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Question: A glass window is to be fit in an aluminum frame. The temperature on the working day is \[40^\circ {...

A glass window is to be fit in an aluminum frame. The temperature on the working day is 40∘C40^\circ {\text{C}} and glass window measures exactly 20 cm×30 cm20\,{\text{cm}} \times {\text{30 cm}}. [αglass=9×10−6 /∘C{\alpha _{{\text{glass}}}} = 9 \times {10^{ - 6}}\,/^\circ {\text{C}},αaluminum=24×10−6 /∘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∘C0^\circ {\text{C}}?

Explanation

Solution

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
L0−L=αL0ΔT{L_0} - L = \alpha {L_0}\Delta T …… (1)
Here, LL is the changed length, α\alpha is the coefficient of linear expansion, L0{L_0} is the original length and ΔT\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 cm20\,{\text{cm}} and 30 cm{\text{30 cm}} respectively.
L0=20 cm{L_0} = 20\,{\text{cm}}
B0=30 cm{B_0} = {\text{30 cm}}
The coefficients of linear expansion for glass and aluminum are
αglass=9×10−6 /∘C{\alpha _{{\text{glass}}}} = 9 \times {10^{ - 6}}\,/^\circ {\text{C}}
αaluminum=24×10−6 /∘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 LL of the glass window due to change in temperature.
L0−L=αglassL0ΔT{L_0} - L = {\alpha _{{\text{glass}}}}{L_0}\Delta T
⇒L=L0(1−αglassΔT)\Rightarrow L = {L_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right)
Rewrite equation (1) for the changed length L′L' of the aluminum frame due to change in temperature.
⇒L′=L0′(1−αaluminumΔT)\Rightarrow L' = {L_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)
Here, L0′{L_0}' is the original length of the aluminum frame.
The change in the temperature is
ΔT=40∘C−0∘C=40∘C\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′L = L'
Substitute L0(1−αglassΔT){L_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right) for LL and L0′(1−αaluminumΔT){L_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right) for L′L' in the above equation.
L0(1−αglassΔT)=L0′(1−αaluminumΔT){L_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right) = {L_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)
⇒L0′=L0(1−αglassΔT1−αaluminumΔT)\Rightarrow {L_0}' = {L_0}\left( {\dfrac{{1 - {\alpha _{{\text{glass}}}}\Delta T}}{{1 - {\alpha _{{\text{aluminum}}}}\Delta T}}} \right)
Substitute 9×10−6 /∘C9 \times {10^{ - 6}}\,/^\circ {\text{C}} for αglass{\alpha _{{\text{glass}}}}, 24×10−6 /∘C24 \times {10^{ - 6}}\,/^\circ {\text{C}} for αaluminum{\alpha _{{\text{aluminum}}}}, 40∘C40^\circ {\text{C}} for ΔT\Delta T and 20 cm20\,{\text{cm}} for L0{L_0} in the above equation.
⇒L0′=(20 cm)[1−(9×10−6 /∘C)(40∘C)1−(24×10−6 /∘C)(40∘C)]\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]
⇒L0′=20.012 cm\Rightarrow {L_0}' = 20.012\,{\text{cm}}
Hence, the length of the aluminum frame should be 20.012 cm20.012\,{\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′B = B'
Substitute B0(1−αglassΔT){B_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right) for BB and B0′(1−αaluminumΔT){B_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right) for B′B' in the above equation.
B0(1−αglassΔT)=B0′(1−αaluminumΔT){B_0}\left( {1 - {\alpha _{{\text{glass}}}}\Delta T} \right) = {B_0}'\left( {1 - {\alpha _{{\text{aluminum}}}}\Delta T} \right)
⇒B0′=B0(1−αglassΔT1−αaluminumΔT)\Rightarrow {B_0}' = {B_0}\left( {\dfrac{{1 - {\alpha _{{\text{glass}}}}\Delta T}}{{1 - {\alpha _{{\text{aluminum}}}}\Delta T}}} \right)
Substitute 9×10−6 /∘C9 \times {10^{ - 6}}\,/^\circ {\text{C}} for αglass{\alpha _{{\text{glass}}}}, 24×10−6 /∘C24 \times {10^{ - 6}}\,/^\circ {\text{C}} for αaluminum{\alpha _{{\text{aluminum}}}}, 40∘C40^\circ {\text{C}} for ΔT\Delta T and 30 cm30\,{\text{cm}} for B0{B_0} in the above equation.
⇒B0′=(30 cm)[1−(9×10−6 /∘C)(40∘C)1−(24×10−6 /∘C)(40∘C)]\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]
∴B0′=30.018 cm\therefore {B_0}' = 30.018\,{\text{cm}}
Hence, the breadth of the aluminum frame should be 30.018 cm30.018\,{\text{cm}}.

Hence, the size of the aluminum frame should be 20.012 cm×30.018 cm20.012\,{\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.