Cubic Expansivity

Cubic or volume expansivity is the increase in volume per unit volume per degree rise in temperature. The S.I unit is k^-1

Volume/cubical expansivity, = γ

\(\scriptsize (γ) = \normalsize \frac {increase \: in \: Volume}{Original \: Volume \: \times \: temperature \: rise}\)

\(\scriptsize \gamma = \normalsize \frac {V_2 \: – \: V_1}{V_1 \: \times \: (\theta_2 \: – \: \theta_1)}\) 

V₂ = New volume, V₁ = Original Volume

 θ₂ = New temperature, θ₁ = Original temperature

From the equation above we have

\( \scriptsize V_2 \: – \: V_1 = \gamma V_1(\theta_2 \: – \: \theta_1)\)

\( \scriptsize V_2 = V_1 \: + \: \gamma V_1(\theta_2 \: – \: \theta_1)\)

or

\( \scriptsize V_2 = V_1(1 \: + \: \gamma (\theta_2 \: – \: \theta_1)) \)

Cubic expansivity = 3 Linear expansivity

\( \scriptsize \gamma = 3 \alpha \) (Proof of The Relationship between Linear Expansivity and Cubic Expansivity)

Example 10.4.1:

A piece of mass 170 kg has its temperature raised from 0°C to 30°C. Calculate its increase in volume, given the densityDensity is the measurement of how tightly a material is packed together i.e. how closely the particles are packed in the material. The tighter the material is packed the more its... More of brass at 0°C as 8.5 × 10³ Kgm^-3 and its cubic expansivity as 57 × 10^-4 K^-1

Solution

 Mass = 170 kg

Original temperature  θ₁ = 0°C

New temperature  θ₂ =  30°C

density of brass at 0°C = 8.5 × 10³ Kgm^-3

cubic expansivity = 57 × 10^-4 K^-1

change in temperature, Δθ = θ₂ – θ₁ = 30°C – 0°C = 30°C

To calculate the increase in volume we need to calculate V₁ and V₂

Step 1: Calculate volume V₁

V₁, volume at temperature 0°C is unknown but the mass and density of the object at temperature 0°C are given.

Use the Values of mass & density to find V₁

Density = \( \frac {mass}{volume}\)

8.5 × 10³ = \( \frac {170}{volume}\)

Volume = \( \frac {170}{8.5 \: \times \: 10^{-3}}\)

V₁ at temperature 0°C = 0.02 m³

Step 2: calculate V₂

⇒ \(\scriptsize γ = \normalsize \frac {V_2 \: – \: V_1}{V_1 \: \times \: (\theta_2 \: – \: \theta_1)}\)  

⇒ \( \scriptsize V_2 \: – \:V_1 = V_1 \times (\theta_2 \: – \: \theta_1)γ \)  

⇒ \( \scriptsize V_2 = V_1 \times (\theta_2 \: – \: \theta_1)γ\: +\: V_1 \) 

⇒ \( \scriptsize V_2 = V_1 ( (\theta_2 \: – \: \theta_1)γ \: + \:1) \) 

⇒ \( \scriptsize V_2 = 0.02 ( 30\: \times\: 57\: \times\: 10^{-4}\: +\: 1) \) 

⇒ \( \scriptsize V_2 = 0.02 ( 0.171 \:+\: 1) \) 

⇒ \( \scriptsize V_2 = 0.02 (1.171) \) 

⇒ \( \scriptsize 0.02342 m^3 \) 

Increase in volume = V₂ – V₁

= 0.02342 – 0.02

= 0.00342 m³

Example 10.4.2:

A solid metal cube of side 15 cm is heated from 5°C to 50°C. if the linear expansivity of the metal is 1.8 × 10^-4 K^-1, calculate the increase in its volume.

Solution

Length of metal cube = 15 cm, initial temperature =  5°C, final temperature = 50°C 
Linear expansivity of the metal is 1.8 × 10^-4 K^-1.

To calculate the increase in Volume we need to calculate the values of V₁ and V₂. These values are not given in the question but we can calculate V₁ using the length of the cube. The material is a cube and we are given only one side of the cube. Since a cube has equal sides, the length, breadth, and height of the cube are the same.  i.e l = b = h

Therefore, Volume of cube at 5°C = l × b × h = 15 × 15 × 15 = 3,375 cm³

Cubic expansivity = 3 Linear expansivity

Cubic expansivity = 3 × 1.8 × 10^-4 K^-1

Cubic expansivity = 5.4 × 10^-4 K^-1

⇒ \(\scriptsize γ = \normalsize \frac {V_2 \: – \: V_1}{V_1 \: \times \: (\theta_2 \: – \: \theta_1)}\) 

Make V₂ the subject of the formula

⇒ \(\scriptsize V_2 = V_1 ( (\theta_2 \: – \: \theta_1)γ \:+\: 1) \) 

⇒ \(\scriptsize V_2 = 3,375 ( (50 \: – \: 5)5.4 \; \times \: 10^{-4} \:+ \:1) \) 

⇒ \(\scriptsize V_2 = 3,375 ( (45)5.4 \: \times \: 10^{-4} \:+\: 1) \) 

⇒ \(\scriptsize V_2 = 3,375 ( 0.0234\:+\: 1) \) 

⇒ \(\scriptsize V_2 = 3,375 ( 1.0234) \) 

⇒ \(\scriptsize V_2 = 3,457 \: cm^3 \) 

 Increase in volume = V₂ – V₁  =  3,457 – 3,375 = 82 cm³