Pen-Write E-Library

SCHEME OF WORK
WEEKS TOPICS
1. Temperature and its measurement

2. Heat capacity and specific heat capacity and verifications

3. Latent Heat- fusion and vaporization and verification

4. Effect of pressure and impurities on boiling and melting points

5. Vapour pressure –saturated and unsaturated vapour; Motion

6. Gas laws

7. Waves- production, propagation, types and wave equation

8. Properties of waves-reflection, refraction, diffraction and interference

9. Polarization-production and uses and uses of polarized waves. Pressure Cooker. Cooling by Evaporation, Cooling and heating curves.

10. Revision

Reference Book: [ 1 ] New School Physics By M.W Anyahoha
[ 2 ] Senior Secondary School Physics By P.N Okeke

TEMPERATURE AND ITS MEASUREMENT
CONTENT:
• Definition of heat and temperature.
• Differences between heat and temperature
• Thermometer and its different types

Heat can be defined as the total internal energy of a body. It is a form of energy .Its measured in joules (J).Temperature is a measure of the degree of hotness or coldness of a body. It is measured by means of a thermometer. The S. I. Unit of temperature is the Kelvin. However, it is also measured in degree Celsius and Fahrenheit.

Difference between Heat and Temperature



HEAT TEMPRATURE
Ii Heat is a form of energy |Temperature is measured of degree of hotness or coldness of a body
iii Heat flows from body at higher temperature to that at lower temperature . |Temperature does not flow
Iiii Heat cannot be measured with any instrument rather it can be calculated |Temperature is measured by an instrument called thermometer
iiv Its unit is in joules |Units of temperature is in Kelvin (K), degree Celsius(0c) and Fahrenheit (0F)
V Heat can do some work |Temperature can do no work


https://youtu.be/LL54E5CzQ-A

SS/N THERMOMETER THERMOMETRICAL SUBSTANCE PHYSICAL PROPERTIES
11 Liquid in glass |Mercury or alcohol |Change in volume with temperature
22 Constant volume gas thermometer |Gas |Change in pressure with temperature
33 Thermoelectric thermometer |Two different metal (iron and copper) |Change in potential difference due to temperature difference
44 Resistant thermometer |Resistant wire |Change in resistant with temperature
55 Bimetallic thermometer |Two dissimilar metal (bras and iron) |Differential expansion of two metals of the bimetallic stripes


EVALUATION:
1. Define heat and temperature
2. State five (05) thermometer, their, thermometric substance and physical properties.



THE THERMOMETER
The thermometer is the instrument used for measuring temperature. There are various types of thermometer, e.g. liquid-in-glass thermometer and the thermocouple. Each one makes use of the change in the physical properties of materials they are made of, to indicate temperature change.

1. THE LIQUID-IN-GLASS THERMOMETER. The liquid-in-glass thermometer depends on the uniform expansivity of the liquid used with temperature change. Any liquid that will be used as a thermometric liquid must be good conductor of heat, be easily seen in glass, have a high boiling point, have a low freezing point, have a low specific heat capacity, must not wet glass and must expand uniformly.

2. THE CLINICAL THERMOMETER has a constriction and it has a short range (350C – 430C). The narrow constriction prevents the mercury from flowing back into the bulb immediately after the thermometer has been removed from the patient’s body.

3. PLATINUM RESISTANCE THERMOMETER. This thermometer depends on the variation in the electrical resistance of a conductor with temperature as expressed below:


4. THE THERMOCOUPLE
A thermocouple consists of two different metals joined together by a circuit containing a galvanometer. The working of a thermocouple depends on the variation of the electromotive force (e.m.f) between junctions. The equation of the relationship is

E = a + bt + ct[sup]2[/sup], (a, b, c are constants)

5 .CONSTANT – VOLUME GAS THERMOMETER
The constant – volume gas thermometer depends on the variation in the pressure of a gas at constant volume with changes in the temperature of the gas.
The equation is:
P[sub]1[/sub] = P[sub]2[/sub]

T[sub]1[/sub] T[sub]2[/sub]

https://youtu.be/SoYOMi8-fTY

ABSOLUTE SCALE OF TEMPERATURE
Temperature has no property of direction but has magnitude or size, which depends on the scale being used.
The absolute scale of temperature is thermodynamic scale because it gives us the idea of the lowest possible temperature or absolute zero with the value of –2730K.
The absolute zero is the lowest possible temperature below which nothing can be cooled since temperature is the measure of the average or mean kinetic energy of the molecules of a substance. It follows that as we subtract heat from a substance, its temperature drops and hence its kinetic energy until it eventually becomes zero under which the molecules remain stationary.
The Celsius seal is the most commonly used scale. This scale is based on two fixed points – the lower fixed point (O0C) and the upper fixed point (1000C). The gap between these points is called the fundamental interval.
To convert from one scale to another, use interpolation technique.


EVALUATION:
1. The length of mercury thread when it is at 00C, 1000C and at an unknown temperature 0 is 25mm, and 175mm respectively. Find the value of 0.

ASSIGNMENT
1. The ice and stem point of a certain thermometer are 200 and 1000 respectively. Calculate the Celsius temperature corresponding to 700 on the thermometer.
(a)84.00C (b)75.00C (c)62.5.00C (d)58.00C

2. A short response time is obtained in a clinical thermometer when the
(a) bulb is large and thick walled (b) bulb is small and thick walled (c) stem is long and thin walled (d) bulb is thin walled and the liquid is a good conductor of heat.

3. HEAT:
I is form of energy II is measured with a thermometer
III cannot be created nor destroyed IV from a hot object is transferred through conduction only
Which of the statement above is correct?
(a) I and II only (b) I and III only (c) I, II and III only (d) II, III and IV only

4. The upper and lower fix point of a thermometer are 30mm and 180mm respectively. Calculate the temperature in degree Celsius when the thermometer reads 45mm.
(a)10.00C (b) 15.00C (c) 20.00C (d)30.00C
5. A liquid in glass thermometer can be made sensitive if …………..
(a) the size of the bulb is reduced (b) the size of the bulb is increased (c) bore of the tube is increased.

THEORY
1. The electrical resistance of the element in a platinum resistance thermometer at 100oC, 0 oC and room temperature are 750Ω, 630Ω and 649Ω respectively. Use the data to determine room temperature.
2. Give two reason why water is considered an unsuitable liquid for a thermometer

CONTENT: HEAT CAPACITY AND SPECIFIC HEAT CAPACITY AND VERIFICATIONS
 Definition of heat, heat capacity and specific heat capacity
 Determination specific heat capacity

MEASUREMENT OF HEAT ENERGY
In order to assess the quantity of heat energy possessed by a body, three quantities are needed. They are:
(i) the change in temperature (∆0)
(ii) the specific heat capacity of the body (C)
(iii) mass of the body (m)

The quantity of heat( Q) of a body is a product of the three quantities above as expressed by the equation.
Q = MC∆0. It is measured in Joules

HEAT CAPACITY
This is the quantity of heat required to raise the temperature of a substance by one degree. It is measured in Joules/0K. H = mC

SPECIFIC HEAT CAPACITY OF MATERIAL
It has been discovered that the quantity of heat (Q) received by a body is directly proportional to its mass(m), change in temperature (θ2 – θ1) and also on nature of the material of which the body is made up of, stating this mathematically we can write
Q α m(θ2 – θ1)
Or
Q = mc(θ2 – θ1)................................................................(1.1)
Where c is the proportionality constant which depends on the material of which the body is made up of, c is called the specific heat capacity of the body.

From equation 1.1 above, we can define c as
Q
m(θ2 – θ1) ...........................................(1.2)
unit of c is Jkg-1K-1. (Because unit of mass is kilogram(kg), temperature in Kelvin (K).

