Course Content
Properties of Matter
OBJECTIVES By the end of this topic , the trainee should be able to: 1.Explain the three states of matter. 2.Explain phenomenon of surface tension. 3.Explain gas laws.
0/3
Introduction
States Of Matter
Gas Laws
Measurement
OBJECTIVES By the end of this topic , the trainee should be able to: 1.State the basic and derived quantities of measurements 2.Describe quantities of measurements 3.State the SI units of quantities of measurements and their sub-multipliers 4.Convert units from one to another
0/4
Introduction
Measurement of Length and Distance
  Measurement of Mass and weight
 Measurement of Volume and Density
Pressure
OBJECTIVES By the end of this topic , the trainee should be able to: 1.Define pressure 2.Determine pressure in solids and fluids 3.Explain the type of instruments used in measurement of pressure of gases 4.Explain transmission of presure in liquids
0/2
Introduction
Measuring Pressure
Force
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define force and its SI units 2.Explain types of force 3.Measure force 4.Work out calculations involving force 5.Explain Hooke’s law
0/3
Force
Friction Force
Centrifugal and Centripetal Force
Hooke’s Law
Hooke's Law  is a law named after 17th century British physicist Robert Hooke, who sought to demonstrate the relationship between the forces applied to a spring and its elasticity.
0/1
Hooke’s Law
Moments
OBJECTIVES By the end of this topic, the trainee should be able to: 1. Define moments of a force 2.Explain principles of moments 3.Explain statics , dynamics and kinematics
0/1
Moments
Velocity and Speed
OBJECTIVES By the end of this topic , the trainee should be able to : 1.Define speed and velocity 2.Explain types of motion 3.Determine Linear Determine Circular Motion
0/1
Velocity and Speed
Work and Power
OBJECTIVES By the end of this topic , the trainee should be able to: 1.Define work and power 2.Work out calculations involving work and power
0/1
Work and Power
Energy
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define energy 2.Explain forms of energy 3.Explain law of conservation of energy
0/1
Energy
Motion
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define motion 2.Explain causes of motion 3.Explain types of motion 4.Describe linear and circular motion
0/3
Introduction
Linear Motion 
Circular Motion
Newton’s Laws of Motion
OBJECTIVES By the end of this topic, the trainee should be able to: 1.State Newton’s laws of motion 2.Explain relationships between mass and inertia 3.Explain significance of Newton’s laws of motion
0/1
Newton’s Laws of Motion
Heat and Heat Transfer
OBJECTIVES By the end of this topic, the trainee should be able to 1.Define temperature and heat 2.Describe instruments used to measure temperature 3.Explain modes of heat transfer 4.Explain factors that influence heat transfer 5.Explain heat capacity, specific heat capacity and specific latent heat of substances 6. Work out calorimetric calculations
0/5
Introduction
Instruments Used For Measuring Temperature  
Heat Transfer
Measuring The Quantity Of Heat
Heat and Changes of State
Light
OBJECTIVES By the end of this topic , the trainee should be able to: 1.Define light and its properties 2.Explain types of mirrors and lenses 3.Explain reflection and refraction of light 4. Explain the lens formulae
0/10
Introduction
Mirrors
Reflection 
Mirror Equation
Image Formation By Convex Mirror
Refraction Of Light
 Refractive index
Thin Lenses
The lens formula
 Uses of lenses on optical devices
WAVES
OBJECTIVES By the end of this topic , the trainee should be able to : 1. Define waves 2.Explain types of waves 3.Explain characteristics and properties of waves 4.Explain rectilinear propagation of waves 5.Describe refraction of waves 6.Describe sound and sound waves
0/6
Introduction
Electromagnetic Wave
Electromagnetic Wave
Properties of Waves
Rectilinear Propagation Of Waves
Sound and Sound Waves
Electrostatics
OBJECTIVES By the end of this topic , the trainee should be able to: 1.Define electrostatics 2. Describe the working of electroscopes 3.Describe different methods of charging a body 4.Explain electric fields 5.Describe charge distribution on conductors’ surface 6. Describe capacitors and capacitance
0/5
Introduction
Gold Leaf Electroscope
Electric Fields
Charge Distribution on Conductors’ Surface
Capacitors And Capacitance
Magnetism and Electromagnetism
OBJECTIVES By the end of this topic, the trainee should be able to: 1.Define magnetism 2.Identify magnetic and non- magnetic materials 3.Explain properties of magnets 4.Describe magnetic field patterns 5.Describe making magnets 6.Explain domain theory of magnets 7.Describe magnetic effect of an electric current. 8. Describe electromagnets and electromagnetism 9.Describe electromagnetic induction
0/6
Introduction
Magnetic Field Patterns.
Making Magnets
Magnetic Effect of An Electric Current.
Electromagnetism 
Electromagnetic Induction
Current Electricity
CURRENT ELECTRICITY OBJECTIVES By the end of this topic, the trainer should be able to: 1. Define current and electricity 2.Electric potential difference 3.Explain use of a meters, voltmeters and resistors 4.Explain ohm’s law 5. Explain Electromotive force and internal resistance 6.Describe production of electric current 7.Describe simple cells, Leclanche’ cell, dry cells and lea-acid accumulators 8.Describe mains electricity 9.Describe domestic wiring
0/6
Introduction
Electric Potential Difference And Electric Current
Resistors and Resistance
Production Of Electric Current:
Heating Effect Of An Electric Current
Mains Electricity
Electromagnetic Spectrum
OBJECTIVES By the end of this topic , the trainee should be able to: 1.Define magnetic spectrum 2. Explain properties of electromagnetic waves 3. Describe applications of electromagnetic radiation
0/2
Introduction
Properties and Application of Electromagnetic Waves
Radioactivity
OBJECTIVES By the end of this topic, the trainee should be able to: 1. Define radioactivity 2.Classify radiations 3.Explain the properties of radiation materials 4.Describe nuclear fission and nuclear fusion 5.Describe methods for detecting nuclear radiation 6.Describe activity and half-life of elements 7. Discuss hazards of radioactivity and their precautions 8. Outline the applications of radioactivity
0/4
Introduction
Nuclear Fission
Activity and Half-Life Of Elements
Applications Of Radioactivity
Cathode and X-Rays
OBJECTIVES By the end of this topic, the trainee should be able to: 1. Define cathode and X- rays 2.Explain production of cathode and X- ray 3.Explain properties of cathode and X- ray 4.Describe cathode ray oscilloscope (CRO) 5. Describe the applications of Cathode and X-rays
0/2
Cathode Rays
X-Rays
Physics Techniques For Science Laboratory Technicians
About Lesson

