How to Calculate Cubic Expansivity with Examples

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 [V₁] x rise in temperature [θ] )

and

γ = ∆V / V₁∆θ

where ∆V = V₂ – V₁

Now,

γ = (V₂ – V₁) / V₁ ( θ₂ – θ₁ )

and

γ = cubic expansivity

V₁ = Original or Initial volume

V₂ = Final volume

θ₁ = Initial temperature

θ₂ = Final temperature

from the above equation, we can see that

γ V₁∆θ = V₂ – V₁ = Change in volume

To make V₂ subject of the formula, we now say

V₂ = γ V₁∆θ + V₁

This is also equals to

V₂ = V₁ (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 1⁰c.

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 60⁰c, if the density at 0⁰c is 15.2g/cm³.

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 0⁰c is heated to 100⁰c. If the sides becomes 101 centimeter long, find the cubic expansivity of it is material.

Solution

Data:

γ = (V₂ – V₁) / V₁ ( θ₂ – θ₁ )

and

γ = cubic expansivity = ?

L₁ = Original or Initial length = 100 cm

L₂ = Final length = 101 cm

θ₁ = 0⁰c

θ₂ = 100⁰c

looking at

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

From L₁ = 100 cm

This shows that

V₁ = 100 x 100 x 100 = 1,000,000 cm³ = 10⁶ cm³

V₂ = 101 x 101 x 101 = 1,030,301 cm³ = 10⁶ cm³

Now we apply

γ = (1,030,301 cm³ – 1,000,000 cm³) / 1,000,000 cm³ ( 100⁰c – 0⁰c )

This is equal to

γ = 30,301 cm³ / 10⁷ 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 cm²

Volume, V = L x b x h = 10 x 10 x 10 = 1000 cm³

Therefore,

Initial area, A_`1 = 100 cm²

Initial volume is V₂ = 1000 cm³

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

A_{`2</sub> = A<sub>`1} ( 1 + βθ)

which is equal to

A_`2 = 100 ( 1 + 24 x 10^-6 x 30) = 100 cm²

To calculate the final volume V₂, we now use the formula

V₂ = V₁ (γ∆θ + 1)

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

V₂ = 100 (36 x 10^-6 x 30 + 1) = 1001 cm³

Example 4

A solid metal cube of side 20 cm is heated from 10⁰c to 100⁰c. 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 V₁∆θ = V₂ – V₁ = Change in volume = increase in volume = ∆V

we now have

∆V = γ V₁∆θ

and ∆θ = θ₂ – θ₁

From the question, our data stands as

Initial temperature = θ₁ = 10⁰c

Final temperature = θ₂ = 10⁰c

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 cm³ (where l = b = h = 10 cm)

Thus V₁ = 1000 cm³

Now, we calculate increase in volume as

∆V = γ V₁ (θ₂ – θ₁)

∆V = 5.4 x 10^-4 x 1000 (100 – 10) = 48.6 cm³

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 30⁰c to 90⁰c. 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, V₁ = 40 cm³

Initial temperature = θ₁ = 30⁰c

Final temperature = θ₂ = 90⁰c

Change in volume, ∆V = ?

∆V = γ V₁∆θ

and ∆θ = θ₂ – θ₁ = (90⁰ – 30⁰) = 60⁰

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 cm³

Example 7

The coefficient of cubical expansion of a cube occupying a volume of 16cm³ is 1.8 x 10^-6 k^-1. At the same temperature and condition, another metal of volume 8cm³ is used. Find the coefficient of the cubical expansion of the metal.

Solution:

Data:

γ₁ = 1.8 x 10^-6 k^-1

γ₂ = ?

V₁ = 16cm³

V₂ = 8cm³

Since the formula for increase in volume is

∆V = γ V₁∆θ

we can use the relation

γ₁V₁ = γ₂V₂

By making γ₂ subject of the formula, we find

γ₂ = γ₁V₁ / V₂

we now input our values into the above equation

γ₂ = (1.8 x 10^-6 k^-1 x 16cm³ ) / 8cm³

This will give us

γ₂ = 0.0000288 / 8

Therefore, our final answer is

γ₂ = 0.0000036 = 3.6 x 10^-6 k^-1

Example 8

A density bottle full of liquid is heated from 0⁰c to 100⁰c 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, M₁ = 6.8g

Also, mass of the remaining liquid, M₂ = 400g

rise in temperature, ∆θ = 100 – 0 = 100⁰c

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 = M₁ / ( M₂ ∆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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