Chapter 4 Heat Teacher's Guide

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

CHAPTER 4: HEAT 4.1 : UNDERSTANDING THERMAL EQUILIBRIUM By the end of this subtopic, you will be able to •

Explain thermal equilibrium



Explain how a liquid-in glass thermometer works

Thermal equilibrium :Keseimbangan terma

Faster. rate of energy transfer A

B

Hot object

Equivalent to

Equivalent to

Cold object No net heat transfer

Slower rate of energy transfer

1. The net heat will flow from A to B until the temperature of A is the ( same, zero as the temperature of B. In this situation, the two bodies are said to have reached thermal equilibrium. 2. When thermal equilibrium is reached, the net rate of heat flow between the two bodies is (zero, equal) 3. There is no net flow of heat between two objects that are in thermal equilibrium. Two objects in thermal equilibrium have the same temperature. 4. The liquid used in glass thermometer should (a) Be easily seen (b) Expand and contract rapidly over a wide range of temperature (c) 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

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

6. ( Heat, Temperature ) is a form of energy. It flows from a hot body to a cold body. 7. The SI unit for ( heat , temperature) is Joule, J. 8. ( Heat , Temperature ) is the degree of hotness of a body 9. The SI unit for (heat , temperature) is Kelvin, K. 10. Lower fixed point (l 0 )/ ice point 11. Upper fixed point( l

: the temperature of pure melting ice/00C

/steam point: the temperature of steam from water that is boiling

100)

under standard atmospheric pressure /1000C

Temperature, θ = l0 l100 lθ

l - l θ

0

x 1000C

l100 - l0

: length of mercury at ice point : length of mercury at steam point : length of mercury at θ point

Exercise 4.1 Section A: Choose the best answer 1. The figure shows two metal blocks. Which the following statement is false?

A. It warms the surroundings B. It warms the water of the tea C. It turns into heat energy and disappears. 3. Which of the following temperature corresponds to zero on the Kelvin scale? A. 2730 C B. 00C C. -2730 C D. 1000 C

A. P and Q are in thermal contact B. P and Q are in thermal equilibrium C. Energy is transferred from P to Q D. Energy is transferred from Q to P

4. How can the sensitivity of a liquid- in – glass thermometer be increased? A. Using a liquid which is a better conductor of heat

2. When does the energy go when a cup of hot tea cools?

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

B. Using a capillary tube with a narrower bore. C. Using a longer capillary tube D. Using a thinner-walked bulb

6. 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.

5. Which instrument is most suitable for measuring a rapidly changing temperature? A. Alcohol-in –glass thermometer B. Thermocouple C. Mercury-in-glass thermometer D. Platinum resistance thermometer

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? Temperature, θ = lθ – l0 x 1000C l100 – l0 θ = 27 – 5 x 1000C 65 - 5 θ = 36.670C

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

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 fixed point. What is the temperature of the water? What is the length of mercury column from the bulb at temperatures i) 300C Temperature, θ = lθ – l0 x 1000C l100 – l0 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

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

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) Alkohol ………………………………………………………………………………………… …. (b) He chooses two fixed points of Celsius scale to calibrate his thermometer. State them (2m) Lower fixed point = freezing point of water. ………………………………………………………………………………………… Upper fixed point = boiling point of water …… ……………………………………………………………………………………………… (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.

100 = 82.5 16-5 x–5 100x – 500 = 907.5 x = 14.08cm

(d) Why should he use a drinking straw of small diameter? To increases the sensitivity of the thermometer ………………………………………………………………………………………… …… (e) What kind of action should he take if he wants to increase the sensitivity of his thermometer? Use a copper can instead of the aluminum can because it is a better thermal conductor 5

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

………………………………………………………………………………………… …… ………………………………………………………………………………………… …… 2. What do you mean by heat and temperature? 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. ……………………………………………………………………………………………… 4.2

: UNDERSTANDING SPECIFIC HEAT CAPACITY By the end of this subtopic, you will be able to •