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Heat and Changes of State

The discussion above and the accompanying equation (Q = m•C•∆T) relates the heat gained or lost by an object to the resulting temperature changes of that object. As we have learned, sometimes heat is gained or lost but there is no temperature change. This is the case when the substance is undergoing a state change. So now we must investigate the mathematics related to changes in state and the quantity of heat.

To begin the discussion, let’s consider the various state changes that could be observed for a sample of matter. The table below lists several state changes and identifies the name commonly associated with each process.

Process

Change of State

Melting

Solid to Liquid

Freezing

Liquid to Solid

Vaporization

Liquid to Gas

Condensation

Gas to Liquid

Sublimation

Solid to Gas

Deposition

Gas to Solid

In the case of melting, boiling and sublimation, energy would have to be added to the sample of matter in order to cause the change of state. Such state changes are referred to as being endothermic. Freezing, condensation and deposition are exothermic; energy is released by the sample of matter when these state changes occur. So one might notice that a sample of ice (solid water) undergoes melting when it is placed on or near a burner.

 When heat is transferred from the burner to the sample of ice; energy is gained by the ice causing the change of state. But how much energy would be required to cause such a change of state?

The amount of energy required to change the state of a sample of matter depends on three things.

  1. It depends upon what the substance is,
  2. on how much substance is undergoing the state change, and
  3. upon what state change that is occurring.

For instance, it requires a different amount of energy to melt ice (solid water) compared to melting iron. And it requires a different amount of energy to melt ice (solid water) as it does to vaporize the same amount of liquid water. And finally, it requires a different amount of energy to melt 10.0 grams of ice compared to melting 100.0 grams of ice.

The substance, the process and the amount of substance are the three variables that affect the amount of energy required to cause a specific change in state.

