## Introduction

In this post, you will learn how to calculate cubic expansivity. I will try and break it down in way you will understand and do it at ease.

## Understanding Cubic Expansivity

To understand how to calculate cubic expansivity, you will need to understand the concept of cubic expansivity.

### What is Cubic Expansivity

**Definition:** Cubic expansivity or otherwise volume expansivity, is the increase in volume of a material per unit volume per degree rise in temperature. Cubic expansivity is denoted by γ.

It is the ratio change in volume to the product of original volume by temperature rise.

The formula for calculating cubic expansivity is

Cubic expansivity (γ) = the change in volume (∆V) / ( original volume [V_{1}] x rise in temperature [θ] )

and

γ = ∆V / V_{1}∆θ

where ∆V = V_{2} – V_{1}

Now,

γ = (V_{2} – V_{1}) / V_{1} ( θ_{2} – θ_{1} )

and

γ = cubic expansivity

V_{1} = Original or Initial volume

V_{2} = Final volume

θ_{1} = Initial temperature

θ_{2} = Final temperature

from the above equation, we can see that

γ V_{1}∆θ = V_{2} – V_{1} = Change in volume

To make V_{2} subject of the formula, we now say

V_{2} = γ V_{1}∆θ + V_{1}

This is also equals to

V_{2} = V_{1} (γ∆θ + 1)

Another formula for cubic expansivity is

γ = 3α

and if α = β / 2,

this implies that

γ = 3 x ( β / 2 )

Therefore, we can also use the formula below to calculate cubic expansivity

γ = 3β / 2

### What is apparent Cubic Expansivity

**Definition:** Apparent cubic expansivity is the ratio of the mass of liquid ejected to the remaining mass when the temperature increases by 1^{0}c.

The formula for calculating apparent cubic expansivity is

Apparent Cubic expansivity, γ_{a} = ( Mass of the ejected liquid / [mass of the remaining liquid x rise in temperature] )

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## Solved Problems on How to Calculate Cubic Expansivity

Here are few examples on cubic expansivity questions to make you understand how to calculate cubic expansivity at ease.

### Example 1

What is the cubical expansivity of brass at 60^{0}c, if the density at 0^{0}c is 15.2g/cm^{3}.

**Solution**

Data:

We can easily extract our data from the above question

We know that linear expansivity of brass (α_{b}) = 1.9 x 10^{-5} K^{-1}

and the formula for cubic or volume expansivity is γ = 3α

Thus γ = 3 x α_{b}

Therefore, we can write

γ = 3 x 1.9 x 10^{-5} K^{-1} = 5.7 x 10^{-5} K^{-1}

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### Example 2

A cube with sides 100 centimeters at 0^{0}c is heated to 100^{0}c. If the sides becomes 101 centimeter long, find the cubic expansivity of it is material.

**Solution**

Data:

γ = (V_{2} – V_{1}) / V_{1} ( θ_{2} – θ_{1} )

and

γ = cubic expansivity = ?

L_{1} = Original or Initial length = 100 cm

L_{2} = Final length = 101 cm

θ_{1} = 0^{0}c

θ_{2} = 100^{0}c

looking at

V = L x b x h = L x L x L

From L_{1} = 100 cm

This shows that

V_{1} = 100 x 100 x 100 = 1,000,000 cm^{3} = 10^{6} cm^{3}

V_{2} = 101 x 101 x 101 = 1,030,301 cm^{3} = 10^{6} cm^{3}

Now we apply

γ = (1,030,301 cm^{3} – 1,000,000 cm^{3}) / 1,000,000 cm^{3} ( 100^{0}c – 0^{0}c )

This is equal to

γ = 30,301 cm^{3} / 10^{7} cm^{3} ^{0}c

γ = 0.0030301 k^{-1}

### Example 3

The linear expansivity of the material of a cube is 12 x 10^{-6} k^{-1}. If the length of each side of the cube is 10 centimeter, find the area of one face of the cube and the volume of the cube when it is temperature is raised by 30 k

**Solution:**

**Data**

Linear expansivity of the material, α = 12 x 10^{-6} k^{-1}

This implies that the cubic expansivity of the material is γ = 3α

Which shows that

γ = 3 x 12 x 10^{-6} k^{-1} = 36 x 10^{-6} k^{-1}

and α = β / 2 which implies that β = 2α

Thus

β = 2 x 12 x 10^{-6} k^{-1} = 24 x 10^{-6} k^{-1}

Length of each side of the cube = 10 cm

and area, A = L x b = 10 x 10 = 100 cm^{2}

Volume, V = L x b x h = 10 x 10 x 10 = 1000 cm^{3}

Therefore,

Initial area, A_{`1} = 100 cm^{2}

Initial volume is V_{2} = 1000 cm^{3}

Now, we apply the formula below to find the final area

A_{`2} = A_{`1} ( 1 + βθ)

which is equal to

A_{`2} = 100 ( 1 + 24 x 10^{-6} x 30) = 100 cm^{2}

To calculate the final volume V_{2}, we now use the formula

V_{2} = V_{1} (γ∆θ + 1)

By substituting our data into the above equation, we now have

V_{2} = 100 (36 x 10^{-6} x 30 + 1) = 1001 cm^{3}

### Example 4

A solid metal cube of side 20 cm is heated from 10^{0}c to 100^{0}c. if the linear expansivity of the metal is 1.8 x 10^{-4} k^{-1}. Calculate the increase in volume of the metal cube.

