperature of the sea decrease more slowly
than temperature of the land because water
has a higher specific heat capacity than earth.
The land become colder than the sea. The hot
air above the sea flows upwards and the cool
air from the land flows towards the sea. A
land breeze is produced due to the movement
of air from the land towards the sea
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
D. body of a kettle
6. The figure below shows a clay pot with the water inside it is still boiling even though the pot has been removed from the sea.
Which statement correctly explains the observation above?A. The temperature of the outer wall of
the clay pot is higher than the temperature of the water.
B. Water has a large specific heat capacity
C. Clay can give out a lot of heatD. Water has a low boiling point
7. 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
heatB. 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 calculation
1. 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 = mcθ
= 4 x 452 x (52-32)
= 36 160J
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 = mcθ
= 0.8 x 400 x (60-35)
= 8 000J
3. 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 = mcθ
= 2.5 x 4200 x (82-32)
= 525, 000J
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
4. 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 = mcθ
= 0.75 x 900 x (120-45)
= 50 625J
5. 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
mcθ = mcθ
0.2 x 4200 x ( 70- θ) = 0.6 x 4200 x (θ - 30)
θ = 400C
6. 900g of water is cooled from 980C to X0C. Given that 2.1 x 105 J of heat is
released, find the value of X. (Specific heat capacity of water = 4200 J kg-1 C-1)
Amount of heat required, Q = Amount of heat released, Q
mcθ = mcθ
2.1 x 105 J = 0.9 x 4200 x (98 – X)
X = 42.450 C
7. 20 475 J of heat is needed to raise the temperature of a block of copper
weighing 3.5kg from 270 C to 420 C. Determine the heat capacity of copper?
Amount of heat required, Q = mcθ
c =
c =
c = 390 J kg-1 C-1
8. Haikal uses a 2.2kW heater to heat 500g of water in a beaker. The specific
heat capacity of water is 4200 J kg-1 C-1and the initial temperature of the water is 290C.
By assuming that there is no heat loss to the surroundings how much time will it take to
boil the water in the beaker.
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
Pt = mcθ
2.2 x 103 t = 0.5 x 4200 x ( 100 – 29 )
t =
t = 67.77s
9. A water temperature at the top of a 380m high waterfall is 150C. What is the
water temperature at the bottom of the waterfall? ( g = 9.8ms-2, c = 4200 J kg-1 C-1)
mgh = mcθ
9.8 x 380 = 4200 x θ
θ = 0.887 C
10. A bullet travelling at a velocity of 100ms-1 about to be stopped by a stationary
sand bag. If the specific heat capacity of the bullet is 160 J kg -1 C- , what is the increases
in temperature of the bullet?
mv2 = mcθ
x 1002 = 160 x θ
θ = 31.250C
SECTION C: Structured questions
1. 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.
AB
1000C 200C
5kg2.25kg
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(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
mcθ = mcθ
5.0x 900 x ( 100- θ) = 2.25 x 400 x (θ - 20)
θ = 86.670C
(b) Find the energy given by A during the process.
Energy given = mcθ
= 5 x 900 x (100 – 86.67
= 60 000J
(c) Suggest one method to reduce the energy loss to the surroundings.
…………………………………………………………………………………………..
2. (a) Define specific heat capacity.
Specific heat capacity is the amount of heat required to raise the temperature of 1 kg of
a substance by 10 C or 1 K.
(b) A container made of copper has a mass of 2.5kg. Calculate the energy required to
increased the temperature of the container by 300C if the specific heat capacity of copper
is 390Jkg-1C-1
Q = mcθ
= 2.5 x 390 x 30
= 29 250 J
4.3 UNDERSTANDING SPECIFIC LATENT HEAT
By 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
Put them in a sealed polystyrene box./ blanket/towel
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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= ml
3. Complete the diagrams below and summarized.
(a) Melting
(b) Boiling
SolidSolidification
Latent heat released
BoilingLatent heat absorbed Condensation
Latent heat released
Liquid
Gas
AB : The heat absorbed by the solid rises its temperature to its melting point.