The specific heat capacity of a substance is the quantity of heat required to raised the temperature of unit mass (1kg) of a substance through a degree rise in temperature (i.e 10 C or 1K)
Specific heat capacity of water is 4200JKg-1 K-1 ( or 4.2Jg-1 K-1 when the mass is in grams), it is observed that the specific heat capacity of water is higher than that of most other substances.
Specific heat capacities of some substances are given intable below
substance C in Jkg-1 K-1
Lead 130
Mercury 140
Brass 380
Zink 380
Copper 400
Iron 450
Glass 670
Aluminium 900
Ice 2100
Methylated spirit 2400
Sea-water 3900
Water 4200
https://youtu.be/_L9IQW6oaQE

THE HEAT CAPACITY(C)
When an entire mass of a body is taken into consideration instead of a unit mass, we talk of heat capacity. When the temperature is raised by 1K that is θ2 – θ1 is unit, then quantity of heat C is given by C = mc, this is called the heat capacity C.
The heat capacity of a body is the quantity of heat required to raise the temperature ot the entire body through one degree rise in temperature (1K).
Unit of C is in Joule per Kelvin (JK-1)

THE HEAT CAPACITY(C)
When an entire mass of a body is taken into consideration instead of a unit mass, we talk of heat capacity. When the temperature is raised by 1K that is θ2 – θ1 is unit, then quantity of heat C is given by C = mc, this is called the heat capacity C.
The heat capacity of a body is the quantity of heat required to raise the temperature ot the entire body through one degree rise in temperature (1K).
Unit of C is in Joule per Kelvin (JK-1)
https://youtu.be/TqJFIBODrjM

EXAMPLE 1
Calculate the mass of iron with heat capacity of 900 Jkg -1 K-1
SOLUTION
Specific heat capacity of iron = 450Jkg-1 K-1 using heat capacity C = mc and substituting yields:
900 = m x 450
m = 900/450
= 2.0kg
Example 2
What quantity of heat is needed to raise the temperature of 20kg of aluminium through 10k ?
Solution
Specific heat capacity of aluminium is 900Jkg -1 K -1
Using Q = mc (θ2 - θ1) and substituting we get,
Q = 20 x 900 x 10
= 180000J
= 18 x 104 J.

EXAMPLE 3
What is the heat capacity of 20kg of brass?
Solution
Specific heat capacity of brass is 380Jkg -1K -1
C = mc = 20 x 380
= 7600Jk-1

SPECIFIC HEAT CAPACITY
Specific heat capacity ( c )of a substance is the heat required to raise the temperature of a unit 0r 1Kg mass of the substance through one degree.
The quantity of heat Q received by a body is proportional to its mass (m), and temperature change
(02 - 01) and on the nature of the material the body is made of.
This Q α m (02 - 01)
Q = mC (02 - 01)
C is a constant of proportionality called the specific heat capacity of the body, which depends on the nature of the body.


specific heat capacity of a body can be determined by using
(i) the method of mixtures
(ii) the electrical method

DETERMINATION OF SPECIFIC HEAT CAPACITY BY METHOD OF MIXTURES.
The solid lead block is weighed on a balance to be ms. A lagged calorimeter is dried and weighed to be mc. It is then reweighed to obtain the mass of water m w ,when half filled with water. The initial temperature of the water is taken to be 01.
The lead block is suspended in boiling water with a temperature 03 after which it is transferred to the calorimeter and the mixture stirred to maintain a uniform temperature 02.
The specific heat capacity of the lead can be calculated using the fact that heat loss by the lead = heat gained by calorimeter and water. Given the specific heat capacity of calorimeter and water to be C[sub]c[/sub] and C[sub]w[/sub] respectively.



https://youtu.be/yhNHJ7WdT8A

DETERMINATION OF SPECIFIC HEAT CAPACITY BY ELECTRICAL METHOD.
To calculate the specific heat capacity Cb of a solid brass block, we make two holes in a weighed brass block into which a thermometer and a heating element connected to a source of power supply are inserted. Oil is poured in the holes to ensure thermal conductivity.


Assuming no heat is lost to the surrounding, the total amount of electrical heat energy supplied by the coil IVt = heat gained by the brass, MC[sub]b[/sub]0

Ivt = MC[sub]b[/sub]0……………(1)

From v = IR (ohms law)

I2 Rt = MC[sub]b[/sub]0………….(2)

V2 t = MC[sub]b[/sub]0………(3)

R


https://youtu.be/HAPmwu7byGM

EXAMPLE 1
250g of lead at 170ºC is dropped into 100g of water at 0ºC. If the final steady temperature is 12ºC, calculate the specific heat capacity of lead.

(Cw = 4.2 x 103 J/kgk)
heat lost lead = heat gained by water
0.25 x c x (170 – 12) = 0.1 x 4200 (12 – 0)

C = [sup]420 x 12[/sup]/[sub]0.25 x 158[/sub]

C = 127.6 J/kgk



EXAMPLE 2
A piece of iron of specific heat capacity,0.04J/KgK and mass 400Kgis quickly dropped into 30kg of water at 100C contained in ac calorimeter of 120Kg mass of C 0.1 J/KgK. If the temperature of the mixture is 30 0C calculate the initial temperature of the hot iron.

SOLUTION
heat lost by hot iron = mi ci (Ǿ3 - Ǿ2 ) = 400 X 0.04 X (Ǿ3 – 30) = 160 Ǿ3- 480
heat gain body = mw cw (Ǿ2 - Ǿ1 ) + mc cc (Ǿ2 - Ǿ1 )
= 30X 4200 X ( 30- 20) + 120 X 0.1 X ( 30-20)
= 2520000 + 240
But heat gain by cold body = heat lost by hot body
2520000 + 240 = 160 Ǿ3- 480
2520000 + 240 + 480 = 160 Ǿ3
Ǿ3 = 2520720/16
Ǿ3 = 157545K


https://youtu.be/hAbNyM8aKhM





SPECIFIC TOPIC: MEASUREMENT OF SPECIFIC HEAT CAPACITY
REFERENCE BOOK: SENIOR SECONDARY PHYSICS,BY OKEKE.
OBJECTIVE: At the end of the lesson, the students should be able to: Enumerate the measurement of specific heat capacity

CONTENT: MEASUREMENT OF SPECIFIC HEAT CAPACITY
Quantity of heat Q is usually obtained using a calorimeter. The most common method used in calorimeter experiment is known as the method of mixtures, another method usually employed is electrical method

TRANSFER OF HEAT
When a hot metallic object is dropped into a copper vessel containing known mass of water at lower temperature, heat continues to pass from the metallic object to the copper vessel and water, until the temperature of the three items (hot metallic object, copper vessel and water) are the same. The hot metallic object loss heat, while the copper vessel and water gain heat. When they finaly come to a steady temperature (same temperature), then we can equate heat lost by hot metallic object to be equal to the heat gained by the copper vessel and water. This is referred as heat transfer.

Photocopiers and Document Reproduction

The photocopiers commonly found in offices and classrooms use electric charges to transfer the image of an original document to a plain piece of paper. The document to be copied is placed face down on the platen and illuminated by a lamp. Its image is directed to a negatively charged metal drum (the electrostatic drum) by a series of mirrors. Where light strikes the drum, the charge disappears, so that dark areas remain charged. Next, positively charged particles of toner powder are brushed onto the drum. These stick only to the charged areas. (The first- and second-erase lamps remove the charge on the drum between different copying tasks.) The image on the drum is then transferred to a piece of paper that has been given a negative charge by the transfer charger. A heater is used to seal the toner to the paper, which is why copies are warm when they emerge.
https://youtu.be/HFxo4j1EWsA
https://youtu.be/NxUbPE8RsiM

MEASUREMENT OF SPECIFIC HEAT CAPACITY OF A SOLID BY METHOD OF MIXTUERS
The apparatus are calorimeter, the stirrer are weighted empty and the weight recorded. Then, the weight of the calorimeter which is then two – thirds filled with water and the stirrer are weighed and recorded, too. The metal block is weighed (mb), the initial temperature of cold water say θ1 is taken. The metal block is put inside a boiling water and the temperature of boiling water is recorded, which is the temperature of the hot metal block say θ2. The metal block is quickly transferred to the lagged calorimeter containing water. The whole content is stirred until all the content come to a final steady temperature. The final temperature of the mixture is taken, say θ f .At this final steady temperature, heat lost by hot metal = heat gained by water and calorimeter.
Hence
mbc(θ2 – θf) = Mccc(θf—θ1) + Mwcw(θf –θ1)
Therefore,
C = Mccc(θf—θ1) + Mwcw(θf –θ1)
Mb(θ2 – θf ) ...............................................................(1.3)
Where c is specific heat capacity of the metal block and Mwcw and McCc are the mass and specific capacity of water and calorimeter respectively.
Consequently heat capacity is mc.

https://youtu.be/LqaThFQyUHI

EVALUATION:
Explain how do we measure the specific capacity of a material
Hot water is added to four times its mass of water at 250C and thoroughly stirred. If the final temperature of the mixture is 400C, calculate the initial temperature of the hot water.