**Solution:**

Data:

Looking at this formula

γ V_{1}∆θ = V_{2} – V_{1} = Change in volume = increase in volume = ∆V

we now have

∆V = γ V_{1}∆θ

and ∆θ = θ_{2} – θ_{1}

From the question, our data stands as

Initial temperature = θ_{1} = 10^{0}c

Final temperature = θ_{2} = 10^{0}c

Linear expansivity = α = 1.8 x 10^{-4} k^{-1}

Cubic or volume expansivity = γ = 3α = 3 x 1.8 x 10^{-4} k^{-1} = 5.4 x 10^{-4} k^{-1}

Also Volume, V = l x b x h = 10 x 10 x 10 = 1000 cm^{3} (where l = b = h = 10 cm)

Thus V_{1} = 1000 cm^{3}

Now, we calculate increase in volume as

∆V = γ V_{1} (θ_{2} – θ_{1})

∆V = 5.4 x 10^{-4} x 1000 (100 – 10) = 48.6 cm^{3}

### Example 5

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10^{-6} k^{-1}. What is the area expansivity of the metal

**Solution**

**Data**

From the above question γ = 6.3 x 10^{-6} k^{-1}

β = ? and we know that β = 2α

But we don’t have alpha from our question

Therefore we need to first find α

And the relationship between linear expansivity α and cubic (volume) expansivity γ is

γ = 3α

we make α subject of the formula

α = γ / 3 = 6.3 x 10^{-6} / 3

Hence,

α = 2.1 x 10^{-6}

β = 2α = 2 x 2.1 x 10^{-6}

Therefore, the area expansivity of the metal is

β = 4.2 x 10^{-6} k^{-1}

### Example 6

A metal of volume 40 centimeter cube is heated from 30^{0}c to 90^{0}c. What is the increase in volume of the metal.[Linear expansivity (α) = 2.0 x 10^{-3} k^{-2} ]

**Solution:**

**Data:**

By extracting our data, we have

Initial or original volume, V_{1} = 40 cm^{3}

Initial temperature = θ_{1} = 30^{0}c

Final temperature = θ_{2} = 90^{0}c

Change in volume, ∆V = ?

∆V = γ V_{1}∆θ

and ∆θ = θ_{2} – θ_{1} = (90^{0} – 30^{0}) = 60^{0}

where γ = 3α = 3 x 2.0 x 10^{-3} = 0.06

Now we can substitute our data into the main equation

∆V = 0.06 x 40 x 60 = 14.4 cm^{3}

### Example 7

The coefficient of cubical expansion of a cube occupying a volume of 16cm^{3} is 1.8 x 10^{-6} k^{-1}. At the same temperature and condition, another metal of volume 8cm^{3} is used. Find the coefficient of the cubical expansion of the metal.

**Solution:**

**Data:**

γ_{1} = 1.8 x 10^{-6} k^{-1}

γ_{2} = ?

V_{1} = 16cm^{3}

V_{2} = 8cm^{3}

Since the formula for increase in volume is

∆V = γ V_{1}∆θ

we can use the relation

γ_{1}V_{1} = γ_{2}V_{2}

By making γ_{2} subject of the formula, we find

γ_{2} = γ_{1}V_{1} / V_{2}

we now input our values into the above equation

γ_{2} = (1.8 x 10^{-6} k^{-1} x 16cm^{3} ) / 8cm^{3}

This will give us

γ_{2} = 0.0000288 / 8

Therefore, our final answer is

γ_{2} = 0.0000036 = 3.6 x 10^{-6} k^{-1}

### Example 8

A density bottle full of liquid is heated from 0^{0}c to 100^{0}c during which 6.8g of a liquid is expelled, leaving 400g of it in the bottle. calculate the apparent cubic expansivity of the liquid.

**Solution:**

**Data:**

Mass of the ejected liquid, M_{1} = 6.8g

Also, mass of the remaining liquid, M_{2} = 400g

rise in temperature, ∆θ = 100 – 0 = 100^{0}c

Apparent Cubic expansivity, γ_{a} = ?

Now we apply the formula below

Apparent Cubic expansivity, γ_{a} = ( Mass of the ejected liquid / [mass of the remaining liquid x rise in temperature] )

γ_{a} = M_{1} / ( M_{2} ∆θ )

after substituting our values into the above equation, we now have

γ_{a} = 6.8 / ( 400 x 100 ) = 1.7 x 10^{-4} k^{-1}

Therefore the apparent cubic expansivity (γ_{a}) is 1.7 x 10^{-4} k^{-1}

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