BC : Heat is absorbed by the solid as it melts, but the temperature remains constant. The solid melts to become a liquid at the same temperature
AB : The heat absorbed by the liquid rises its temperature
BC : Heat is absorbed by the liquid as it melts, but the temperature remains constant. The heat energy use to breakdown the bonding. The liquids boils to become a liquid at the same temperature
Temperature
Time0
A
B CMelting point
Temperature
Time0
A
B CBoiling point
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(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.
Latent heat of fusion
temperature
water ice
Latent heat absorbed( melting)
heat lost/release( freezing)
PQ : Heat is released by the liquid as it cools to its freezing point.
QR : Heat is given out by the liquid as it solidifies but the temperature remains constant. The liquid solidifies to become a solid at the same temperature.
Temperature
Time0
Q R
S
P
Freezing point
Temperature
Time0
Q R
S
P
Boiling point
PQ : Heat is given out by the gas as it cools to its boiling point.
QR : Heat is given out by the gas as it condenses but the temperature remains constant. The gas condenses to become a liquid at the same temperature.
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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 ( minuman )
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
(e) Preservation of Food
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
(f) Steaming Food
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
When ice melts, its large latent heat is absorbed from surroundings. This property
makes ice a suitable substance for use as a coolant (penyejuk) 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/food.
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.
Latent heat of vaporisation
temperature
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
………………………………………………………………………………………………
(g)
(h)
(i)
(j) Killing of Germs and Bacteria
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
………………………………………………………………………………………………
EXERCISE 4.3
Section 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. PQB. QRC. RSD. ST
2. Whenever matter changes from one phase to another, some energy is either released or absorbed without change of temperature. The energy is calledA. heat capacityB. latent heatC. thermal energy
3. Which of the following changes occurs when water is freezing?A. Temperature of
water risesB. Temperature of
water decreaseC. Temperature of
water is constant
4. The boiling point of pure water at the summit is less than 1000C .it is due to
Steam that releases a large quantity of heat is used in the autoclave (pressure
cooker) to kill germs and bacteria on surgery equipment in hospitals.
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
A. the lower temperature at the summitB. the lower atmosphericC. the windy condition at the summit
5. Which of the following changes in phase involves the release of latent heat?A. Freezing of waterB. Vaporisation of waterC. Melting of ice
6. The following graph shows the change of temperature with time as a solid is heated.
Which of the following statement is not correctA. The energy being absorbed in
BC is latent heat of fusionB. The energy being absorbed in
DE is latent heat of vaporisation.C. R is the boiling point of the
liquid.D. P is the melting point of the
solid
7. Which of the following is an application using the cooling effect of evaporation?A. ThermometerB. RadiatorC. Air conditionerD. Pressure cooker
8. The special latent heat of fusion of a substance is the quantity of heat required toA. separate the molecules in the
solid substance so that they are free to move
B. increase the kinetic energy of the molecules in the solid substance
C. increase the temperature of the substance
9. 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 meltB. The voltage of the electricity supplyC. The mass of water produced by
melting iceD. The temperature change of the ice.
10. 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
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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
11. Which of the following is not a characteristics of water that makes it widely used as a cooling agent?A. Water is readily availableB. Water does not react with many
other substanceC. Water has a large specific heat
capacityD. Water has a large density
12. 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
13. Scalding of the skin by boiling water is less serious then by steam. This is because…A. the boiling point of water is less than
the temperature of steamB. the heat of boiling water is quickly
lost to the surroundingsC. steam has a high specific latent
heat.D. Steam has a high specific heat
capacity.