The values for the specific heat of fusion and the specific heat of vaporization are reported on a per amount basis. For instance, the specific heat of fusion of water is 333 J/gram. It takes 333 J of energy to melt 1.0 gram of ice. It takes 10 times as much energy – 3330 J – to melt 10.0 grams of ice. Reasoning in this manner leads to the following formulae relating the quantity of heat to the mass of the substance and the heat of fusion and vaporization.

For melting and freezing: Q = m•ΔHfusion
For vaporization and condensation: Q = m•ΔHvaporization

Where

 Q represents the quantity of energy gained or released during the process, 

m represents the mass of the sample, 

ΔHfusion represents the specific heat of fusion (on a per gram basis) and

 ΔHvaporization represents the specific heat of vaporization (on a per gram basis).

Similar to the discussion regarding Q = m•C•ΔT, the values of Q can be either positive or negative. Values of Q are positive for the melting and vaporization process; this is consistent with the fact that the sample of matter must gain energy in order to melt or vaporize. Values of Q are negative for the freezing and condensation process; this is consistent with the fact that the sample of matter must lose energy in order to freeze or condense.

As an illustration of how these equations can be used, consider the following two example problems.

Example Problem 3
Elise places 48.2 grams of ice in her beverage. What quantity of energy would be absorbed by the ice (and released by the beverage) during the melting process? The heat of fusion of water is 333 J/g.

The equation relating the mass (48.2 grams), the heat of fusion (333 J/g), and the quantity of energy (Q) is Q = m•ΔHfusion. Substitution of known values into the equation leads to the answer.

Q = m•ΔHfusion = (48.2 g)•(333 J/g)
Q = 16050.6 J
Q = 1.61 x 104 J = 16.1 kJ (rounded to three significant digits)

Example Problem 3 involves a rather straightforward, plug-and-chug type calculation. Now we will try Example Problem 4, which will require a significant deeper level of analysis.

Example Problem 4
What is the minimum amount of liquid water at 26.5 degrees that would be required to completely melt 50.0 grams of ice? The specific heat capacity of liquid water is 4.18 J/g/°C and the specific heat of fusion of ice is 333 J/g.

In this problem, the ice is melting and the liquid water is cooling down. Energy is being transferred from the liquid to the solid. To melt the solid ice, 333 J of energy must be transferred for every gram of ice. This transfer of energy from the liquid water to the ice will cool the liquid down. But the liquid can only cool as low as 0°C – the freezing point of the water. At this temperature the liquid will begin to solidify (freeze) and the ice will not completely melt.

We know the following about the ice and the liquid water:

Given Info about Ice:

m = 50.0 g
ΔHfusion = 333 J/g

Given Info about Liquid Water:

C = 4.18 J/g/°C
Tinitial = 26.5°C
Tfinal = 0.0°C
ΔT = -26.5°C (Tfinal – Tinitial )

The energy gained by the ice is equal to the energy lost from the water.

Qice = -Qliquid water

The – sign indicates that the one object gains energy and the other object loses energy. We can calculate the left side of the above equation as follows:

Qice = m•ΔHfusion = (50.0 g)•(333 J/g)
Qice = 16650 J

Now we can set the right side of the equation equal to m•C•ΔT and begin to substitute in known values of C and ΔT in order to solve for the mass of the liquid water. The solution is:

16650 J = -Qliquid water
16650 J = -mliquid water•Cliquid water•ΔTliquid water
16650 J = -mliquid water•(4.18 J/g/°C)•(-26.5°C)
16650 J = -mliquid water•(-110.77 J/°C)
mliquid water = -(16650 J)/(-110.77 J/°C)
mliquid water = 150.311 g
mliquid water = 1.50×102 g (rounded to three significant digits)

Latent Heat:

Latent heat is the quantity of heat energy that is required to change a substance from one state to another at constant temperature and pressure. The unit of latent heat is Joule (J), and the symbol used for latent heat is Q. The quantity of heat that is required to change a substance from one state to another is directly proportional to the mass of the substance.

Mathematically:

  • Quantity of heat energy ∞ mass of substance
  •                                         Q ∞ m
  • Introducing the constant of proportionality:   Q = L X m
  • L is the constant of proportionality. It is called  the specific latent heat of the substance.
  • M is the mass of the substance measured in kilogram.
  • Q is the quantity of heat measured in Joule.

