# Lesson Notes By Weeks and Term - Senior Secondary 1

Thermal expansions in solids

Term: 2nd 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
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.

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

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.
• αl 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.
• αA 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.
• αV 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 = 60o, 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 = 100oC

V1 = original volume = 10cm2

γ=0.18210×100

=0.1821000

=0.00082/k

=1.82×10−4K−1

EVALUATION:   1. Define thermal expansion