SECTION B: Answer the question by showing the calculation
1. 300g of ice at 00C melts. How much energy is required for this
Q = ml
= 0.3 x 330 000 kJ kg-1
= 99 000kJ
Question 2-6 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
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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-1
2. An immersion heater rated at 500 W is fitted into a large block of ice at 0 0C. 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 = 1002s
3. 300 g of water at 350C is mixed with x g of water at 900C. The final temperature
of the mixture is 700C. Find the value of x
mcθ = mcθ
0.3 x c x (70- 35) = x c x (90 – 70)
10.5 = 0.02x
x = 525g
4. Calculate the amount of heat released when 2 kg of ice at 00C is changed into
water at 00C.
Q = ml
= 2 x (3.34 x 105)J
= 6.68 X 105 J
5. Calculate the amount of heat needed to convert 3 kg of ice at 00C to water at 300C.
Q = ml + mcθ
= ( 3 x 3.34 x 105) + (3 x 4200 x 30)
= 1.38 x 106 J
6. Find the amount of heat needed to convert 0.5 kg of ice at —150C into steam at
1000C
Q = mcθ + ml + mcθ + ml
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
= ( 0.5 x 2100 x 15) + (0.5 x 3.34 x 105) + (0.5 x 4200 x 100)+ (0.5 x 3.34 x 105)
= 5.0575 x 105J
7. Calculate the amount of heat needed to convert 100 g of ice at 00C into steam at
1000C.
Q = ml + mcθ + ml
= ( 0.1 x 3.34 x 105) + (0.1 x 4200 x 100)+ (0.1 x 2.26 x 106 )
=3.014 x 105J
8. A 500W immersion heater is used to heat a liquid which is boiling at 900C.
Given that 5g of the liquid is charged into steam in 12 seconds, find the specific latent
heat of vaporisation of the liquid.
Pt = ml
l =
= 1.2 x 106J
SECTION C
1. Figure below shows the apparatus for an experiment to determine the specific latent heat
of vaporisation of water. Water is boiled using a 500 W immersion heater. The quantity
of water in 5 minutes is 60g.
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(a). Calculate the value of the specific latent heat of vaporisation
ml = Pt
l =
= 2.5 x 10 6 Jkg-1
(b) State two factors which affect the accuracy of this experiment?
…………………………………………………………………………………………………
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…………………………………………………………………………………………………
(c) Why does the water boil at a constant temperature even though heat energy is continuously
being supplies?
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(d) State two precautions in order to obtain a more accurate result.
…………………………………………………………………………………………………
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2. Figure below show the apparatus and materials for the experiment to determine the
specific latent heat of fusion of ice. The initial mass of ice is 180g and the ice is heated
with a 56W immersion heater. The time taken for the ice to melt completely is 30
minutes.
1. The portion of the heat energy supplied by the heater is lost to the surroundings.
2. The water vapour that condenses flows back into the beaker
During the boiling process, the heat energy supplied does not cause the average
velocity of the water molecules to increases but is absorbed in order to overcome the
forces at attraction between the molecules of the liquid as it changes into the gas
phase
1. To ensure that the immersion heater does not touch the bottom of
the beaker
2. To ensure that the heater element is completely immersed in the
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(a). What is the purpose of using the beaker B in this experiment?
…………………………………………………………………………………………………
(b). State the function of the thermometer in the experiment
…………………………………………………………………………………………………
(c) Calculate the value of the specific latent heat of fusion of ice
Q = ml
l =
=
= 5.6 x 106 J
(e) The standard value of the specific latent heat of fusion of ice is 3.36 x 105 Jkg-1. Give
one reason for the different between the experimental and standard value of the specific
latent heat of the fusion of ice.
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4.4 UNDERSTANDING THE GAS LAW
By 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 gas Explanation
To prevent the ice from being melted by the heat from the surroundings
To ensure that the heating does not continue after all the ice has melted at 00C
The time of heating is greater than that required to melt the ice completely
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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 Boyle’s Law
Small volume molecules hit wall more often, greater pressure
P α 1 V
That is PV = constantOr P1V1 = P2V2
Relationship between pressure and volume
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
1. Boyle’s 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. Boyle’s 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 250cm 3. 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 Charles’s Law
0
P
V
(a) P inversely proportional to V
0
P
1/V
(b) P directly proportional to 1/V
V α Tthat is V = constant
T
PI=50m + 10m
V1=250cm3
P2= 10m P1V1 = P2V2
60m (250 x 10-6)m3 = 10m x V2
1.5 x 10-3 m3 = V2
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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.
