## What is Cubic Expansivity

**Definition:** Cubic expansivity or otherwise volume expansivity, is the increase in the volume of a material per unit volume per degree rise in temperature. Cubic expansivity is denoted by γ. It is the ratio of the change in volume to the product of the original volume by temperature rise. Now, let us have a look at how to calculate cubic expansivity with examples:

## Understanding Cubic Expansivity

Cubic expansivity, often referred to as the coefficient of volume expansion, is a fundamental concept in physics that characterizes how the volume of a substance changes with variations in temperature. It specifically applies to isotropic materials like liquids and solids. When a substance undergoes a temperature change, it generally expands when heated and contracts when cooled.

The cubic expansivity is denoted by the symbol β (beta) and is defined as the fractional change in volume (ΔV/V) per unit change in temperature (ΔT). Mathematically, it is expressed as β = (1/V) * (ΔV/ΔT).

The concept is particularly relevant in understanding thermal expansion phenomena, such as the expansion of substances in response to increased temperature. Different materials have different cubic expansivity values, which influence their thermal behaviour. Solids like metals often have relatively low values of β, making them less prone to drastic volume changes with temperature fluctuations, while liquids like water have higher values of β, resulting in more significant volume changes when heated or cooled.

In practical applications, knowing the cubic expansivity of a material is crucial for designing systems and structures that can withstand temperature variations without detrimental effects. To understand how to calculate cubic expansivity, you will need to know its formula:

### Cubic Expansivity Formula

The cubic expansivity formula is:

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

and can be written mathematically as:

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 equal 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 a few examples of cubic expansivity questions to make you understand how to calculate cubic expansivity with 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 the 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 centimeters, 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 the 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 the 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

The 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 an 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, the 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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