Question

A glass window is to be fit in an aluminum frame. The temperature on the working day is \[40^\circ {\text{C}}\] and glass window measures exactly \[20\,{\text{cm}} \times {\text{30 cm}}\]. [\[{\alpha _{{\text{glass}}}} = 9 \times {10^{ - 6}}\,/^\circ
{\text{C}}\],\[{\alpha _{{\text{aluminum}}}} = 24 \times {10^{ - 6}}\,/^\circ {\text{C}}\]]. The size of the aluminum frame is - - -, so that there is no stress on the glass in winter even if the temperature drops to \[0^\circ {\text{C}}\]?

Answer 1

Hint:Use the formula for the linear thermal expansion of an object. This formula gives the relation between the changed length, original length, coefficient of linear expansion and change in temperature of the object. For no stress condition, the changed lengths of the glass window and the aluminum frame should be equal.

Formula used:
The expression for linear thermal expansion is given by
\[{L_0} - L = \alpha {L_0}\Delta T\] …… (1)
Here, \[L\] is the changed length, \[\alpha \] is the coefficient of linear expansion, \[{L_0}\] is the original length and \[\Delta T\] is the change in temperature.

Complete step by step answer:
We have given that the length and breadth of the glass window are \[20\,{\text{cm}}\] and \[{\text{30 cm}}\] respectively.
\[{L_0} = 20\,{\text{cm}}\]
\[{B_0} = {\text{30 cm}}\]
The coefficients of linear expansion for glass and aluminum are
\[{\alpha _{{\text{glass}}}} = 9 \times {10^{ - 6}}\,/^\circ {\text{C}}\]
\[{\alpha _{{\text{aluminum}}}} = 24 \times {10^{ - 6}}\,/^\circ {\text{C}}\]
Let us determine the length of the aluminum frame so that there is no tension on the glass even in winter.
Rewrite equation (1) for the changed length \[L\] of the glass window due to change in temperature.
\[{L_0} - L = {\alpha _{{\text{glass}}}}{L_0}\Delta T\]
\[ \Rightarrow L = {L_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right)\]
Rewrite equation (1) for the changed length \[L'\] of the aluminum frame due to change in temperature.
\[ \Rightarrow L' = {L_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)\]
Here, \[{L_0}'\] is the original length of the aluminum frame.
The change in the temperature is
\[\Delta T = 40^\circ {\text{C}} - 0^\circ {\text{C}} = 40^\circ {\text{C}}\]
There will not be any stress on the glass window if the changed lengths of the glass window and the aluminum frame are the same.
\[L = L'\]
Substitute \[{L_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right)\] for \[L\] and \[{L_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)\] for \[L'\] in the above equation.
\[{L_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right) = {L_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)\]
\[ \Rightarrow {L_0}' = {L_0}\left( {\dfrac{{1 - {\alpha _{{\text{glass}}}}\Delta T}}{{1 - {\alpha _{{\text{aluminum}}}}\Delta T}}} \right)\]
Substitute \[9 \times {10^{ - 6}}\,/^\circ {\text{C}}\] for \[{\alpha _{{\text{glass}}}}\], \[24 \times {10^{ - 6}}\,/^\circ {\text{C}}\] for \[{\alpha _{{\text{aluminum}}}}\], \[40^\circ {\text{C}}\] for \[\Delta T\] and \[20\,{\text{cm}}\] for \[{L_0}\] in the above equation.
\[ \Rightarrow {L_0}' = \left( {20\,{\text{cm}}} \right)\left[ {\dfrac{{1 - \left( {9 \times {{10}^{ - 6}}\,/^\circ {\text{C}}} \right)\left( {40^\circ {\text{C}}} \right)}}{{1 - \left( {24 \times {{10}^{ - 6}}\,/^\circ {\text{C}}} \right)\left( {40^\circ {\text{C}}} \right)}}} \right]\]
\[ \Rightarrow {L_0}' = 20.012\,{\text{cm}}\]
Hence, the length of the aluminum frame should be \[20.012\,{\text{cm}}\].
We can determine the breadth of the aluminum frame in the same way.
There will not be any stress on the glass window if the changed breadths of the glass window and the aluminum frame are the same.
\[B = B'\]
Substitute \[{B_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right)\] for \[B\] and \[{B_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)\] for \[B'\] in the above equation.
\[{B_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right) = {B_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)\]
\[ \Rightarrow {B_0}' = {B_0}\left( {\dfrac{{1 - {\alpha _{{\text{glass}}}}\Delta T}}{{1 - {\alpha _{{\text{aluminum}}}}\Delta T}}} \right)\]
Substitute \[9 \times {10^{ - 6}}\,/^\circ {\text{C}}\] for \[{\alpha _{{\text{glass}}}}\], \[24 \times {10^{ - 6}}\,/^\circ {\text{C}}\] for \[{\alpha _{{\text{aluminum}}}}\], \[40^\circ {\text{C}}\] for \[\Delta T\] and \[30\,{\text{cm}}\] for \[{B_0}\] in the above equation.
\[ \Rightarrow {B_0}' = \left( {30\,{\text{cm}}} \right)\left[ {\dfrac{{1 - \left( {9 \times {{10}^{ - 6}}\,/^\circ {\text{C}}} \right)\left( {40^\circ {\text{C}}} \right)}}{{1 - \left( {24 \times {{10}^{ - 6}}\,/^\circ {\text{C}}} \right)\left( {40^\circ {\text{C}}} \right)}}} \right]\]
\[ \therefore {B_0}' = 30.018\,{\text{cm}}\]
Hence, the breadth of the aluminum frame should be \[30.018\,{\text{cm}}\].

Hence, the size of the aluminum frame should be \[20.012\,{\text{cm}} \times {\text{30}}{\text{.018 cm}}\].

Note:The students may think that the units of the length and breadth of the glass window are not converted into the Si system of units. But there is no need for unit conversion as the ultimate answer is in centimeters. The students should be careful while writing the formula for linear expansion that the change in length is equal to the right hand side quantity and the changed length.