ASSIGNMENT:
.A tap supplies water at 150 C while another supplies water at 900C. If a man wishes to bathe with water at 300C, calculate the ratio of the mass of cold water to the mass of hot water.

EVALUATION
1. A metal of mass 0.5kg is heated to 100ºC, transferred to a well lagged calorimeter of heat capacity 80 J/k containing water of heat capacity 420 J/k at 15ºC. If the final steady temperature of the mixture is 25ºC, find the specific heat capacity of the metal.

3. Heat is supplied to a liquid of mass 500g contained in a can by passing a current of 4A through a heating coil of resistance 12.500 immersed in the liquid. The initial temperature of the liquid is 240C. The liquid reaches its boiling point 10min after the current is switched on, calculate the specific heat capacity of the liquid.

3. Calculate the mass of glass with heat capacity of 670Jkg-1K-1


ASSIGNMENT
1. A body of mass 40g losses 80J of heat. If the specific heat capacity of the body is 400J/KgK. Calculate the change in temperature of the body. (a)0.2K (b)0.4K (c) 2.0K (d)5.0K

2. A 50W electric heater is used to heat a metal block of mass 5Kg. if in 10min a temperature rise of 120C is obtained, the specific heat capacity of the metal is?
(a)130 J/KgK (b) 390 J/KgK (c)400 J/KgK (d) 500J/KgK

3. A water fall 420m is high. Calculate the difference in temperature of the water between the top and bottom of the water fall. Negelect heat loss. [g=10m/s2, specific heat capacity of water= 4200 J/KgK] (a)0.10C (b) 1.0 0C (c) 4.20C (d) 420C

4. A good calorimeter should be (a) low specific heat capacity and low heat conductivity
(b) low specific heat capacity and high heat conductivity
(c) high specific heat capacity and low heat conductivity
(d) high specific heat capacity and low heat conductivity

5. The temperature of a piece of metal of mass 9g is raised from 100C to 1100C when it absorbs 108J of heat energy. Determine the specific heat capacity of the metal in J/KgK (a) 120 (b) (c) (d)

THEORY
1. A copper block of mass 24g at 2300C is placed in a copper calorimeter of mass 60g containing water of mass 54g at 310C,assuming heat loss are negligible, calculate the final steady temperature of the mixture. [specific heat capacity of water= 4200J/KgK specific heat capacity of copper =400J/KgK]

2. A hot body of mass 3g at a temperature θ is transferred into a calorimeter of negligible heat capacity containing water of mass 30g at 15oC. if the mixture is observed to be at 55oC. find the temperature of the hot body. [ specific heat capacity of water 4200J/KgK, specific heat capacity of body= 350 J/KgK]

LATENT HEAT

CONTENT:
 Latent heat
 Latent heat of fusion
 Latent heat of vaporization
Sometimes when heat energy is given to a substance, it does not increase its temperature, rather, it changes the state of the substance either from solid to liquid ,or liquid to gaseous state, such heat is referred to as latent heat. There are two types of latent heat:
(i) Latent heat of fusion
(ii) Latent heat of vaporization

Latent heat of fusion: This is the quantity of heat energy needed to change solid to liquid without any change in temperature. Its unit is J/Kg

Latent heat of vaporization: This is the amount of heat energy needed to change liquid to gases (vapour) or steam at constant temperature. Its unit is J/Kg

SPECIFIC LATENT HEAT
This is the quantity of heat required to change a unit mass of a substance from one state to another without a change in its temperature.

When a small piece of solid (ice) is placed in a beaker and the beaker is heated slowly, a temperature is reached where the ice starts to melt.
As it melts, the temperature remains constant or steady.
The heat supplied is used to change the state of ice from solid to liquid.
Similarly, when heat is supplied to a liquid, there is a rise in temperature up to a point where the liquid starts boiling and the temperature here is called saturation temperature.

The temperature reading which is constant leads to vapourization of gas, no matter the increase in heat energy. This heat is hidden because there is no increase in temperature, and is called latent heat.

Latent heat is defined as the quantity of the heat required or needed to change the state of a substance at constant temperature and pressure.
It is denoted by Q, measured in joules (J) and it is a scalar quantity.
Q α m
Q = Ml
M = mass (kg)
L = specific latent heat (J/kg or J/g)

Latent heat depends on:
(i) nature of the material
(ii) the mass of the material
(iii) the energy supplied
(iv) volume and area of the material
(v) the state of the material

(a) SPECIFIC LATENT HEAT OF FUSION (L[sub]f[/sub])

This is the quantity of heat required to change a unit mass of a solid without a change in its temperature. The specific latent heat of fusion of ice is 33600 J/kg.
Q = m Lf

Specific latent heat of fusion (ice) is defined as the quantity of heat required or change a unit mass of solid (ice) into a liquid (water) at constant temperature (melting point) measured in J/kg or J/g. It is also a scalar quantity.
Thus, the experimental value of Lice = 336J/g

(b) SPECIFIC LATENT HEAT OF VAPORISATION (L[sub]v[/sub])

This is the quantity of heat required to convert a unit of mass of liquid from liquid state to vapour (gaseous state) without any change in temperature. For water the value is 2.26 x 106 J/kg
Q = mLv

Specific latent heat of vapourization (steam), is defined as the quantity of heat required or needed to change a unit of mass of liquid to gaseous state at constant temperature ( boiling point) and pressure.

It is denoted by Lsteam , measured in J/kg or J/g and a scalar quantity.
Thus , the experimental value of Lsteam = (2260 – 2268)J/g
Hence, Lice = 1/7 of Lsteam
The temperature – time sketched graph of ice at – θ0 C Changed to steam at 100ºC

https://youtu.be/6lAxBTLgYfU

EVALUATION:
1. Define the following terms (a) latent heat (b) specific latent heat of fusion (c) specific latent heat of vaporization

EXAMPLE 1:
1. How much heat energy is needed to change 3g of ice at 00C to steam at 1000C.
(Lf = 336 KJ / kg, Cw = 4200 J/kgk, Lv = 2.26 x 106 J/kg)
First stage:
Heat required for the ice to melt at 00C
Q1 = mLf
= 0.003 x 336000
Second Stage
Heat required to raise the temperature of the melted ice from 00C to 1000C
Q2 = mC0
= 0.003 x 4200 x 100
Third Stage
Heat required to convert the liquid to steam
Q3 = mLv
= 0.003 x 2260000
The total energy, Q = Q1 + Q2 + Q3
= 0.003 (336000 + 420000 + 2260000)
= 9048J


EXAMPLE 2: 0.5kg of water at 100cis completely converted to ice at 00c by extracting 188000J of heat from it. If the specific heat capacity of water is 4200J/kgk, calculate the specific latent heat of fusion of ice.
Solution
Q = MLice + MC∆θ
Therefore, 188,000 = 0.5L + 0.5 x 4200 x 10
188,000 = 0.5L + 21000
0.5L = 188,000 – 21000
0.5L = 167,000
L = 167,000
6.5
L = 334,000J/kg
= 333KJ/Kg

https://youtu.be/oc0ypeDELb0

EVALUATION
1. Calculate the total energy required to evaporate completely 1kg of ice that is initially at – 100C
Given that:
Specific capacity of Ice = 2.2 x 10[sup]3[/sup] J/kgk

