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# Linear Expansivity: Definition and Calculations

## What is Linear Expansivity?

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

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).

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.

## 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:

## 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 0C-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.

### Question 6

If an iron rail of 8 meters long are laid close up end to end when the temperature is 300C. What gap will be provided between consecutive rails when the temperature rises to 600C?

(Take linear expansivity of iron = 1.2 x 10-5 K-1)

#### Solution

Linear expansivity of iron, α = 1.2 x 10-5 K-1

Change in Temperature, Δθ = (60 K – 30 K) = 30 K

Original length of rail, L1 = 8 m

The increase in length due to an increase in temperature, = αL1Δθ

and by inserting our data into the above formula, we will have

The increase in length due to increase in temperature, = 1.2 x 10-5 x 8 x 30 = 0.003 m = 0.3cm

Hence, the gap that will be provided between consecutive rails is 0.3 centimeters

### Question 7

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-1. Calculate the safety gap that must be left between successive bars if the highest temperature expected is 410C.

The final answer to the above question is 3.6 x 10-4 m

Explanation

We will make ΔL subject of the formula from the equation α = ΔL /L0Δθ to obtain

ΔL = αL0Δθ

where ΔL = >

α = 1.0 x 10-5 K-1

L0 = 3m

Δθ = 2 – θ1) = (41 – 29) = 120C

Hence by inserting our data into the main formula ΔL = αL0Δθ, we will have

ΔL = αL0Δθ = 1.0 x 10-5 x 3 x 12 = 0.00036 m = 3.6 x 10-4 m

Therefore, the safety gap that must be left is 3.6 x 10-4 m.

### Question 8

A metal rod of length 50 centimeters is heated from 400C to 800C. If the linear expansivity of the material is α, calculate the increase in length of the rod (in meters) in terms of α.

The final answer to the above question is 20α

Explanation

We will use the formula ΔL = αL02 – θ1) to solve the problem

Therefore, since

L0 = 50 cm = 0.5 m

θ1 = 400C

θ2 = 800C

Thus,

ΔL = αL02 – θ1) = α x 0.5 x (80 – 40) = α x 0.5 x 40 = 20α

Therefore, the increase in length of the rod is 20α

### Question 9

A metal rod of length 100 cm, is heated through 1000C. Calculate the change in length of the rod. (Linear expansivity of the material of the rod is 3 x 10-5 K-1).

The final answer to the question is 0.3 cm

Explanation

We will apply the formula ΔL = αL0Δθ to find the change in length

ΔL = αL0Δθ = 3.0 x 10-5 x 100 x 100 = 0.3 cm

### Question 10

A bridge made of steel is 600 m long. What is the daily variation in its length if the night-time and day-time temperature are 100C and 350C respectively. The linear expansivity of steel is 0.0000120C-1.

The final answer to the above question is 18 cm

Explanation

We will also use the same formula in question 9 (ΔL = αL0Δθ ) to solve this problem

ΔL = αL0Δθ = 0.000012 x 600 x 25 = 0.18 m = 18 cm

### Question 11

A metal rod of length 40.00 cm at 200C is heated to a temperature of 450C. If the new length of the rod is 40.05 cm. Calculate its linear expansivity.

The final answer to this question is 5 x 10-5 K-1

Explanation

The formula α = (L – L0 )/L02 – θ1) will help us to find the linear expansivity of the metal rod.

Data

L = 40.05 cm

L0 = 40.00 cm

θ1 = 200C

θ2 = 450C

Thus, by putting our data into the formula

α = (L – L0 )/L02 – θ1) = (40.05 – 40.00) / 40.00 (45 – 20)

The above expression will become

α = 0.05 / (40 x 25) = 0.00005 = 5 x 10-5 K-1

Therefore, the linear expansivity of the metal rod is 5 x 10-5 per kelvin

### Question 12

A brass rod is 2 m long at a certain temeperature. What is the length for a temperature rise of 100K, if the expansivity of brass is 18 x 10-6 K-1

The final answer to the above question is 2.0036 meters

Explanation

We will make L subject of the formula from the formula α = (L – L0 )/L0Δθ

Thus, by inserting our numerical information from the question into the formula, we will get

18 x 10-6 = (L – 2)/ 2 x 100

Will give us

L – 2 = 18 x 10-6 x 200

The above expression will now become

L – 2 = 36 x 10-4

By collecting like terms, we will now have

L = 2 + 36 x 10-4 = 2.0036 m

Therefore, the length for a temperature rise of 100k will be 2.0036 meters.

### Question 13

The linear expansivity of a metal P is twice that of another metal Q. When these materials are heated through the same temperature change, their increase in length is the same. Calculate the ratio of the original length of P to that of Q.

The final to the above question is Q = 1:2

Explanation

Data

The linear expansivity of Q = α

Linear expansvity of P = 2α

Original length for P = Lp

The original length at Q = LQ

Formula

We will use the formula α = ΔL /L0Δθ

Solution

At P, we will have

2α = ΔL /LPΔθ

At Q, we will have

α = ΔL /LQΔθ

The ratio of LP to LQ will be

LP : LQ

2α = ΔL /LPΔθ : α = ΔL /LQΔθ

Hence, the above expression will left us with

ΔL /2αΔθ : ΔL /αΔθ

Thus, we now have

1/2 : 1

After multiplying both sides by 2, we will be left with

Ratio of P to Q = 1:2

### Question 14

The ratio of the coefficient of linear expansion of two metals α12 is 3:4. When heated through the same temperature change, the ratio of the increase in lengths of the two metals, L1/L2 is 1:2, the ratio of the original lengths L1/L2 is

The final answer to the above question is 2:3

Explanation

Data

α1 = 3

α2 = 4

The change in temperature for the two rods is the same = Δθ = θ

L1 = 1

L2 = 2

α = ΔL /L0Δθ

Therefore,

L0 = ΔL /αΔθ

Therefore

At rod 1

L1 = 1 /3θ

At rod 2

L2 = 2 /4θ

The ratio of L1 to L2 = (1 /3θ) : (2 /4θ)

Therefore, our final answer is 2/3

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

How to Calculate the Relative Density of a Liquid

Linear Expansivity Definition and Calculations

Sources:

Wikipedia