Define specific heat capacity



State that c = Q/MCθ



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

Heat capacity Muatan haba Specific heat capacity Muatan haba tentu

1. The heat capacity of a body is the amount of heat that must be supplied to the body to increase its temperature by 10C. 2. The heat capacity of an object depends on the Temperature of the body (a) ………………………………………………………………………………………. Mass of the body (b) ………………………………………………………………………………………. Type of material (c) ……………………………………………………………………………………… 3. The specific heat capacity of a substance is the amount of heat that must be supplied to increase the temperature by 1 0C for a mass of 1 kg of the substance. Unit Jkg-1 K-1 Specific heat capacity , c =

Q__ m∆θ

4. The heat energy absorbed or given out by an object is given by Q = mc∆O.

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

5. High specific heat capacity absorb a large amount of heat with only a small temperature increase such as plastics.

6. Conversion of energy Heater

Electrical energy Electrical energy

Potential energy

Kinetic energy

Power = P

Object falls from A high position

Moving object stopped due to friction

Heat energy Pt = mcθ

Heat energy mgh= mcθ

Heat energy ½ mv2= mcθ

7. Applications of Specific Heat Capacity Faster increase in temperature Small value of c

Slower increase in temperature Two object of equal mass

Big value of c

Equal rate of heat supplied

Explain the meaning of above application of specific heat capacity: (a) Water as a coolant in a car engine (i)

Water is a good example of substance with a high specific capacity. It is used as a cooling agent to prevent overheating of the engine .Therefore, water acts as a heat reservoir as it can absorb a great amount of heat before it boils.

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

(b) Household apparatus and utensils ………………………………………………………………………………………... ………………………………………………………………………………………... ………………………………………………………………………………………... ………………………………………………………………………………………... (c) Sea breeze …………………………………………… …………………………………………… …………………………………………… …………………………………………… …………………………………………… …………………………………………… …………………………………………… ………………… (d) Land breeze …………………………………………… …………………………………………… …………………………………………… …………………………………………… …………………………………………… …………………………………………… ………………

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

Exercise 4.2 SECTION A : Choose the best answer 1. The change in the temperature of an object does not depend on A. the mass of the object B. the type of substance the object is made of C. the shape of the object D. the quantity of heat received

The temperature of the ethanol rises faster. This is because the ethanol.. A. is denser than water B. is less dense than water C. has a larger specific heat capacity than water D. has a smaller specific heat capacity than water

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.

4. In the experiment to determine the specific heat capacity of a metal block, some oil is poured into the hole containing thermometer. Why is this done? A. To ensure a better conduction of heat B. To reduce the consumption of electrical energy C. To ensure the thermometer is in an upright position. D. To reduce the friction between the thermometer and the wall of the block.

3. Heat energy is supplied at the same rate to 250g of water and 250g of ethanol.

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) 9

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

= 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 32 0C 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 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 SECTION C: Structured questions

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

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. 5kg 2.25kg A

B

1000C

200C

Assume that there is no energy loss to the surroundings. (a) Find the final temperature of A and B if they are in thermal equilibrium. Given the specific heat capacity of A and B are 900 Jkg-1 C-1 and 400 Jkg-1 C-1 respectively. Amount of heat required, Q = Amount of heat released, Q 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. Put them in a sealed polystyrene box. …………………………………………………………………………………………..

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JPN Pahang Teacher’s Guide

4.3

Physics Module Form 4 Chapter 4: Heat

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



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. Gas Boiling Latent heat absorbed

Solid

Solidification Latent heat released

Condensation Latent heat released

Liquid

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

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

………………………………………

Temperature

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

Time (b) Boiling

………………………………………

Temperature

……………………………………… ……………………………………… ……………………………………… ……………………………………… ……………………………………… Time

(c) Solidification Temperature

……………………………………….

……………………………………… ……………………………………… ……………………………………… ……………………………………… ……………………………………… ……………………………………… Time

……………………………………….

(d) Condensation ………………………………………

Temperature

……………………………………… ……………………………………… ……………………………………… ……………………………………… 13 Time

……………………………………… ……………………………………….

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

Latent heat of fusion 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 temperature melting point without any increase in ……………………….. The S.I unit of the specific latent heat of fusion is Jkg-1.

