Linear Expansivity: Definition and Calculations

Introduction

In this post, you will learn about linear expansivity: definition and calculations. I will also acquaint you with the symbols, and units of linear expansivity.

When heat is applied to metal, the change in temperature would cause the object to expand. Different metals have different expansion rates when the same quantity of heat is applied to them. Some of these metals tend to expand more the others at the same temperature.

Linear expansivity is also known as the coefficient of linear expansion.

The coefficient of linear expansion is the reason these metals have different rates of expansion at the same temperature.

Definition: Linear expansivity of a substance can be defined as the fractional increase in length per unit length per degree rise in temperature.

The symbol for linear expansivity is α.

Also, the unit of linear expansivity is per degree celsius (⁰C^-1) or per kelvin (K^-1).

The Formula For Calculating Linear Expansivity and Increase in Length (Expansion)

The formula for calculating linear expansivity is α = e/L₀Δθ

But since e = extension = L – L₀

Hence,

α = (L – L₀ )/L₀Δθ

We can also write the above formula as

α = (L – L₀ )/L₀(T₂ – T₁) or α = e/L₀(T₂ – T₁)

and Δθ = T₂ – T₁

The formula for calculating the increase in length is L = L₀(1 + αΔθ).

Where

α = Linear expansivity

L = change or increase the length

Also, L₀ = original length

e = extension

Δθ = Change in temperature = T₂ – T₁

Table for Substances and their Linear Expansivity

Here is a table for materials, their symbols, and linear expansivity:

The material	Symbol	Linear expansivity	Unit
Aluminium	Al	0.000023	k-1
Iron	Fe	0.000012	k-1
Copper	Cu	0.000017	k-1
Brass		0.000018	k-1
Lead	Pb	0.000029	k-1
Platinium	Pt	0.000009	k-1
Zinc	Zn	0.000030	k-1
Alloy (Invar)		0.000001	k-1
Glass		0.0000085	k-1
Silica	SiO2	0.0000004	k-1

Linear Expansivity Calculations

Here are a few solved problems to help you understand how to calculate linear expansivity:

Question 1:

An invar of length 50 meters increases to 50.5 meters when heated from 20⁰C to 70⁰C. Calculate its linear expansivity.

Data

Original length, L₀ = 50 m

The increase in length, L = 50.5 m

Initial temperature, T₁ = 20⁰C

Final temperature, T₂ = 70⁰C

Unknown value to find

Linear expansivity, α =?

Formula

α = (L – L₀ )/L₀(T₂ – T₁)

Solution

α = (L – L₀ )/L₀(T₂ – T₁) = (50.5 – 50)/50(70-20) = 0.5/(50 x 50) = 0.5/2,500 = 0.0002 K^-1

Which implies that

α = 2 x 10^-4 K^-1

Therefore, the linear expansivity of the invar is 2 x 10^-4 per kelvin.

Question 2:

A wire of length 5 meters is heated from a temperature of 20⁰C to 70⁰C. If it undergoes a change in length of 2 centimeters, find the linear expansivity of the wire.

Data

Original length, L₀ = 5 m

Change in length = extension = 2 cm = (2/100) m = 0.02 m

Initial temperature, T₁ = 20⁰C

Final temperature, T₂ = 70⁰C

Unknown value to find

Linear expansivity, α =?

Formula

α = e/L₀(T₂ – T₁)

Solution

α = e/L₀(T₂ – T₁) = 0.02/5(70-20) = 0.02/(5 x 50) = 0.02/250 = 0.00008 K^-1

Which implies that

α = 8 x 10^-5 K^-1

Therefore, the linear expansivity of the invar is 8 x 10^-5 per kelvin.

Question 3:

When 15 meters of a metallic rod is heated from 10⁰C to 110⁰C, its length becomes 15.03 meters. What is the linear expansivity of the metal?

Data

Original length, L₀ = 15 m

The increase in length, L = 15.03 m

Initial temperature, T₁ = 10⁰C

Final temperature, T₂ = 110⁰C

Unknown value to find

Linear expansivity, α =?

Formula

α = (L – L₀ )/L₀(T₂ – T₁)

Solution

α = (15.03 – 15)/15(110 – 10) = 0.03/15(100) = 0.03/150 = 0.0002 K^-1

Which implies that

α = 2 x 10^-4 K^-1

Therefore, the linear expansivity of the metal rod is 2 x 10^-4 per kelvin.

Question 4:

The length of copper is 50 meters when it’s 20⁰C. By how much will it expand when the temperature increases to 30⁰C? [Take the linear expansivity of a copper to be 1.7 x 10^-5 K^-1]

Data: Extract your data from the above question.

Original length, L₀ = 50 m

Linear expansivity, α = 1.7 x 10^-5 K^-1

Initial temperature, T₁ = 20⁰C

Final temperature, T₂ = 30⁰C

Unknown value to find

The increase in length, L =?

The formula we need to apply

L = L₀(1 + αΔθ)

Solution

L = L₀(1 + αΔθ) = L₀[1 + α (Δθ = T₂ – T₁)] = 50 [1 + 1.7 x 10^-5 (30 – 20 )]

And we will now have

L = 50 [ 1 + 0.000017 x 10 ]

Which will become

L = 50 [ 1 + 0.00017]

After adding 1 to 0.00017 you will then have

Increase in length, L = 50 x 1.00017 = 50.0085 m

Thus,

L = 50.0085 m

Therefore, the increase in length of the copper rod is 50.0085 meters. Thus, the rod added 0.0085 of length to its original size.

Question 5:

Compute the increase in length of 500 meters of a copper wire, when its temperature changes from 12⁰C to 32⁰C. The linear expansivity of a copper wire is 17 x 10^-5 per degree celsius.

Data: Always extract your data from the question before you start solving the problem.

Original length, L₀ = 500 m

Linear expansivity, α = 17 x 10^{-5 0}C^-1

Initial temperature, T₁ = 12⁰C

Final temperature, T₂ = 32⁰C

Unknown value to find

The increase in length, L =?

The formula to solve the problem is

L = L₀(1 + αΔθ)

Solution

L = L₀(1 + αΔθ) = L₀ [1 + α (T₂ – T₁)] = 500 [1 + 17 x 10^-5 (32 – 12)]

The above expression will give us

L = 500 [ 1 + 0.00017 x 20 ]

We will then obtain

L = 50 [ 1 + 0.0034]

we will then add 1 to 0.0034 to get

Increase in length, L = 500 x 1.0034 = 50.0034 m

Therefore,

L = 50.0034 m

Therefore, the increase in length of the copper wire is 50.0034 meters. Thus, the rod added 0.0034 of length to its original size.

I hope you are now familiar with linear expansivity: definition and calculations. Drop a comment if there is anything you would like me to explain further.

You may also like to read:

How to Derive the Formula For Increase in Volume

Also How to Calculate Cubic Expansivity with Examples

How to Calculate the Relative Density of a Liquid

Sources:

Wikipedia