Temperature Celsius scale (0C) Kelvin Scale(K)
Absolute zero -273 0
Ice point 0 273
Steam point 100 373
Unknown point θ ( θ + 273 )
4. Complete the diagram below.
(a) (b)
Relationship between volume and temperature
Lower temperature
Higher temperature, faster molecules, larger volume to keep the pressure constant
θ/ 0C-273
Vcm3
V cm3
T/K
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
5. A balloon is filled with 180 cm3 of gas at 370C.The balloon is immersed in a beaker of
water and the water is heated so that the pressure in the balloon remains constant. What is
the volume of the gas when its temperature reaches 570C?
V1 = 180 cm3 , V2 = ? , T1 = 273 + 37 =310K , T2 = 273 + 57
=330K
V2 =
V2 =191.6 cm3
4.4.3 Pressure’s 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.
2. Complete the graphs below.
P α TThat is P = constant
T
Relationship between pressure and temperature
Higher temperature molecules move faster, greater pressure
θ/ 0C-273
P/PaP/Pa
T/K
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(a) (b)
3. A closed flask contains gas at a temperature of 90OC and a pressure of 165 kPa. If the
temperature rises by 250C, what is the new pressure of the gas?
P1 = 165kPa , P2 = ? , T1 = 273 + 90 =363K , T2 = 273 + 25+90
=388K
P2 =
=176.4 kPa
EXERSICE 4.4
SECTION A
1. Absolute zero is the temperature at which gas molecules theoreticallyA. occupy no volume at allB. move the fastestC. will read 2730C in
temperatureD. contain twice the number of
molecules at room temperature
2. Absolute zero is the temperature at whichA. ice meltB. the particles of a
substance have their minimum amount of energy
C. a gas changes to a liquidD. a salt solution freezes
3. When the gas in an closed container is heated at fixed pressure, what will happen to the frequency of collision between the gas molecules?A. It decreasesB. It increasesC. It remains constant
4. When the fixed mass of a gas expands slowly at fixed temperature, which of the following quantities increases?
A. Kinetic energyB. Average speed of gasC. Gas pressureD. Volume of gas
5. The closed vessel contains a gas. The pressure of the gas is constant because the gas moleculesA. are in random motionB. travel at a constant
average velocityC. collide with the wall of the
container
6. What will happen when a fixed mass of gas in an closed container is compress slowly without any change in temperature?A. The gas pressure
increasesB. The volume of gas
remains constantC. The kinetic energy of gas
increasesD. The distance between gas
molecules increases
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
7. The air pressure in a car tyre is increased after the car has travelled a long distance becauseA. the average velocity of the air
molecules in the tyre has decreased
B. the temperature of the air in the tyre has increased
C. the size of the air molecules in the tyre has increased
SECTION B: 1. A mixture of air and petrol vapour is injected into the cylinder of a car engine
when the cylinder volume is 150 cm3. Its pressure is then 1.0 atm. The valve closes and
the mixture is compressed to 25 cm3. Find the pressure now.
P1 V1 = P2V2
1x 150 = P2 x 25
P2 = 6 atm
2. The volume of an air bubble at the base of a sea of 60 in deep is 210 cm 3. If the
atmospheric pressure is 10 in of water, find the volume of the air bubble when it reaches
the surface of the sea.
P1 V1 = P2V2
( 60 + 10) x 210 = 10 V2
V2 = 1470 cm3
3. The volume of an air bubble is 5 mm3 when it is at a depth of h in 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)
P1 V1 = P2V2
( 10 + h) x 5 = ( 10 + 2) x 15
h = 26m
4. An air bubble has a volume of V cm3 when it is released at a depth of 55m from
the water surface. Find its volume (V) when it reaches the water surface. (Atmospheric
pressure = 10 m of water)
P1 V = P2V2
( 10 + 55) x V = 10 x V2
V2 = 6.5cm3
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
5. A gas of volume 20m3 at 370C is heated until its temperature becomes 870C at
constant pressure. What is the increase in volume?