Latent heat absorbed ( melting)

water

ice heat lost ( freezing)

Latent heat of vaporisation 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 temperature into gas or vapour of its boiling point without any change in …………………….. The S.I unit is Jkg-1. Latent heat absorbed ( boiling)

gas

water heat lost 14 ( condensation)

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

6. Explain the application of Specific Latent Heat above: : (d) Cooling of beverage When ice melts, its large latent heat is absorbed from surroundings. This property …………………………………………………………………………………………… makes ice a suitable substance for use as a coolant to maintain other substance at a … low temperature. Beverage can be cooled by adding in several cubes of ice. When the …………………………………………………………………………………………… ice melts a large amount of heat (latent heat) is absorbed and this lowers the … temperature of the drink. …………………………………………………………………………………………… … …………………………………………………………………………………………… … …………………………………………………………………………………………… … (e) Preservation of 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. … …………………………………………………………………………………………… … …………………………………………………………………………………………… …

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

…………………………………………………………………………………………… … (f) Steaming Food 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. …………………………………………………………………………………………… … …………………………………………………………………………………………… … …………………………………………………………………………………………… … (g) Killing of Germs and Bacteria Steam that releases a large quantity of heat is used in the autoclave to kill germs …………………………………………………………………………………………… and bacteria on surgery equipment in hospitals. … …………………………………………………………………………………………… … …………………………………………………………………………………………… … …………………………………………………………………………………………… …

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?

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JPN Pahang Teacher’s Guide

A. B. C. D.

Physics Module Form 4 Chapter 4: Heat

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

PQ QR RS ST

2. Figure show a joulemeter used for measuring the electrical energy to melt some ice in an experiment. To find the specific latent heat of fusion of ice, what must be measured?

4. Which of the following is not a characteristics of water that makes it widely used as a cooling agent? A. Water is readily available B. Water does not react with many other substance C. Water has a large specific heat capacity D. Water has a large density

A. The time taken for the ice to melt B. The voltage of the electricity supply C. The mass of water produced by melting ice D. The temperature change of the ice. 5. Figure below shows the experiment set up to determine the specific latent heat of fusion of ice. A control of the experiment is set up as shown in Figure (a) with the aim of

3. It is possible to cook food much faster with a pressure cooker as shown above. Why is it easier to cook food using a pressure cooker?

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

6. 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 steam B. the heat of boiling water is quickly lost to the surroundings C. steam has a high specific latent heat. D. Steam has a high specific heat capacity.

A. determining the rate of melting of ice B. 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

SECTION B: Answer the question by showing the calculation 2. 300g of ice at 00C melts. How much energy is required for this Q = ml = 0.3 x 330 000 kJ kg-1 = 99 000kJ Question 2-7 are based on the following information •

Specific heat capacity of water = 4 200 J kg-1 C-1



Specific heat capacity of ice = 2 100 J kg-1 C-1



Specific latent heat of fusion of ice = 3.34 X 105J kg-1



Specific latent heat of vaporization of water = 2.26 X 106 J kg-1

3. An immersion heater rated at 500 W is fitted into a large block of ice at 00C. How long does it take to melt 1.5kg of ice? Q = ml Pt = 1.5 x 3.34 x 105 500 x t t

= 501 000 = 1002s

4. 300 g of water at 400C is mixed with x g of water at 800C. The final temperature of the mixture is 700C. Find the value of x

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

5. Calculate the amount of heat released when 2 kg of ice at 00C is changed into water at 00C.

6. Calculate the amount of heat needed to convert 3 kg of ice at 00C to water at 300C.

7. Find the amount of heat needed to convert 0.5 kg of ice at —150C into steam at 1000C

8. Calculate the amount of heat needed to convert 100 g of ice at 00C into steam at 1000C.

9. The specific latent heat of vaporization of water is 2300 kJ kg-’. How much heat will be absorbed when 3.2 kg of water is boiled off at its boiling point.

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.

