# A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at 27.0 ^oC. What is the...

## Question:

A hole is drilled in a copper sheet. The diameter of the hole is 4.24 cm at {eq}27.0 ^oC. {/eq} What is the change in the diameter of the hole when the sheet is heated to {eq}227^oC? {/eq} Coefficient of linear expansion of {eq}copper =1.70 \times 10^{-5} K^{-1}. {/eq}

## Coefficient of Linear Expansion:

The coefficient of linear expansion is a material property, in which change in length of a specimen one unit long when its temperature is changed by one degree. It is different for different types of material.

Given data

• Initial temperature of copper sheet: {eq}{T_i} = 27^\circ {\rm{C}} {/eq}
• Final temperature of copper sheet: {eq}{T_f} = 227^\circ {\rm{C}} {/eq}
• Initial diameter of hole: {eq}{d_i} = 4.24\;{\rm{cm}} {/eq}
• Coefficient of linear expansion of copper: {eq}{\alpha _l} = 1.7 \times {10^{ - 5}}\;{{\rm{K}}^{ - 1}} {/eq}

The expression for the Initial area of the hole is given as:

{eq}{A_i} = \dfrac{\pi }{4}d_i^2 {/eq}

The expression for the coefficient of superficial expansion is given as:

{eq}{\alpha _s} = 2 \times {\alpha _l} {/eq}

The expression for the area of hole drilled in copper sheet is given as:

{eq}\begin{align*} {A_f} &= {A_i}\left( {1 + {\alpha _s}\left( {{T_f} - {T_i}} \right)} \right)\\ {A_f} &= {A_i}\left( {1 + 2{\alpha _l}\left( {{T_f} - {T_i}} \right)} \right)\\ \dfrac{\pi }{4}d_f^2 &= \dfrac{\pi }{4}d_i^2\left( {1 + 2{\alpha _l}\left( {{T_f} - {T_i}} \right)} \right)\\ d_f^2 &= d_i^2\left( {1 + 2{\alpha _l}\left( {{T_f} - {T_i}} \right)} \right) \end{align*} {/eq}

Substitute the value in the above expression:

{eq}\begin{align*} d_f^2 &= {4.24^2}\left( {1 + 2 \times 1.7 \times {{10}^{ - 5}} \times \left( {227 - 27} \right)} \right)\\ {d_f} &= 4.254\;{\rm{cm}} \end{align*} {/eq}

The expression for change in the diameter of the hole when the sheet is heated is:

{eq}\Delta d = {d_f} - {d_i} {/eq}

Substitute the value in the aboive expression.

{eq}\begin{align*} \Delta d &= 4.254 - 4.24\\ \Delta d &= 0.0143\;{\rm{cm}}\\ \Delta d &= 0.143\;{\rm{mm}} \end{align*} {/eq}

Thus, change in the diameter of the hole when the sheet is heated is {eq}0.143\;{\rm{mm}} {/eq}.