## 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 Î³. 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 Î² (beta) and is defined as the fractional change in volume (Î”V/V) per unit change in temperature (Î”T). Mathematically, it is expressed as Î² = (1/V) * (Î”V/Î”T).

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 Î², making them less prone to drastic volume changes with temperature fluctuations, while liquids like water have higher values of Î², 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 (Î³) = the change in volume (âˆ†V) / ( original volume [V _{1}] x rise in temperature [Î¸] )**

and can be written mathematically as:

Î³ = âˆ†V / V_{1}âˆ†Î¸

where âˆ†V = V_{2} â€“ V_{1}

Now,

Î³ = (V_{2} â€“ V_{1}) / V_{1} ( Î¸_{2} â€“ Î¸_{1} )

and

Î³ = cubic expansivity

V_{1} = Original or Initial volume

V_{2} = Final volume

Î¸_{1} = Initial temperature

Î¸_{2} = Final temperature

from the above equation, we can see that

Î³ V_{1}âˆ†Î¸ = V_{2} â€“ V_{1} = Change in volume

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

V_{2} = Î³ V_{1}âˆ†Î¸ + V_{1}

This is also equal to

V_{2} = V_{1} (Î³âˆ†Î¸ + 1)

Another formula for cubic expansivity is

Î³ = 3Î±

and if Î± = Î² / 2,

this implies that

Î³ = 3 x ( Î² / 2 )

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

Î³ = 3Î² / 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, Î³_{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 (Î±_{b}) = 1.9 x 10^{-5} K^{-1}

and the formula for cubic or volume expansivity is Î³ = 3Î±

Thus Î³ = 3 x Î±_{b}

Therefore, we can write

Î³ = 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:

Î³ = (V_{2} â€“ V_{1}) / V_{1} ( Î¸_{2} â€“ Î¸_{1} )

and

Î³ = cubic expansivity = ?

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

L_{2} = Final length = 101 cm

Î¸_{1} = 0^{0}c

Î¸_{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

Î³ = (1,030,301 cm^{3} â€“ 1,000,000 cm^{3}) / 1,000,000 cm^{3} ( 100^{0}c â€“ 0^{0}c )

This is equal to

Î³ = 30,301 cm^{3} / 10^{7} cm^{3} ^{0}c

Î³ = 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, Î± = 12 x 10^{-6} k^{-1}

This implies that the cubic expansivity of the material is Î³ = 3Î±

Which shows that

Î³ = 3 x 12 x 10^{-6} k^{-1} = 36 x 10^{-6} k^{-1}

and Î± = Î² / 2 which implies that Î² = 2Î±

Thus

Î² = 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 + Î²Î¸)

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} (Î³âˆ†Î¸ + 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

Î³ V_{1}âˆ†Î¸ = V_{2} â€“ V_{1} = Change in volume = increase in volume = âˆ†V

we now have

âˆ†V = Î³ V_{1}âˆ†Î¸

and âˆ†Î¸ = Î¸_{2} â€“ Î¸_{1}

From the question, our data stands as

Initial temperature = Î¸_{1} = 10^{0}c

Final temperature = Î¸_{2} = 10^{0}c

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

Cubic or volume expansivity = Î³ = 3Î± = 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

âˆ†V = Î³ V_{1} (Î¸_{2} â€“ Î¸_{1})

âˆ†V = 5.4 x 10^{-4} x 1000 (100 â€“ 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 Î³ = 6.3 x 10^{-6} k^{-1}

Î² = ? and we know that Î² = 2Î±

But we donâ€™t have alpha from our question

Therefore we need to first find Î±

And the relationship between linear expansivity Î± and cubic (volume) expansivity Î³ is

Î³ = 3Î±

we make Î± subject of the formula

Î± = Î³ / 3 = 6.3 x 10^{-6} / 3

Hence,

Î± = 2.1 x 10^{-6}

Î² = 2Î± = 2 x 2.1 x 10^{-6}

Therefore, the area expansivity of the metal is

Î² = 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 (Î±) = 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 = Î¸_{1} = 30^{0}c

Final temperature = Î¸_{2} = 90^{0}c

Change in volume, âˆ†V = ?

âˆ†V = Î³ V_{1}âˆ†Î¸

and âˆ†Î¸ = Î¸_{2} â€“ Î¸_{1} = (90^{0} â€“ 30^{0}) = 60^{0}

where Î³ = 3Î± = 3 x 2.0 x 10^{-3} = 0.06

Now we can substitute our data into the main equation

âˆ†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:**

Î³_{1} = 1.8 x 10^{-6} k^{-1}

Î³_{2} = ?

V_{1} = 16cm^{3}

V_{2} = 8cm^{3}

Since the formula for an increase in volume is

âˆ†V = Î³ V_{1}âˆ†Î¸

we can use the relation

Î³_{1}V_{1} = Î³_{2}V_{2}

By making Î³_{2} subject of the formula, we find

Î³_{2} = Î³_{1}V_{1} / V_{2}

we now input our values into the above equation

Î³_{2} = (1.8 x 10^{-6} k^{-1} x 16cm^{3} ) / 8cm^{3}

This will give us

Î³_{2} = 0.0000288 / 8

Therefore, our final answer is

Î³_{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, âˆ†Î¸ = 100 â€“ 0 = 100^{0}c

Apparent Cubic expansivity, Î³_{a} = ?

Now we apply the formula below

Apparent Cubic expansivity, Î³_{a} = ( Mass of the ejected liquid / [mass of the remaining liquid x rise in temperature] )

Î³_{a} = M_{1} / ( M_{2} âˆ†Î¸ )

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

Î³_{a} = 6.8 / ( 400 x 100 ) = 1.7 x 10^{-4} k^{-1}

Therefore the apparent cubic expansivity (Î³_{a}) is 1.7 x 10^{-4} k^{-1}

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