## 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 (^{0}C^{-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/L_{0}Δθ

But since e = extension = L – L_{0}

Hence,

**α = (L – L _{0} )/L_{0}Δθ**

We can also write the above formula as

**α = (L – L _{0} )/L_{0}(T_{2} – T_{1})**

**or**

**α = e/L**

_{0}(T_{2}– T_{1})

and **Δθ** = **T _{2} – T_{1}**

**The formula for calculating the increase in length is L = L _{0}(1 + αΔθ). **

Where

α = Linear expansivity

L = change or increase the length

Also, L_{0} = original length

e = extension

Δθ = Change in temperature = T_{2} – T_{1}

## 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 | SiO_{2} | 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 20^{0}C to 70^{0}C. Calculate its linear expansivity.

**Data**

Original length, L_{0} = 50 m

The increase in length, L = 50.5 m

Initial temperature, T_{1} = 20^{0}C

Final temperature, T_{2} = 70^{0}C

**Unknown value to find**

Linear expansivity, α =?

**Formula**

α = (L – L_{0} )/L_{0}(T_{2} – T_{1})

**Solution**

α = (L – L_{0} )/L_{0}(T_{2} – T_{1}) = (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 20^{0}C to 70^{0}C. If it undergoes a change in length of 2 centimeters, find the linear expansivity of the wire.

**Data**

Original length, L_{0} = 5 m

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

Initial temperature, T_{1} = 20^{0}C

Final temperature, T_{2} = 70^{0}C

**Unknown value to find**

Linear expansivity, α =?

**Formula**

α = e/L_{0}(T_{2} – T_{1})

**Solution**

α = e/L_{0}(T_{2} – T_{1}) = 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 10^{0}C to 110^{0}C, its length becomes 15.03 meters. What is the linear expansivity of the metal?

**Data**

Original length, L_{0} = 15 m

The increase in length, L = 15.03 m

Initial temperature, T_{1} = 10^{0}C

Final temperature, T_{2} = 110^{0}C

**Unknown value to find**

Linear expansivity, α =?

**Formula**

α = (L – L_{0} )/L_{0}(T_{2} – T_{1})

**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 20^{0}C. By how much will it expand when the temperature increases to 30^{0}C? [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, L_{0} = 50 m

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

Initial temperature, T_{1} = 20^{0}C

Final temperature, T_{2} = 30^{0}C

**Unknown value to find**

The increase in length, L =?

**The formula** **we need to apply**

L = L_{0}(1 + αΔθ)

**Solution**

L = L_{0}(1 + αΔθ) = L_{0}[1 + α (**Δθ** = **T _{2} – T_{1}**)] = 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 12^{0}C to 32^{0}C. 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, L_{0} = 500 m

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

Initial temperature, T_{1} = 12^{0}C

Final temperature, T_{2} = 32^{0}C

**Unknown value to find**

The increase in length, L =?

**The formula** **to solve the problem is**

L = L_{0}(1 + αΔθ)

**Solution**

L = L_{0}(1 + αΔθ) = L_{0} [1 + α (T_{2} – T_{1})] = 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:**