chapter 4 heat teacher's guide

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


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


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


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?


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?


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


l100 – l0

65cm l1005cm l0

27 cm (Q) lθ

0 cm l0

24.5 cm (Q) lθ

28 cm l100


θ = 16 – 0 x 1000C

25 - 0

θ = 64.00C


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


(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


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


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


(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


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).


= 2.5 x 4200 x (82-32)

= 525, 000J


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)


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?


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.


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


(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.


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


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


(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.


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


………………………………………………………………………………………………

(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.


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


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


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


= ( 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.


(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?

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

(c) Why does the water boil at a constant temperature even though heat energy is continuously

being supplies?

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

………………………………………………………………………………………………….

(d) State two precautions in order to obtain a more accurate result.

…………………………………………………………………………………………………

…………………………………………………………………………………………………

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


(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


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


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


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


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


(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


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


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


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


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;


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)

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(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


(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)

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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


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.

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(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


(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


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).


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


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 teacher's guide

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