Linear Expansivity

Expansivity is the increase in the length of a substance when its temperature increases by one kelvin or degree Celsius.

Linear expansivity (α) of a substance is defined as the increase in length of a substance, per unit length of the substance, per degree rise in temperature.

Mathematically,

linear expansivity  = α

\(\scriptsize α = \normalsize \frac {Change \: in \: Length}{Original \: Length \: \times \: temperature \: rise}\)

If L₁ = the original or initial length

θ₁ = Original or initial temperature

L₂  = New Length

θ₂ = New temperature

α = Linear expansivity

\( \scriptsize Then \;  α = \normalsize \frac {L_2\: – \:L_1}{L_1 \: \times \: (\theta_2\: – \: \theta_1\ ) }\)

The S.I unit is per kelvin \( \scriptsize K^{-1} \) or per degree Celsius \( \scriptsize ^{०} C^{-1} \)

For a change in length,

L₂ – L₁ = α L₁ (θ₂ – θ₁)

New length, L₂ = L₁ + α L₁ (θ₂ – θ₁)

Or

L₂ = L₁ (1 + α (θ₂ – θ₁))

The value of linear expansivity (α) differs from substance to substance and is highest for metals.

Substance	Linear Expansivity (K-1)
Platinum	0.000009
Iron	0.000012
Copper	0.000017
Brass	0.000018
Aluminium	0.000023
Lead	0.000029
Zinc	0.000030
Invar (alloy)	0.000001
Glass	0.0000085
Silica	0.0000084

Example 10.2.1:

A copper rod whose length at 30ºC is 10.0 cm is heated to 60ºC. Find its new length if the Linear expansivity (α) for copper = 0.000017 k^-1.

Solution

L₁ = 10.0 cm,    θ₁ =   30ºC,   α = 0.000017 k^-1

L₂ = ?               θ₂ =   60ºC

\( \scriptsize α = \normalsize \frac {L_2\: – \:L_1}{L_1 \: \times \: (\theta_2\: – \: \theta_1\;) }\)

\(\scriptsize L_2 = L_1 \: + \: \alpha L_1 \left( \theta_2 \: -\: \theta_1 \right) \)

        = 10 + 0.000017 × 10 × (60 – 30)

        = 10 + 0.0051

       = 10.0051 cm

Example 10.2.2:

An iron rod of length 12.01 m is heated from 10ºC to 70ºC. If its new length is 12.06 m, calculate its linear expansivity.

Solution

L₂ = 12.06 m             L₁ = 12.01 m, 

θ₂ =   70ºC                 θ₁ =   10ºC,

 α = ?      

\( \scriptsize α = \normalsize \frac {L_2\: – \:L_1}{L_1 \: \times \: (\theta_2\: – \: \theta_1\;) }\)

\( \scriptsize α = \normalsize \frac {12.06\: – \:12.01}{12.01 \: \times \: 70\: – \: 10 } \)

\( \scriptsize α = \normalsize \frac {0.05}{12.01 \: \times \: 60 } \scriptsize = 0.000069 \: K^{-1} \)

\( \scriptsize α = 0.000069\: K^{-1}\)

Example 10.2.3:

A metal rod has a length of 100 cm at 200ºC. At what temperature will its length be 99.4 cm if the linear expansivity of the material of the rod is 0.00002/K?

Solution 

L₂ = 99.4 cm

L₁ = 100 cm

 Initial temperature  θ₁ = 200ºC

Final temperature  θ₂ = ?

α = 0.00002/K

\( \scriptsize α = \normalsize \frac {L_2\: – \:L_1}{L_1 \: \times \: (\theta_2\: – \: \theta_1\;) }\)

\( \scriptsize 0.00002 = \normalsize \frac {99.4\: – \:100}{100 \: \times \: (\theta_2\: – \: 200\;) }\)

\( \scriptsize 0.00002 = \normalsize \frac {0.6}{100 \: \times \: (\theta_2\: – \: 200\:) }\)

0.00002  (100θ₂ –  20000 ) = 0.6

 0.002 θ₂ + 0.4 = 0.6

 0.002 θ₂  = 0.6 – 0.4

0.002 θ₂  = 0.2

θ₂ = \( \frac {0.2}{0.002}\)

θ₂ = 100ºC