Specific heat that capacity of water = 4.2 x 10[sup]3[/sup] J/kgk

Specific latent heat of fusion of Ice = 3.36 x 10[sup]5[/sup] J/kgk

Specific latent heat of vaporization = 2.26 x 10[sup]6[/sup] J/kgk

2. Describe briefly, with two reasons, two ways by which a quantity of liquid may be made to evaporate more quickly.

3. What amount of current would pass through a 10Ω coil, if it takes 21s for the coil to melt a lump whose ice of mass is 10g at 00c, if there are no heat losses?(Lice = 336J/g)

https://youtu.be/dxtz2POUTJE

ASSIGNMENT
1. A block of ice of mass 2Kg at 00C absorbs heat energy from its surroundings and changes to water at the same temperature. The heat absorbed is referred to as (a) expansivity (b) heat capacity (c) latent heat (d) specific latent heat
2. How much heat energy is required to change 10g of ice at 00C to water at 100C [ latent heat of fusion= 330J/g specific heat capacity of water = 4.2J/KgK] (a)3720J (b) 3460J (c) 3402J (d)3360J
3. Calculate the quantity of heat heat released when 100g of steam at 1000C condenses to water. Take specific latent heat of vaporization of water =2.3X106J/Kg] (a)2.3X101J (b) 2.3X102J (c) 2.3X104J (d) 2.3X105J
4. The amount of heat given out or absorbed when a substance changes it state at a constant temperature is known as (a)latent heat (b) heat capacity (c) specific latent heat (d)
Specific heat capacity
5. A quantity of steam at 1000C condenses to water at the same temperature by releasing 6.9X104J of heat. Calculate the mass of the steam condensed.(a) 3.0X10-1Kg (b) 3.0X10-2Kg (c) 3.0X10-3Kg (d) 3.0X10-4Kg

THEORY
1. Define the following terms (a) latent heat (b) specific latent heat (c) specific latent heat of vaporization (d) specific latent heat of fusion

2. An electric heater immersed in some water rises the temperature from 400C to 1000C in 6min. after another 25min it’s noticed that half of the water has boiled away. Ignoring heat losses to the surrounding calculate the specific latent heat of vaporization of the water.

3. Explain specific heat capacity and describe a method of finding the latent heat of steam.

4. Calculate the quantity of heat required to melt an ice of o.7kg which has latent heat capacity of 3.4 x 105 J/kg

CONTENT:
 Effect of pressure on boiling and melting point
 Effect of impurities on boiling and melting point
 Differences between boiling and evaporation.

THE EFFECT OF PRESSURE AND IMPURITIES ON BOILING AND MELTING POINT
As heat is being added to a liquid, its temperature increases steadily until at a certain temperature when rapid evaporation is seen to occur in every parts of the liquid with the bubbles of vapour escaping to the surface. This process is known boiling.

Boiling point of a liquid can be defined as the temperature at which the saturated vapour pressure of the liquid is equal to external atmospheric pressure.

As a matter of distinction, evaporation occurs only at the surface of the liquid while boiling occurs throughout the entire mass of the liquid. Also evaporation takes place at all temperature but boiling occurs at a particular temperature called boiling point.

To demonstrate the effect of pressure on the boiling point of a liquid, a round-bottomed flask is half filled with water and heated above the boiling point of the water to drive out all the air. The flask is now tightly corked and left in an inverted position to cool until the water stops, boiling is noticed to resume and stop immediately we stop pouring the cold water over the flask.

The poured cold water causes condensation of the vapour inside the flask, which consequently reduced the pressure on the surface of the water. This reduced pressure lowers the boiling point of the liquid and make it boils again. Hence reduced pressure always lowers the boiling point of a liquid. Dissolved impurities raises the boiling point of liquid.
Boiling is defined as the change from liquid to vapour when the saturation vapour pressure is equal to the atmospheric pressure, or the vapourization of liquid molecules throughout the entire liquid.
When a liquid is boiled to a particular temperature, it starts giving up its molecules as gases since the kinetic energy absorbed increases with energy impact per unit time, and in doing so, the molecules escape. The temperature at which the liquid boils to become vapour is known as boiling point.
Boiling is affected by the following factors: (i) impurities (ii) pressure (iii) mass of the liquid (iv) nature of the liquid (v) area of the surface exposed.

EFFECT OF IMPURITIES ON BOILING
Impurities affect the boiling point of a liquid because their presence increases the boiling point of the liquid, compared to the boiling point of a pure solvent.

EFFECT OF PRESSURE ON BOILING
Pressure affects the boiling point of a liquid, because an increase in pressure will lead to a decrease in boiling point. This explanation has a practical application in pressure cooker which is a sauce – pan with lid that can be held down.

APPLICATION OF PRESSURE ON BOILING
1. It is used in the principle of pressure cooker.
2. The aircraft flying at high altitude, where air is of lower pressure than normal, needs to be pressurized so that the people can be at their normal pressure.
3. Astronauts must wear space suits not only for breathing but for them to be at righ pressure.

EFFECT OF EVAPORATION
One of the important properties possessed by evaporation is the way it causes cooling. Everybody requires cooling to lower the body temperature.
Example, when a piece of either is dropped on the skin, after some seconds, the either dries up and at the point of application, coldness is felt on the skin because when the liquid is converted to vapour, heat is absorbed from the liquid or any object in contact with it. This is the latent heat used in causing transformation from liquid to vapour or gas.
The absorption of the heat causes a fall in its temperature. The faster the evaporation, the greater the fall in temperature. Water stored in earthen pot is cooler than that stored in a glass bottle because evaporation takes place through the pore on the earthen pitchers which causes the water to cool. It also explains why either is evaporated on the skin before a person is given an injection. As the either evaporates, it exerts some cooling effect on the skin.

SUBLIMATION
Sublimation is a process by which a solid is heated straight to vapour without passing through the intermediate liquid state. The solid molecules acquire some kinetic energy which is very great and instead of the solid melting into a liquid phase, it is converted into gaseous or vapour state at a very high temperature, e.g. dry ice when changed to vapour.


https://youtu.be/AFARWpoaSJA

https://youtu.be/3TPxSbGA798

https://youtu.be/wX4iBnIn9ZQ

DIFFERENCES BETWEEN BOILING AND EVAPORATION
Evaporation--------------------------------------------------------------------------------------------Boiling
This is the change from liquid to vapour at the temperature below boiling point ---This is the change from liquid to vapour at the boiling point
Occurs at all temperature---------------------------------------------------------------------Occurs at fixed temperature at a given pressure
Temperature during evaporation is not steady---------------------------------------------Temperature remains steady during boiling
Occurs only at the surface of the liquid-----------------------------------------------------Occurs throughout the entire liquid




EVAPORATION ------------------------- BOILING
(I) It occurs at all temperature.---------- (i) It occurs at the boiling point of the liquid
(ii) It causes cooling ----------(iii) It does not cause cooling
(iv) It occurs only at the surface.---------- (i) It occurs in every part of the liquid.
(ii) It is not affected by the mass of the liquid exposed.---------- (v) It is affected by the mass of the liquid exposed.
(vi) It does not depend on the container of the liquid----------(vii) It depends on the container because they absorb their own energy first.

SIMILARITIES BETWEEN BOILING AND EVAPORATION
(1) They both involve the escape of molecules into the atmosphere.
(2) They both depend on the nature of the liquid
(3) They both depend on pressure
(4) They both depend on the area of the surface
(5) They both depend on impurities.
https://youtu.be/nZXoe0fgJ8Q

APPLICATION IN PRESSURE COOKER
The fact that increased pressure raises the boiling point is put into a useful application in the pressure cooker. The increased pressure of the trapped gas above the liquid raises the boiling of the liquid inside the cooker. This provides a high cooking temperature needed to conserve fuel and save time.
https://youtu.be/TWV3FbgPPXo

APPLICATION IN REFRIGERATORS
Refrigerators make use of the cooling effect of evaporation. The volatile liquid such as liquid ammonia or freon evaporates inside copper coil surrounding the freezing compartment, supported by electric pump which reduced the pressure. As the volatile liquid evaporates in those coils, it absorb heat from the surround air, consequently and cooling the inside of the refrigerator and its content.