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JPN Pahang Teacher’s Guide



Physics Module Form 4 Chapter 4: Heat

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 Volume,V



m3

Temperature,T •



Explanation 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



The molecules are in continuous random motion and have an

K (Kelvin)

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

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JPN Pahang Teacher’s Guide

4.4.1

Physics Module Form 4 Chapter 4: Heat

Boyle’s Law

Pα1 V That is PV = constant Or P V = P V 1

1

2

Small volume molecules hit wall more often, greater pressure

2

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 P

P

0

V

(a) P inversely proportional to V

0

1/V

(b) P directly proportional to 1/V

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. P2= 10m

P1V1 = P2V2 60m (250 x 10 )m3 = 10m x V2 1.5 x 10-3 m3 = V2 -6

PI=50m + 10m

4.4.2

Charles’s Law V1=250cm3

VαT that is V = constant T Relationship between 21 volume and temperature Lower temperature

Higher temperature, faster molecules, larger volume to keep the pressure constant

JPN Pahang Teacher’s Guide

Physics Module Form 4 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 Absolute zero Ice point Steam point Unknown point

Celsius scale (0C) -273 0 100 θ

Kelvin Scale(K) 0 273 373 ( θ + 273 )

4. Complete the diagram below.

-273

4.4.3

100

θ/0C

Pressure’s Law

PαT That is P = constant T 22

Higher temperature molecules move faster, greater pressure

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

Relationship between pressure and temperature

1. The pressure law states that for a fixed mass of gas, the pressure of the gas is directly proportional to its absolute temperature when its volume is kept constant.

EXERSICE 4.4Gas Law 1. A mixture of air and petrol vapour is injected into the cylinder of a car engine when the cylinder volume is 100 cm3. Its pressure is then 1.0 atm. The valve closes and the mixture is compressed to 20 cm3. Find the pressure now.

2. The volume of an air bubble at the base of a sea of 50 in deep is 200 cm3. If the atmospheric pressure is 10 in of water, find the volume of the air bubble when it reaches the surface of the sea.

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 in, find the value of h. (Atmospheric pressure = 10 m of water)

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

4. An air bubble has a volume of V cm3 when it is released at a depth of 45m from the water surface. Find its volume (V) when it reaches the water surface. (Atmospheric pressure = 10 m of water)

5. A gas of volume 20m3 at 370C is heated until its temperature becomes 870C at constant pressure. What is the increase in volume?

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?

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.

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

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)

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.

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?

D. A iron sphere will have a temperature of 273K E. The copper sphere will have a temperature of 500C. F. Both the sphere have the same temperature. 25

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

G. The temperature of the iron sphere will be lower than 500C

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.

2. In the process to transfer heat from one object to another object, which of the following processes does not involve a transfer to material? A. Convection B. Vaporisation C. Radiation D. Evaporation

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.

3. When we use a microwave oven to heat up some food in a lunch box, we should open the lid slightly. Which of the following explanations is correct? A. To allow microwave to go inside the lunch box B. To allow the water vapors to go out, otherwise the box will explode C. To allow microwave to reflect more times inside the lunch box D. To allow microwave to penetrate deeper into the lunch box.

7. Water molecules change their states between the liquid and gaseous states A. only when water vapour is saturated B. at all times because evaporation and condensation occur any time C. only when the vapour molecules produce a pressure as the same as the atmospheric pressure D. only when the water is boiling

4. Water is generally used to put out fire. Which of the following explanation is not correct? A. Water has a high specific heat capacity B. Steam can cut off the supply of oxygen C. Water is easily available D. Water can react with some material

8. Based on the kinetic theory of gas which one of the following does not explain the behaviour of gas molecules in a container? A. Gas molecules move randomly B. Gas molecules collide elastically with the walls of the container

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.

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

C. Gas molecules move faster as temperature increases D. Gas molecules collide inelastically with each other

10. A plastic bag is filled with air. It is immersed in the boiling water as shown in diagram below.

9. A cylinder which contains gas is compressed at constant temperature of the gas increase because A. the average speed of gas molecules increases B. the number of gas molecules increases C. the average distance between the gas molecules increases D. the rate of collision between the gas molecules and the walls increases

Which of the following statements is false? A. The volume of the plastic bag increases. B. The pressure of air molecules increases C. The air molecules in the bag move faster D. The repulsive force of boiling water slows down the movement of air molecule

PART B; 1. A research student wishes to carry out an investigation on the temperature change of the substance in the temperature range -500C to 500C. The instrument used to measure the temperature is a liquid in glass thermometer.

