# Consider a 20mm long brass wire aligned with a 10mm long aluminum wire. The wires are separated...

## Question:

Consider a 20mm long brass wire aligned with a 10mm long aluminum wire. The wires are separated by a distance of 30{eq}\displaystyle \mu m {/eq} at a temperature of 22 {eq}\displaystyle ^{\circ}C {/eq}. If the far ends of the wires are restrained, at what temperature will the two wires touch?

Take the following values to be the coefficients of linear expansion:

{eq}\displaystyle \alpha _{brass}=19 \ X \ 10^{-6} {/eq}

{eq}\displaystyle \alpha _{aluminum}=22 \ X \ 10^{-6} {/eq}

## Coefficient Of Linear Expansion :

When matter is heated it undergoes changes in its dimensions. At the molecular level, the heating causes higher amplitude vibrations of the molecules about their mean positions. The average bond length increases because of the enhanced vibrations. This will manifest as a change in the dimension of the material. For small increases in the temperature, the length will increase linearly with the temperature. The proportionality constant is called the coefficient of linear expansion. Similarly, the area and the volume will also increase linearly but with different proportionality constants.

The length of the aluminium wire is {eq}\displaystyle {L_a=10 \ mm } {/eq}

The length of the brass wire is {eq}\displaystyle {L_b=20 \ mm } {/eq}

Given that the coefficient of linear expansion of brass is,

{eq}\displaystyle {\alpha _{brass}=19 \times 10^{-6} \ /^{\circ}C} {/eq}

The coefficient of linear expansion of aluminium is,

{eq}\displaystyle {\alpha _{aluminium}=22 \times 10^{-6} \ /^{\circ}C} {/eq}

The change in length of the brass wire due to a temperature change {eq}\displaystyle { \Delta T } {/eq} is given by,

{eq}\displaystyle {\Delta L_a =L_a \alpha _{aluminium} \Delta T } {/eq}

The change in length of the aluminium wire due to a temperature change {eq}\displaystyle { \Delta T } {/eq} is given by,

{eq}\displaystyle {\Delta L_b =L_b \alpha _{brass} \Delta T } {/eq}

Then,

{eq}\displaystyle {\Delta L_a+\Delta L_b=(L_a \alpha _{aluminium}+L_b \alpha _{brass}) \Delta T \quad.......(1) } {/eq}

Given that initially the two wires are at a separation of {eq}\displaystyle {d=30 \ \mu m =30 \times 10^{-3} \ mm } {/eq}

For the wires to meet we require the condition,

{eq}\displaystyle {\Delta L_a+\Delta L_b=d \quad........(2) } {/eq}

Comparing equations (1) and (2) we get,

{eq}\displaystyle {(L_a \alpha _{aluminium}+L_b \alpha _{brass}) \Delta T=d } {/eq}

Or,

{eq}\displaystyle {\Delta T=\dfrac{d}{(L_a \ \alpha _{aluminium}+L_b \ \alpha _{brass})} } {/eq}

{eq}\displaystyle {\Delta T=\dfrac{30\times 10^{-3} \ mm}{10 \ mm\times (22 \times 10^{-6} \ /^{\circ}C) +20 \ mm \times (19 \times 10^{-6} \ /^{\circ}C)} } {/eq}

{eq}\displaystyle {\Delta T=50^{\circ}C } {/eq}

Originally the wires are at an ambient temperature, {eq}\displaystyle { T_i=22^{\circ}C } {/eq}

Thus the wires will meet at a temperature,

{eq}\displaystyle {T_f=T_i+\Delta T=50+22=\color{blue}{72^{\circ}C }} {/eq} 