The vapour produced is pumped off into the condenser, where it is compressed by the pump and condenses back to liquid. The latent heat given out during this condensation is quickly dissipated by an arrangement of cooling fins at the back of refrigerator.

Heat is eliminated by convection and radiation to the surroundings and by conduction into fins. The liquid is again passed into the evaporator coil and thus the level of cooling is regulated by a thermostat connected to the switch.
https://youtu.be/h5wQoA15OnQ

EVALUATION:
1. State the effects of the following on boiling point and the melting point of a liquid
(i) presence of impurities
(ii) increase in pressure
(iii) decrease in pressure
2. Define the boiling point of a liquid

Assignment
1. The boiling point of a liquid depends on the following except………………. (a) nature of the liquid (b) external pressure (c) volume of the liquid (d) impurities present in the liquid
2. Which of the following statements is not correct?
(a) evaporation occurs only at the surface of the liquid (b) boiling takes place throughout the volume liquid (c) evaporation takes place at all temperature (d) the boiling point of a liquid is not affected by impurities
3. Which changes can alter the boiling point of a liquid?
I increasing the volume of the liquid
II increasing the external pressure
III increasing the quantity of impurities in the water
(a) I only(b) II only (c)I and II only (d)II and III only
4. Water in an open container boils at a lower temperature when heated at the top of a mountain than at sea level because at the top of the mountain the
(a) relative humidity is higher than at sea level (b) rays of the sun add more heat to the water (c) temperature is lower than at sea level (d) pressure is lower than at sea level
5. Which of the following does not reduce heat lost from a liquid in a calorimeter? (a) lagging the calorimeter (b) using an insulating lid (c) shielding the calorimeter from draught (d) constantly stirring the liquid


THEORY
1. Differentiate between boiling and evaporation.
2. State four (04) factors which affect evaporation of a liquid.

VAPOUR PRESSURE
CONTENT:
 Saturated vapour pressure
 Humidity and relative humidity
 Dew point

When a liquid evaporates in a closed container, the vapour formed above the liquid exerts a pressure. According to kinetic molecular theory, the molecules of the vapour are in constant motion and will hence exert a pressure just like the molecules of a gas. This pressure is called the vapour pressure of the liquid.

SATURATED VAPOUR
A saturated vapour is a vapour that is in contact with its own liquid within a confined space. When the enclosed space above a liquid is saturated with vapour molecules and can hold no more molecules, the pressure exerted by this saturated vapour is said to be the saturated vapour pressure ( s. v. p ) of the liquid. The vapour is said to be saturated when the number of molecules escaping from the liquid per unit is equal to the number returning to the liquid per unit time. The saturated vapour is thus said to be in a state of dynamic equilibrium with its own liquid. Saturated vapour pressure increases with temperature.
https://youtu.be/7E7lvcuC9eQ

https://youtu.be/CfagHzOtIDM

RELATIVE HUMIDITY AND DEW POINT.
There is always some water vapour present in the atmosphere. This is due to evaporation from oceans, rivers, lakes, etc. The water vapour content of the atmosphere is known as its Humidity. If there is a high percentage of water in the air, the air is said to be very humid. Relative humidity is used to describe how humid or moist the air is.
Relative humidity is the ratio of the mass of water vapour present in a certain volume of air to the mass of water vapour required to saturate the same volume of air at the same temperature. It is usually expressed in percentage and the values are used by meteorologist in weather forecasts. An instrument known as the hygrometer is use to measure relative humidity of air.
Dew point is the temperature at which the water vapour present in the air just sufficient to saturate it
Relative humidity=s.v.p at dew point
https://youtu.be/Qsl5yQsinlY
https://youtu.be/OZh9ksAy9kc

EVALUATION:
Define the following (i) dew point (ii) relative humidity (iii) saturated vapour pressure (iv) boiling point

Assignment
1. Which of the following physical quantities affect affects the saturated vapour pressure of liquid? (a)temperature (b) volume (c) mass (d) density
2. Humidity is used to described the amount of (a) air in water vapour (b) cloud available in the atmosphere (c) air in equal volumes of cloud (d) water vapour in the atmosphere.
3. The temperature at which the vapour present in the air is just sufficient to saturate it is called the (a) boiling point (b) steam point (c) dew point (d) saturation point
4. Which of the following instrument may be used to measure relative humidity? (a)hydrometer (b) barometer (c) hypsometer (d) hygrometer
5. The temperature at which the saturated vapour pressure is equal to the external atmospheric pressure is known as its (a) dew point (b) boiling point (c) lower fixed point (d) triple point

Theory
1. 1Kg of boiling water at 1000C changes to ice at 0oC . if the specific latent heat of fusion of ice is 3.36X 10 5 Jk-1 and the specific heat capacity of water is 4200J/kgK the heat extracted from the water is?
2. A heating coil rated at 1KW is used to evaporate 700g of water at 1000C. If the specific latent heat of vaporization of water is 2.3X106 J/Kg. The time taken to evaporate all the water is?








MAIN TOPIC: MOMENT
SPECIFIC TOPIC: moment of c.g about a given axis
REFERENCE BOOK: Igcse Physics by Richard Woodside

OBJECTIVE: At the end of the lesson, the students should be able to:
Explain the moment of c.g about a given axis

CONTENT: MOMENT OF C.G ABOUT A GIVEN AXIS
An object, e.g. a cylinder with a high C.G held on an inclined plane when slightly displaced, moves in an anti-clockwise moment about its axis so that it is triples over. The object is said to be in unstable equilibrium. Similarly, an object with a low C.G held on an inclined plane when displaced, moves in a clockwise moment about its axis will rather be restored to its original position. Thus, the object is said to be stable equilibrium. But if an object, e.g cylinder is placed with its curve surface on a horizontal plane when, slightly displaced, there exists no moment about the axis where the cylinder touches the plane. The object rolls to a few position and the object (cylinder) is said to be neutral equilibrium.

Examples of equilibrium body include:
1. An animal with four legs
2. The bottom of the ship which is made heavy to keep the centre of gravity as low as possible.
3. A truck loaded with iron.


(a) Stable (b) unstable

Example 1:
A metre rule is placed horizontally at a fixed point while masses are suspended at both right and left hands of the ruler.
From the centre of the ruler a mass, M, is hung 20.5cm away from the centre and mass of 0.1kg is hung at the other side from the centre of the ruler distance 45cm. Find the mass required to balance the ruler.
Solution

Applying the principles of moment,
Mg x 20.5 = 0.1g x 45
M = 0.1g x 45
20.5g
M = 0.22kg

Example 2: In order to weigh a boy in the laboratory, a uniform plank of wood AB 3.0m long, having a mass of 8.0kg is pivoted about a point 0.5m from A. The boy stands 0.3m from A and a mass is placed 0.5m from B in order to balance the plank horizontally. Calculate the mass of the boy.
Solution

Applying the principles of moment,
mg x 0.2 = (8g x 1) + (2g x 2)
mg x 0.2 = 8g + 4g = 12g
m = 12g/0.2 x g
m = 60kg

https://youtu.be/RQVoeWx0dz4
https://youtu.be/7y07CWvPTZc
https://youtu.be/qAJnsT9uMUk

EVALUATION:
A uniform half-meter scale AB is balanced horizontally across a knife edge placed 15cm from A. A mass of 30g is hung from the end. What is (a) the mass of the scale? (b) the force exerted on the scale by the knife edge?

ASSIGNMENT:
It is found that a uniform wooden lath 100cm long and of mass 95g can be balanced on a knife-edge when a 5g mass is hung 10cm from one end. How far is the knife - edge from the centre of the lath?






SPECIFIC TOPIC: MOMENT
REFERENCE BOOK: SENIOR SECONDARY PHYSICS,BY OKEKE.

