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
INSTRUCTIONAL TECHNIQUES: Identification, explanation, questions and answers, demonstration, videos from source
INSTRUCTIONAL MATERIALS: Videos, loud speaker, textbook, pictures
INSTRUCTIONAL PROCEDURES
PERIOD 12
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
Advantages of Expansion
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,
Area Expansion:
ΔA = αA.ΔT
A
where,
Volumetric Expansion:
ΔV =αV.ΔT
V
where,
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 60^{0}C. If its new length is 15.05m, calculate its linear expansivity.
Solution:
L_{1} = 15.01m, L_{2} = 15.05, θ_{2} – θ_{1} = 60^{o}, L_{2} – L_{1} = 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 3.45m^{2} at 0^{0}C 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 10cm^{3} of mercury when the temperature rises by 100^{0}C is 0.182cm^{3}. What is cubic expansivity of mercury.
Solution
γ=Increase in Volumeoriginal volume×temperature rise=V2–V1V1(θ2–θ1)
V_{2} – V_{1} = Increase in volume = 0.182cm^{2}
θ_{2} – θ_{1} = Temperature rise = 100^{o}C
V_{1} = original volume = 10cm^{2}
γ=0.18210×100
=0.1821000
=0.00082/k
=1.82×10−4K−1
EVALUATION: 1. Define thermal expansion
CLASSWORK: As in evaluation
CONCLUSION: The teacher commends the students positively