Lesson Notes By Weeks and Term - Senior Secondary 1

Term: 2^nd Term

Week: 1

Class: Senior Secondary School 1

Age: 15 years

Duration: 40 minutes of 5 periods each

Date:       

Subject:      Physics

Topic:-       Thermal expansions in solids

SPECIFIC OBJECTIVES: At the end of the lesson, pupils should be able to

1. Explain the meaning of thermal expansion
2. State the advantages and disadvantages of thermal expansion
3. State the types of thermal expansion
4. Solve exercises on thermal expansion

INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source

INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures

INSTRUCTIONAL PROCEDURES

PERIOD 1-2

PRESENTATION	TEACHER’S ACTIVITY	STUDENT’S ACTIVITY
STEP 1 INTRODUCTION	The teacher reviews the previous lesson on viscosity	Students pay attention
STEP 2 EXPLANATION	He states the advantages and disadvantages of thermal expansion	Students pay attention and participates
STEP 3 DEMONSTRATION	He states the types of thermal expansion and solves some exercises on it	Students pay attention and participate
STEP 4 NOTE TAKING	The teacher writes a summarized note on the board	The students copy the note in their books

 

NOTE

THERMAL EXPANSIONS IN SOLID

This is the increase in size or volume of a substance due to heat applied.

 

Disadvantages of Expansion

1. It deforms the bridge structure.
2. It can make thick glass to break.
3. It causes rail way line to buckle.
4. It affects oscillation of the pendulum clock and the balance wheel of watc

Advantages of Expansion

1. It can be applied in riveting two metal plates.
2. It can be applied in construction of automatic fire alarms.
3. It can be applied in bimetallic strips in electric iron.

 

Thermal Expansion Formula

Thermal expansion can result in linear expansion or area expansion or volumetric expansion. The corresponding formula is given below along with relevant terms. Thus, the formula used to represent the thermal expansion in a body is

Linear Expansion:

Δl      =     αl.ΔT

l

where,

• l is the initial length of the solid
• Δl is the change in length.
• α_l length expansion coefficient
• ΔT is the temperature difference

 

Area Expansion:

ΔA = αA.ΔT

  A

where,

• A is the initial area of the solid
• ΔA is the change in the area.
• α_A Area expansion coefficient
• ΔT is the temperature difference

 

Volumetric Expansion:

ΔV    =αV.ΔT

  V

where,

• V is the initial volume of the solid
• ΔV is the change in volume.
• α_V volume expansion coefficient
• ΔT is the temperature difference

Here, the alpha represents the coefficient of linear expansion and it is a characteristic of the substance the body is made of. For example, for the same temperature rise, copper expands almost five times more than glass. Generally, metals expand more and have higher values of alpha compared to gases and liquids

 

Question 1.

What is meant by the statement, the linear expansivity of copper is 0.000017/k?

 

Solution:

It means that the increase in length per unit length per degree rise in temperature of copper is 0.000017m.

 

Question 2.

A brass is 2 meters long at a certain temperature. What is its length for a temperature rise of 100k, if expansivity of brass is  1.8 x 10^-5/k?

 

Solution

α =Increase in length original length ×temperature rise= L2 – L1 L1(θ2–θ1)

L2–L1= αL1(θ2–θ1)L2=L1{α(θ2–θ1)+1}L2

=2{1.8×10−5(100)+1}L2=2{0.0018+1}L2

=0.0036+2

=2.0036m

 

Question 3.

A metal of length 15.01m is heated until its temperature rises to 60⁰C. If its new length is 15.05m, calculate its linear expansivity.

Solution:

L₁ = 15.01m, L₂ = 15.05, θ₂ – θ₁ = 60^o, L₂ – L₁ = 0.04

α=Increase in lengthoriginal length×temperature rise=L2–L1L1(θ2–θ1)α

=15.05–15.0115.01×60o

=0.04900.6

=0.000044

=4.4×10−5/k

 

Relationship between Linear Expansivity and Area Expansivity: β = 2α

 

Question 1: A metal cube of cross sectional area 3.45m² at 0⁰C is heated at a temperature rise of  70K, when the final length of the cube is 3m. Find the:

(i) coefficient of superficial expansivity.

(ii) coefficient of linear expansivity.

 

Solution

(i) β=Increase in Area original area×temperature rise=A2–A1A1(θ2–θ1)A2=L2

=3×3=9m2θ2–θ1

=70kA1

=3.45m2

β=9–3.453.45×70

=5.55241.5

=0.023/k

=2.3×10−2/k

 

(ii) β=2αα

=β2α

=2.3×10−22

=1.15×10−2K−1

 

Relationship between Linear Expansivity and Cubic Expansivity: γ = 3α

Question 2: The increase in the volume of 10cm³ of mercury when the temperature rises by 100⁰C is 0.182cm³. What is cubic expansivity of mercury.

 

Solution

γ=Increase in Volumeoriginal volume×temperature rise=V2–V1V1(θ2–θ1)

V₂ – V₁ = Increase in volume = 0.182cm²

θ₂ – θ₁ = Temperature rise = 100^oC

V₁ = original volume = 10cm²

γ=0.18210×100

=0.1821000

=0.00082/k

=1.82×10−4K−1

EVALUATION:   1. Define thermal expansion

2. State two advantages and disadvantages of thermal expansion
3. What is meant by the statement that the linear expansivity of copper is 0.000017/k.
4. Steel bars each of length 3m at 29⁰c are to be used for constructing a rail line. If the linear expansivity of steel is 1.0 x 10^-5/k. Calculate the safety gap that must be kept  between successive bars, if the highest temperature expected is 40⁰ 

5. The linear expansivity of a metal is 0.000019 per k. What will the area of 400mm²be if its temperature is raised by 10⁰?

CLASSWORK: As in evaluation

CONCLUSION: The teacher commends the students positively