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JPN Pahang Teacher’s Guide

Thermometer Liquid Freezing point of liquid (0C) Boiling point of liquid (0C) Diameter of capillary tube Cross section

Physics Module Form 4 Chapter 4: Heat

A Mercury -39 360 Large

B Mercury -39 360 Small

C Alcohol -112 360 Large

D Alcohol -112 360 Small

Table 1 (a) (i) State the principle used in a liquid- in –glass thermometer.(1m) Principle of thermal equilibrium ........................................................................................................................................ (ii)

Briefly explain the principle stated in (a)(i) (3m) 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 ………………………………………………………………………………………….

(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) ………………………………………………………………………………………… …………………………………………………………………………………………… 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

28

JPN Pahang Teacher’s Guide

Physics Module Form 4 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) ……………………………………………………………………………………………… • Change of volume of gas with temperature • Change of electrical resistance with temperature ……………………………………………………………………………………………… ………………………………………………………………………………………………

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.

Figure 2

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 29

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

(b) the rate of heat supplied to the water Rate of heat supplied to the water = 12 600J 120s = 105 Js-1 (c) the specific heat capacity of the metal block P Heat supplied by metal block P = heat gained by water 0.500 x c x(100 -42)

= 12 600J c

= 434 J kg-1 C-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).

Figure 3.1

Figure 3.2

(a) State the energy change each time the tube is inverted. Gravitational potential energy → kinetic energy → heat energy ………………………………………………………………………………………….. ………………………………………………………………………………………….. (b) What is the average distance taken by the lead shot each time the tube is inverted? 45.0 cm (c) Calculate the time taken by the lead shot to fall from the top to the bottom of the tube.

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

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.

……………………………………………………………………………………………... No heat loss to the surroundings/All the gravitational potential energy is converted ……………………………………………………………………………………………… into heat energy ………………………………………………………………………………………………. PART C: EXPERIMENT 1. Before travelling on a long journey, Luqman measured the air pressure the tyre of his car as shown in Figure (a) He found that the air pressure of the tyre was 200 kPa. After the journey, Luqman measured again the air pressure of the tyre as shown in Figure (b) He found that the air pressure had increase to 245 kPa. Luqman also found that the tyre was hotter after the journey although the size of the tyre did not change. Using the information provided by Luqman and his observations on air pressure in the tyre of his car:

31 Figure (a)

Figure (b)

JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

(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).

Choose suitable apparatus such as pressure gauge, a

round-bottomed flask and any other apparatus that may he necessary. In your description, state clearly the following: i.

Aim of the experiment,

ii.

Variables in the experiment,

iii.

List of apparatus and materials,

iv.

Arrangement of the apparatus,

v.

The procedure of the experiment including the method of controlling the manipulated variable and the method of measuring the responding variable,

vi.

The way you would tabulate the data,

vii. The way you would analyse the data. [10 marks]

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat

Inference Hypothesis

At constant volume, the air pressure depends on the temperature At constant volume, the air pressure increase as the temperature

Aim

increases To investigate the relationship between the air pressure and the

Variable

temperature at constant volume. Constant variable : Air temperature Manipulate variable : Air pressure

Material and Apparatus

Responding variable : Volume of air Round-bottom flask, rubber tube, Bourdon gauge, beaker, stirrer, thermometer, wire gauze, tripod stand and Bunsen burner.

Arrangement of apparatus

Procedure



The apparatus is set up as shown in the diagram above.



The beaker is filled with ice-cold water until the flask is completely immersed.



The water is stirred and the initial temperature reading taken. The pressure reading from the bourdon gauge is also taken.



The water is heated and constant stirred. When the water temperature increases by 100C, the Bunsen burner is removed and the stirring of water is continued. The temperature and pressure readings of the trapped air are recorded in the table

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JPN Pahang Teacher’s Guide

Physics Module Form 4 Chapter 4: Heat



The above procedure is repeated until the water temperature almost reaches boiling point.

Tabulation of Data

Analysis of Data

34

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