Question

A hole is drilled in a copper sheet. The diameter of the hole is 4.24cm at ${27.0^o}{\text{C}}$. What is the change in the diameter of the hole when the sheet is heated to ${277^o}{\text{C}}$?
Coefficient of the linear expansion of copper $ = 1.70 \times {10^{ - 5}}{{\text{K}}^{ - 1}}$

Answer 1

Hint: To find the change in diameter of the hole can be determined by using the relation Change in area $\left( \Delta \right)$/ Original area $\left( {\text{A}} \right) = \beta \Delta {\text{T}}$, where $\beta $ is coefficient of superficial expansion, change in temperature $\Delta {\text{T}}$. To find the value of coefficient of superficial expansion, we have a relation that is $\beta = 2\alpha $where ${\alpha _{{\text{Cu}}}} = 1.70 \times {10^{ - 5}}{{\text{K}}^{ - 1}}$that is co-efficient of linear expansion of copper. Therefore the final equation we use to solve is \[\left( {{{\text{d}}_2}^2 - {{\text{d}}_1}^2} \right)/{{\text{d}}_1}^2 = 2\alpha \Delta {\text{T}}\].

Complete step-by-step answer:
Given, Initial Temperature applied on copper sheet, ${{\text{T}}_1} = {27.0^o}{\text{C}}$
Diameter of the hole in a copper sheet at temperature${{\text{T}}_1},{{\text{d}}_1} = 4.24{\text{cm}}$
Final Temperature applied on copper sheet, ${{\text{T}}_2} = {277^o}{\text{C}}$
Diameter of the hole drilled in a copper sheet at temperature${{\text{T}}_2} = {{\text{D}}_2}$
Coefficient of linear expansion of copper, ${\alpha _{{\text{Cu}}}} = 1.70 \times {10^{ - 5}}{{\text{K}}^{ - 1}}$
The coefficient of linear is the change in length of a specimen one unit long when its temperature is changed by one degree. Different materials expand by different amounts.
The dimension of coefficient of linear expansion will be $\left[ {{{\text{M}}^0}{{\text{L}}^0}{{\text{T}}^0}{{\text{K}}^{ - 1}}} \right]$
The coefficient of superficial expansion is defined as the ratio of increase in area to its original area for every degree increase in temperature.
For coefficient of superficial expansion$\beta $, and change in temperature$\Delta {\text{T}}$, we have the relation:
Change in area$\left( \Delta \right)$/ Original area $\left( {\text{A}} \right) = \beta \Delta {\text{T}}$
$\left[ {\left( {\pi {{\text{d}}_2}^2/4} \right) - \left( {\pi {{\text{d}}_1}^2/4} \right)} \right]/\left( {\pi {{\text{d}}_1}^2/4} \right) = \Delta {\text{A/A}}$
$\therefore \Delta {\text{A/A = }}\left( {{{\text{d}}_2}^2 - {{\text{d}}_1}^2} \right)/{{\text{d}}_1}^2$
But $\beta = 2\alpha $
\[\therefore \left( {{{\text{d}}_2}^2 - {{\text{d}}_1}^2} \right)/{{\text{d}}_1}^2 = 2\alpha \Delta {\text{T}}\]
\[\left( {{{\text{d}}_2}^2{\text{/}}{{\text{d}}_1}^2} \right) - 1 = 2\alpha \left( {{{\text{T}}_2} - {{\text{T}}_1}} \right)\]
\[{{\text{d}}_2}^2{\text{/}}{\left( {4.24} \right)^2} = 2 \times 1.7 \times 1{0^{ - 5}}\left( {227 - 27} \right) + 1\]
\[{{\text{d}}_2}^2 = 17.98 \times 1.0068 = 18.1\]
\[\therefore {{\text{d}}_2} = 4.2544{\text{ cm}}\]
Change in diameter$ = {\text{ }}{{\text{d}}_2} - {{\text{d}}_1}$
$ \Rightarrow 4.2544 - 4.24 = 0.0144{\text{ cm}}$
Hence, the diameter increases by $1.44 \times {10^{ - 2}}{\text{ cm}}$

Note: We can calculate change in length due to temperature by the dependence of thermal expansion on temperature, substance, and the length is summarized in the equation$\Delta {\text{L = }}\alpha {\text{L}}\Delta {\rm T}$, where $\Delta {\text{L}}$is the change in length, $\Delta {\rm T}$is the change in temperature and $\alpha $is the coefficient of linear expansion, which varies slightly with temperature.