OBJECTIVE: At the end of the lesson, the students should be able to:
Solve problems on moment

CONTENT: MOMENT
Example 1: A uniform wooden rod AB, 120cm long and weighing 120N rest on two sharp-edged supports C and D placed 10cm from each end of the rod respectively. A 20N weight hangs from a loop of thread 30cm from A and a 90N weight hangs similarly 40cm from B. Find the reactions at the supports.
Solution
Taking moment about D,

CLOCKWISE ANTI-CLOCKWISE
RN x 1m = (20 x 0.80) + (120 x 0.50) + (90 x 0.30)
RN x 1m = 16 + 60 + 27 = 103Nm
R = 103N
Therefore, R = 103N and 230 - R = 127N
Answer: Reaction at C = 103N
Reaction at D = 127N

EVALUATION: It is found that a uniform wooden lath 100cm long and of mass 95g can be balanced on a knife-edge when a 5g mass is hung 10cm from one end. How far is the knife - edge from the centre of the lath?

GAS LAW
CONTENT:
 Boyle law, its equation and graphical representation.
 Charles law, its equation and graphical representation.
 Pressure law, its equation and graphical representation.
 General gas law equation.
 Ideal gas equation.

In an attempt to study the behavior of gases in relation to volume, temperature and pressure, the following conditions are investigated
(a) variation of volume with pressure at constant temperature, Boyle’s law (Pv = constant)
(b) variation of volume with temperature at constant pressure,
Charles law (V/T = constant)
(c) variation of pressure with temperature at constant volume, pressure law (P/T = K)
https://youtu.be/robEY-idcLU

A. BOYLES LAW states that the volume of a fixed mass of gas varies inversely as the pressure acting on it provided the temperature remains constant.


https://youtu.be/N5xft2fIqQU

(B) CHARLES LAW states that for a fixed mass of gas at constant pressure, the volume is proportional to its absolute or thermodynamic temperature.


https://youtu.be/GcCmalmLTiU

(C) PRESSURE LAW: states that the pressure of a fixed mass of gas at constant volume is proportional to its absolute temperature.

https://youtu.be/K0V1Lcy2AnE

(D) ABSOLUTE ZERO OF TEMPERATURE
When the graphs of volume – temperature or pressure – temperature are extrapolated backwards they cut the temperature axis at -2730C. This temperature is called absolute zero, the temperature at which the volume of the gas theoretically becomes zero as it is being cooled. This temperature is a mere assumption, as gases are known to liquefy more often than not before such a temperature is reached.

(E) GENERAL GAS LAW
The general gas law is the combination of the Boyle’s, Charles and Pressure Law.



(F)Ideal gas equation:



EVALUATION:
1. Define the following gas laws give their graphical representation
(i) Boyle law (ii) Charles law (iii) pressure law

Assignment
1. The volume of a given mass of gas is 40cm3 at 270C. what is its volume at 900C if its pressure remains constant? (a)32.83 cm3 (b) 40.02 cm3 (c) 48.40 cm3 (d) 57.44 cm3

2. Which of the following gas laws is equal to workdone ? (a) charles law (b) pressure law (c) van er waals law (d) boyle law

3. The pressure of a fixed mass of gas is 2.0X105Nm at a known temperature, assuming that the temperature remains constant. What will be the pressure of the gas if it volume is havled? (a) 1.0X105Nm (b) 2.0X105Nm (c) 3.0X10-2Nm (d) 4.0X10-2Nm

4. A gas occupies a certain volume at 270C. at what temperature will its volume be three the original volume assuming that its pressure remains constant? (a) 810C (b) 1620C (c) 3540C (d) 6270C

5.when the pressure of a fixed mass of gas is doubled at constant temperature , the volume of the gas is (a) increase four times (b) doubled (c) unchanged (d) halved

THEORY
1. State Charles law and show its graphical representation in for temperature in Kelvin and degree Celsius.
2. Explain Boyle law using kinetic theory of matter.

GENERATION AND PROPAGATION OF WAVES
CONTENT:
 Definition of waves
 Types of waves and Classes of waves
 Terms used to explain a wave motion.
 Equations of waves.

WAVES AND PROPERTIES OF WAVES
A wave is a disturbance which travels through a medium and transfers energy from one point to another without causing any permanent displacement of the medium itself.
If a stone is dropped into a pound or swimming pool, ripples or waves are seen spreading on the surface of the water from the point where the stone was dropped. The water itself does not move in the direction of the ripples, but the wave transfers energy from one point to another.
Waves are also encountered in other branches of physics. For example, we can generate a wave along a string fixed at both ends by plucking the string (i.e pulling it vertically and releasing it). Light and sound waves can also be shown to be wave motions.
A wave is a temporary disturbance of a medium when energy is transferred from one point to the other without causing permanent displacement of the medium.

WAVE MOTION
The particles of the medium which transfer energy move to and fro, or vibrate, about a mean position as the wave passes. The vibrations are passed on from one particle of the medium to the next. The direction in which this vibration takes place is significant in classifying the type of wave.


CLASSES OF WAVES
There are two classes of waves
 Electromagnetic waves
 Mechanical waves

(1) Mechanical waves : Mechanical waves are those waves which require a material medium for their propagation (movement). These mediums could be solids (rope), liquids( water) or gases (air). E.g water waves, sound waves, string waves.

(2) Electromagnetic waves : Electromagnetic waves are those waves which don’t require a material medium for their propagation (movement) E.g x-rays, gamma rays, radio waves, infra-red, ultraviolet rays

TYPES OF WAVES
There are also two types of waves which are
 Transverse waves
 Longitudinal waves
Transverse waves are those waves whose direction of travel is perpendicular to the direction of vibration or travel of the medium. E.g water waves, string waves, radio waves, x-rays.

Longitudinal waves are those waves whose direction of travel is the same or parallel to the direction of travel of the material medium. E.g sound waves. It travels through empty spaces in form of oscillating electric and magnetic forces which is at right angles to one another.

TRANSVERSE AND LONGIDUTINAL WAVE FRONT


https://youtu.be/aCu4VRKMstA

https://youtu.be/AUBAMlMoI1g

TERMS USED TO DESCRIBED A WAVE
AMPLITUDE A, is the maximum displacement of a particle from its rest or mean position. It is measured in metre (m).

PERIOD (T) is the time required for a particle to perform one complete cycle or oscillation. FREQUENCY (F) is the number of complete cycles made in one seconds. It is measured in Hertz (Hz)

WAVELENGTH (λ) is the distance covered by the waves after one complete oscillation.
For transverse waves, it is the distance between successive crests or troughs while for longitudinal wave,it is the distance between successive compressions or rare factions. It is measured in metre (m).

WAVE-VELOCITY is the distance traveled by the waves in one second. The S.I unit is m/s

Depending on the direction of particle vibration with respect to the direction of travel of the wave, we can distinguish two types of wave

A wave is said to be a transverse wave if the direction of travel of the wave is perpendicular to the direction of vibration of the medium.

For example, water waves and waves generated by plucking a string a string are transverse waves.

A wave is said to be longitudinal if the direction of travel of the wave is the same as the direction of vibration of the medium.

For example, sound waves are longitudinal waves.

AMPLITUDE (a)
As the wave progresses, the particles of the medium vibrate about a mean position. The maximum displacement of particles from their mean (or rest) position is called the amplitude a of the wave. It is measured in metres

PERIOD(T)
The time required for a particle to perform one complete cycle or to complete an oscillation is called the period T of the wave. It is measured in seconds.
The period is also the time for the wave to travel one wave length, i.e from 0 to A or c to c.

FREQUENCY (f)
The number of cycles which the wave completes in one second is called the frequency, f. The SI unit of frequency is the hertz (Hz). Larger units are kilohertz(kHz) and megahertz (MHz). (1kHz = 103Hz, 1MHz = 106Hz)

WAVELENGHT(λ)
The distance along the x-axis between successive crests or successive troughs is called the wavelength λ. It is the distance covered by the wave after one complete cycle. It is measured in metres. In longitudinal waves the wavelength is the distance between successive compressions or rarefactions.