=
=
V2 = 23.2m3
Increase volume = 23.2 m3 – 20 m3 = 3.2m3
6. The air pressure in a container at 330C is 1.4 X 1O5 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?
=
1.4 x 10 5 = P2 33 + 273 55 + 273 P2 = 1.5 x 105 N m2
7. The volume of a gas is 1 cm3 at 150C. The gas is heated at fixed pressure until the
volume becomes triple the initial volume. Calculate the final temperature of the gas.
1 = 3 15 + 273 T2
T2 = 864 K
8. 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 rn2)
=
1 x 10 5 = P2
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
25 + 273 98 +273P2 = 1.25 x 105N rn2
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.
=
1.2 x 10 5 = 9 x 10 4 T1 40 +273
T1 = 417 K
PART A: CHAPTER 4
1. 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?A. A iron sphere will have a
temperature of 273KB. The copper sphere will have a
temperature of 500C.C. Both the sphere have the
same temperature.D. 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. ConvectionB. VaporisationC. RadiationD. 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 boxB. 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
capacityB. Steam can cut off the supply of
oxygenC. Water is easily availableD. Water can react with some
material
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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
saturatedB. at all times because evaporation
and condensation occur any timeC. 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 containerC. Gas molecules move faster as
temperature increasesD. 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 increasesB. the number of gas molecules
increasesC. the average distance between the
gas molecules increasesD. 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 fasterD. The repulsive force of boiling
water slows down the movement of air molecules
PART B;
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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.
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)
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…..
Thermometer A B C D
Liquid Mercury Mercury Alcohol Alcohol
Freezing point of liquid (0C) -39 -39 -112 -112
Boiling point of liquid (0C) 360 360 360 360
Diameter of capillary tube Large Small Large Small
Cross section
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
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(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 = mcθ
= 0.250 x 4200 x (42-30)
= 12 600J
(b) the rate of heat supplied to the water
Figure 2
Change of volume of gas with temperature Change of electrical resistance with temperature
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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-1
3. 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
Figure 3.1 Figure 3.2
Gravitational potential energy → kinetic energy → heat energy
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
(c) Calculate the time taken by the lead shot to fall from the top to the
bottom of the tube.
S = ut + ½ at2
0.45 = 0 + ½ (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 θ = 225 J
c = 225 (0.500 x 3) = 150 Jkg-1 C-1
iii. State the assumption used in your calculation in (d)ii.
……………………………………………………………………………………………...
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……………………………………………………………………………………………….
PART C: EXPERIMENT
No heat loss to the surroundings/All the gravitational potential energy is
converted into heat energy
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
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,
Figure (a) Figure (b)
(a) State one suitable inference that can be made. [1 mark]
(b) State appropriate hypothesis for an investigation. [1 mark]
(c) Design an experiment to investigate the hypothesis stated in (b).
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
vi. The way you would tabulate the data,
vii. The way you would analyse the data. [10 marks]
Inference the air pressure depends/influence/affect on the temperature
Hypothesis the air pressure increase as the temperature increases
the temperature increases , the air pressure also increase
Aim To investigate/study/find/ the relationship between the air pressure
and the temperature at constant volume.
Variable Constant variable : Volume of air
Manipulate variable : Air temperature
Responding variable : Air pressure
Material and Apparatus Round-bottom flask, rubber tube, Bourdon gauge, beaker, stirrer
(pengacau), 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
JPN Pahang Physics Module Form 4Teacher’s Guide Chapter 4: Heat
recorded in the table
The above procedure is repeated until the water temperature
almost reaches boiling point.
Tabulation of Data
Analysis of Data
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