Factors that determine the quantity of heat required to change the state of a substance:

The following factors determine the quantity of heat energy that is required to change a substance from one state to another:

  1. The nature of the substance.
  2. The mass of the substance.
  3. The quantity of energy supplied.
  4. The volume of the substance
  5. The area of the substance
  6. The state of the substance as  the time the heat energy is supplied.

Specific Latent Heat of A Substance:

Specific latent heat of a substance is the quantity of heat energy that is required to change 1kg mass of the substance from one state to another at constant temperature.

Specific Latent Heat of Fusion of Ice:

Specific latent heat of fusion of ice is the quantity of heat energy that is required to change 1kg mass of ice block into liquid at constant temperature Or melting point of the ice block. It is a scalar quantity.

  • Specific Latent Heat Of Fusion of Ice L ice = 336 J/g
  • Formula Of Specific Latent Heat Of Fusion Of Ice:
  • Quantity of Heat energy Q = mass of ice block * specific latent heat of ice block         Q = mXLice
  •               Specific latent beat of ice  Lice = Q/m

Examples:

  1. Calculate the quantity of heat energy that is required to change 25.6 kg mass of ice to liquid at constant temperature.

Solution:

  • Mass of ice block m = 25.6 kg = 25.6 * 1000 = 25600g, Lice = 336J/g,                    Q = ?
  • Formula:     Q = m X Lice
  • Substitution:  Q = 25600 X 336
  • Q = 8601600 Joules
  1. The quantity of heat that is required to change X g of ice to liquid is 235 J. If the Lice is 336J/g, calculate the mass of the object.

Solution:

  • Mass = ?, Q = 235 J, Lice = 336J/g
  • Formula:     Q = m*Lice
  • Substitution:       235 = m X 336
  • Make m the subject:     m = 235/336
  •   M = 0.699 g

 Effect Of Expansion And Contraction On Fusion:

ü Glass bottle of water crack on freezing because water expands on freezing and because of unequal expansion of the bottle.

ü When water freezes into ice, it expands, become less dense and float on water with nine – tenth (9/10 ) of its volume submerged in water. Ice contracts on melting.

 Effect Of Pressure on Freezing Point of a Substance:

  1. Increase in pressure lower the melting point of ice or the freezing point of water. This is called regelation of ice.
  2. An increase in pressure lowers the freezing point of any liquid which expands on solidifying. E.g water.

III. Increase in pressure increases the freezing point of any substance which contract on solidifying. Eg paraffin-wax.

 Effect Of Impurities On Freezing Point Of Substance

The presence of impurities in a substance lower the melting point of pure substance

Specific Latent Heat Of Vaporization

Specific latent heat of vaporization of steam is the quantity of heat energy that is required to change 1kg mass of a liquid  into steam / gas at constant temperature ( boiling point) and pressure. It is a scalar quantity. The unit of specific latent heat of vaporization of steam is J/Kg or J/g.

Relationship Between Specific Latent Heat Of Fusion Of Ice and Specific Latent Heat Of Vaporization Of Steam

Specific latent heat of fusion of ice Lice  = 1/7 of specific latent heat of vaporization of steam

                                     L ice  =     1/7   X Lsteam

 Example 1:

  1. If the latent heat of fusion of ice is 336J/g, what is e specific latent heat vaporization of steam?
  • Specific latent heat of fusion of ice Lice = 336j/g.
  •                                                           Lsteam  = ?

specific latent heavy of fusion of ice Lice = 1 /7  Specific latent heat of steam

  •                                     Lice = 1/7 of Lsteam
  • Substitution:            336 = 1/7 of Lsteam
  • Make Lsteam the subject:    Lsteam = 7*336  
  • Lsteam = 2352J/g
  1. What is the specific latent heat of fusion of ice if that of steam is 2352J/g?

Specific latent heat of vaporization of steam Lsteam = 2352J/g,

  •                           Lice = ?
  • Formula :       Lice  = 1/7 of L steam
  • Substitution:     L ice = 1/7 * 2352.  

 L ice = 336 J/g

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