WAVE SPEED (v)
The distance which the wave travels in one second, is called the wave speed,v. It is measured in metres per second(ms-1). If the wave covers a distance x metre in t seconds the speed of the wave is given by
V= x/t ms-1 ......................................... (i)

RELATIONSHIP BETWEEN T, f, λ and v

MATHEMATICAL RELATIONSHIP
v = wave – velocity
f = frequency (Hz)
γ = wavelength (m)
+T = period (s-1)
v = f λ --------------- (1)
v/γ = --------------- (2)
v/f = λ --------------- (3)
f = 1/t ---------------- (4)
T = 1/f --------------- (5)
Therefore v = λ /T ------------- (6)
= VT = λ --------------- (7)


From the definition of frequency (f) and period (T), it follows that the two quantities are related by the equation f = 1/T. The argument leading to the formula is simple. In T second, 1/T (i.e. one period) 1 cycle is performed. In 1 second, 1/T cycles will be described.
This number is the frequency f.
Therefore, f = 1/T ............................................... .........(ii)
The speed v, the wavelength λ and frequency fare related by a very important equation
V= fλ ............................................................(iii)

Example 1: A wave travels a distance of 80m in 4s. The distance between successive crests of the wave is 50cm. Calculate the frequency of the wave.
Speed of the wave v = 80/4 = 20ms-1
λ= 50cm = 0.5m
f = v/λ = 20/0.5
f = 40Hz
Hence, f = 40Hz

https://youtu.be/TsQL-sXZOLc

https://youtu.be/Zj8JQJtVl0c

EVALUATION
1. Define a transverse waves and a longitudinal wave. Give two examples each.
2. Define the following terms (a) wavelength (b) frequency (c) period ( d) wave speed.
3. A radio station broadcasts at a frequency of 200kHz.If the speed of the wave is 3 x 108ms-1,
Calculate
(a) The period;
(b) The wavelength of the wave.

Worked example
A radio station broadcasts at frequency of 300KHz. If the speed of the wave is 3 x 108 ms-1, calculate the period and wavelength of the wave?



STATIONARY WAVES
A stationary wave is formed when two traveling waves of equal amplitude and frequency traveling in opposite direction are combined together.


Example:
A plane progressive wave is given by the equation y= Asin (2000 [sup]λt - λx[/sup]/[sub]17[/sub])

Calculate:
(1) The wavelength of the wave
(2) The speed
(3) The frequency
(4) The period
Solution:
By comparing the given equation with the standard equation
Y = Asin (2000 λ t - λ x)
17
2 λ vt = 2000 λ t
λ
= v = 1000Hz
ג
f = 1000Hz
2 גx = λ x
ג 17
2 x 17 = λ
34m = λ
v = f λ
= 1000 x 34
= 3.4 x 104 ms-1
T = 1/f
= 1 / 1000
= 103 s-1


https://youtu.be/Aucu7YshyQ0
Assignment
1. A periodic pulse travels a distance of 20m in 1.0sec. if the frequency is 2.0X103Hz.calculate the wavelength. (a) 1.0X10-3m (b) 1.0X10-2m (c) 2.0X10-2m (d) 1.0X102m

2. The distance between successive crest of a wave travelling at 20m/s is 25cm. calculate the frequency of the wave (a) 0.8Hz (b) 5.0Hz (c) 50.0Hz (d) 80.0Hz

3. A slinky spring fixed at one end is placed horizontally on a table. The free end is displaced parallel to the table and then released. The resulting waveform is (a) transverse (b) longitudinal (c) stationary (d) electromagnetic

4. In the wave equation y= E0 sin (200t - ∏x) E0 represents the (a) amplitude (b) frequency (c) period (d)wavelength

5. The distance between two points in phase on a progressive wave is 5cm. if the speed of the wave is 0.20m/s. calculate its period. (a) 4.00s (b) 2.50s (c) 0.25s (d) 0.04s

THEORY
1. The equation y= 5sin (3x – 4t) where y is in meter, x is in meter and t in sec. determine the
(i) Frequency (ii) period (iii) speed of the wave
2. Define the following terms (i) amplitude (ii) period (iii) frequency (iv) wavelength

PROPERTIES OF WAVE
Content:
• Properties of waves
• Polarization of waves

All waves exhibit the following properties
(1) Reflection
(2) Refraction
(3) Diffraction
(4) Interference
(5) Polarization [ peculiar to transverse waves only]
Apart from the properties listed above transverse waves has another properties called polarization.

REFLECTION
This is a property that is exhibited by all waves. In case of water waves generated in a ripple tank, if the waves were made to incident normally on a plane strip, the wave will be reflected back along their original course.
If the waves are incident at a particular angle, it will be observed that the angle of incidence is equal to the angle of reflection in line with the laws of reflection


LAWS OF REFLECTION
(a) The incident ray, the reflected ray and the normal, at point of incidence, all the three lies in the same plane.
(b) The angle of incidences is equal to the angle of reflection


REFRACTION
This is the change in the direction of waves as it passes from one medium to another. It is caused by the change in the speed of the wave.
When plane waves pass from deep to shallow water, their wavelength becomes shorter and thereby travels slowly. A change in the wavelength and speed produce a change in the direction of travel of waves when they cross the boundary. It is important to note that during refraction, the wavelength remains constant.


https://youtu.be/xobSuCkaiU8

Example
The wavelength and velocity of a set of plane waves traveling in a medium are 60cm and 320cm/s respectively. It meets a plane refracting surface at an angle of 600. Its velocity after refraction is 280cm/s, calculate the wavelength of the waves in the second medium and also the angle of refraction.
Solution
n = velocity in deep water
velocity in shallow water

n = wavelength in deep water
wavelength in shall water

Sin 600 = 320 = 60
Sin r 280 λ2
n = sin i = v1 = λ 1
sin r v2 λ 2
320 = 60
280 λ2
λ2 = 280 x 60
320
λ2 = 52.5cm
sin 600 = 320
sin r 280
sin r = 0.8660
1.143
Sin r = 0.7577
r = sin-1 (0.7577)

r = 49.260


DIFFRACTION
This is the spreading out of a wave on passing through a narrow shit. If waves are directed towards a large gap compared with the wavelength of the waves, slightly bent or beam of waves are formed on passing through the gap.
If the barriers are placed closer to leave a narrow gap waves forms a spherical wave fronts on passing through a narrow shit.


Diffraction occurs when the wavelength of the wave is longer than the width of the opening or the size of the obstacles.

INTERFERENCE
This is a phenomenon which occurs when two similar waves traveling in the same direction cross each other. If the waves are in phase or step so that they travel the same distance at equal time and the crest or trough of the two waves arrive simultaneously or one is a complete wavelength ahead of the other. The resulting wave will build up to twice the amplitude of the two waves, this is called constructive or additive interference.
If the crest of one wave arrives with the troughs of the waves, and vice-versa, the waves cancel each other out to give zero resultant, this is called destructive interference.


https://youtu.be/ITe6snlZBp8

Assignment
1.The ripple tank experiment can be used to demonstrate the following properties except
(a) diffraction (b) interference (c) polarization (d) refraction
2. which of the following are transverse waves?
I Ripples II sound waves in air
III light waves from the sun
(a) II only (b) I and II only (c) II and III only (d) I and III only
3. Which of the following wave characteristics can be used to distinguish a transverse wave from a longitudinal wave? (a) reflection (b) refraction (c) diffraction (d) polarization
4. the period of a wave is 0.02sec . calculate its wavelength if its speed 330m/s (a)6.6m (b) 5.0m (c) 4.0m (d) 3.3m
5. a wave travelling from water to glass suffers a change in its speed at the common boundary. Which of the following properties explains this observation (a) dispersion (b) refraction (c) interference (d) diffraction
THEORY
1. State four (04) characteristics of a progressive wave.
2. Explain with aid of diagrams the term “constructive interference” and “destructive interference” of waves.





SPECIFIC TOPIC: PROPERTY OF WAVES
OBJECTIVE: At the end of the lesson, the students should be able to:
Analyse the properties of waves

CONTENT: PROPERTIES OF WAVES

REFLECTION: This is the change in the direction of waves when they hit an obstacle. The type of waves formed depends on the type of they hit or meet. For instance, straight and parallel waves are set up from a plane metal strip standing upright in the water of a ripple tank. Similarly, when the plane metal is replaced by curved metal strips, the reflection rays form a spherical wave.


Reflection on a plain surface in a ripple tank


Reflection on a concave surface in a ripple tank


Reflection on a convex surface in a ripple tank

2. REFRACTION : This occurs between two media, when wave direction of propagation changes as it enters a different medium. When straight waves pass from deep to shallow water, their wavelength becomes shorter. During this process, the frequency remains the same, but the wave-length varies. It is clear from figure below that wavelength has changed from λ1 to λ2 .



Refraction in shallow and deep waters

NB: Waves travel more slowly in shallow water (same frequency shorter wavelength)
Therefore, the refractive index for water passing from deep to shallow water.
n = sin i/sin r = λ1/AB = λ1
λ2/AB λ2
Velocity in deep water = V1 = fλ1
Velocity in shallow water = V2 = fλ2
Thus , refraction index(n), V1/V2 = Velocity in deep water/velocity in shallow water

DIFFRACTION: This is a phenomenon whereby waves bend round obstacles. It is also the spreading of waves after passing through tiny opening, aperture, a hole or a slit. The smaller the width of the aperture, the smaller the wavelength, the greater will be the spreading of the waves. Similarly, the bigger the width of the aperture, the longer the wavelength, the smaller will be the spreading of the waves.

Example 1: The progressive wave is represented by the equation
Y = 4sin(100∏t - 50∏x)
Find the
(i) Amplitude
(ii) Frequency
(iii) Wavelength
(iv) Velocity
Solution
(i) Y = Asin(2∏ft - 2∏/λ)x
= A sin100∏t - 50∏x
Comparing coefficients, (i) A = 4m
(ii) 2∏f = 100∏
F = 100∏/2∏ = 50Hz
(iii) 2∏/λ = 50∏
λ= 2∏/50∏ = 2/50 = 0.04m (4cm)
(iv) V= fλ
= 50 x 0.04
= 2m/s
https://youtu.be/eW5VGGJuWtQ

EVALUATION: The lesson is evaluated as the students are asked to solve the following problems.
A source of the sound produces waves in air of wavelength 3.42m. If the speed of sound in air is 330m/s, what is the period of vibration.

ASSIGNMENT:
A transverse wave travels a distance of 80cm in 10s. If the distance between successive troughs of the wave is 4cm, calculate the frequency of the wave.

POLARISATION
This is an exclusive property of transverse waves only. It is the production of transverse vibration in only one plane. A transverse wave which vibrates in only one plane is said to be plane-polarized.
Polarized light can be produced by passing an ordinary light through a polarizer called polaroids or crystal of calcite, tourmaline or quartz. The arrangement of molecules within this polarizer will only permit the passage of light in a particular plane and then absorb light due to other vibration. Thus, when an unpolarized light is passed through a polarizer, the emergent light consists in only one plane.

Experiment evidence also shows that if an unpolarised light is incident on a polished glass glate at about 570, the reflected light is plane polarized. To show whether a light is polarized or not, the polarized is rotated while passing the light through it.
(a) If the incident light produce no change in intensity, the incident light is not polarized.
(b) If these is a little variation in intensity twice per revolution, the incident light is partially plane – polarized.
(c) If the incident light disappears twice per revolution, the light is plane polarized.
Application of polaroids The Polaroid is used in sunglasses to reduce the intensity of incident light and to eliminate reflected light glare.



https://youtu.be/DjnDX28l4xA

https://youtu.be/eW5VGGJuWtQ


CONTENT: POLARIZATION- PRODUCTION AND USES OF POLARISED WAVE.
The word polarization comes from Latin word ‘polus’ which means axis or direction. Therefore, if the vibrations of a transverse wave remains parallel to a fixed line in space, the wave is said to be linearly polarized. We thus define polarization of light as the direction of its electric field vector. For a plane polarized light , the electric field vector always remains in a given plane called the plane of polarization.It is the general agreement to speak of the polarization of electric field vector e and not magnetic field B because vision and other types of detection of e-m waves depend primarily on the sensitivity of the eye and the vast majority of other detection equipment, to the oscillation of E, rather than those of B.

PRODUCTION OF POLARIZED LIGHT
To obtain a polarised light we pass an ordinary (or un-polarised) light through a light polarizing filter.
Example of such filters or polarizers are crystals such as tourmaline, calcite and quartz. Another polarizer is a manufactured material known as Polaroid. This artificial crystalline material is made in the form of very thin films, which have the general appearance of the more common substance,cellophane and is made from small needle shaped crystals of an organic compound iodosulphate of quinine.
These polarizers, because of their internal molecular structures, have the property of allowing light vibrations in only a particular direction. They transmit only those vibrations of light in a particular plane and absorb light due to other vibrations. Therefore, when we pass ordinary light through these polarizers, the light emerging through them consists of vibrations in one transverse direction and is said to be plane- polarized.
Light is said to be polarized it its vibrations are in one plane only. This plane is called the plane of polarization .

USES AND PRACTICAL APPLICATIONS OF POLAROIDS.
The Polaroid finds practical applications particularly in areas where light glare is not desirable. Light glare is reflected, for example, from the window panes, the road ahead of a driver, on a sunny day, a polished table top or an open book. This reflected light is polarised, and the glaring can thus be eliminated by the use of a Polaroid.
The Polaroid is therefore used in a sunglasses to reduce the intensity of incident sunlight and to eliminate reflected light or glare.

EVALUATION:
Explain the difference between transverse and longitudinal waves. In which of these two can plane polarization occur?
Explain the uses of polaroids.

ASSIGNMENT:
How can you demonstrate a mechanical analogue of polarization?
Explain what is meant by the statement that a beam of light is plane – polarized
How can a plane- polarized light be produced and detected?



SPECIFIC TOPIC: Pressure Cooker. Cooling by Evaporation, Cooling and heating curves.
REFERENCE BOOK: SENIOR SECONDARY PHYSICS,BY OKEKE.
OBJECTIVE: At the end of the lesson, the students should be able to:
(i) Explain cooling by evaporation
(ii) Sketch the cooling and heating curves
CONTENT:
PRESSURE COOKER.COOLING BY EVAPORATION, COOLING AND HEATING CURVES
Newton’s law of cooling states that for a small difference of temperature between a body and the surroundings, the rate of gain or loss of heat by the body is proportional to the difference in temperature between the body and the surroundings.

This law holds only within small range of temperature. It is important to note that the rate of cooling of liquid depends on:
(i) Its temperature
(ii) The temperature of the enclosure,
(iii) The area of the exposed surface,
(iv) The nature and extent of the surface of the containing vessel.

FINDING THE SPECIFIC HEAT CAPACITY OF A LIQUID USING COOLING CURVE METHOD
The mass of a copper calorimeter, weighed when empty is recorded. It is then half filled with the liquid and heated to about 200c above room temperature.


A stopwatch and a thermometer are used in measuring the temperature of the liquid at regular intervals of about half a minute.
The liquid is continuously stirred during the period of cooling. The calorimeter and the contents is reweighed at the end of the experiment. A graph of temperature is plotted against time for the liquid, and this represents the cooling curve of the liquid.

https://youtu.be/ia0QATYc5jk

https://youtu.be/kBmU13D1sA0

EVALUATION:
Describe an experiment to obtain the cooling curve for naphthalene or paraffin wax when cooled from liquid to solid state..

ASSIGNMENT:
Describe an experiment to demonstrate the effect of impurity on the boiling point of water.



SPECIFIC TOPIC REVISION
OBJECTIVE: At the end of the lesson, the students should be able to:
Answer some questions on treated topics
(1) CONTENT: REVISIONS Define the following
(i) Sublimation
(ii) Evaporation
(iii) Boiling
1bi. Differentiate between boiling and evaporation
bii what is saturated temperature
biii. Explain with the aid of diagram the application of cooling effect of evaporation…..
2a. what is simple harmonic motion
b. A particle of mass 0.25kg vibrates with a period of 2.0s. If its greatest displacement is 0.4m, what is its maximum kinetic energy?
c. Differentiate between the specific latent heat of fusion and the specific latent heat of vaporization.