Thermal Expansion: Glass rectangle

In summary, the problem involves finding the initial area, length, and width of a glass plate that will fit a frame at a different temperature. Using the area expansion formula and the given values, the initial area is calculated to be 2.9988 ft2. The initial length and width are then found using the relationship L=3W, resulting in values of 2.9944 ft and 0.9998 ft respectively. Alternatively, the values can be checked using linear expansion/contraction.
  • #1
paulie
13
0

Homework Statement


A rectangular windshield is to be assembled by installing a glass plate on a 3 ft by 1 ft frame at 60°C. The glass plate is cut at 68°F in such a way that its length is three times its width. The linear expansivity of glass is 5 x 10-6 /C°. (a) What area of the glass plate at 68°F will exactly fit the frame at 60°C? (b) What length of the glass plate at 68°F will exactly fit the frame at 60°C? (c) What width of the glass plate at 68°F will exactly fit the frame at 60°C?

Homework Equations


Af = Ao(1+2αΔT)
Where:
Af = Final Area
Ao = Initial Area
α = coefficient of expansion
ΔT = Change in Temperature

L = 3W
Where:
L = Length
W = Width

The Attempt at a Solution


From what I understand, I am suppose to find the initial area of the glass plate to be fitted in the frame.
First, manipulate the area expansion formula to find Ao:
Ao = Af / (1+2αΔT)
Substituting the values, I will arrive with:
Ao = 2.9988 ft2 → (a)
Next, find the length of the initial area by using L=3W, where:
Ao = L*W
Ao = L*L/3
L = √(3*Ao)
Substituting the values, I will arrive with:
L = 2.9944 ft → (b)
Finally, using L=3W to find the width:
L=3W
W=L/3
W=0.9998 ft → (c)

Is this the right answer? The initial values seems too close to the final.
 
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  • #2
Hi paulie,

Welcome to Physics Forums!

I see a slightly different value for L, perhaps due to rounding or truncation? How many decimal places did you maintain for intermediate values in your calculations?

Another approach you might try in order to check your results is to go directly for the length and width values. You know that the length and width at 60°C are 3 ft and 1 ft respectively. What are those values at the cutting temperature? This would be a linear contraction, not an area.
 
  • #3
Did you get 2.9994 ft for the length? My solution paper gives that value, it seems I typed it incorrectly.

Tried checking using linear expansion/contraction and it gives the same value, never knew I can use this to check.
 
  • #4
paulie said:
Did you get 2.9994 ft for the length? My solution paper gives that value, it seems I typed it incorrectly.
Yes, that's what I got.
Tried checking using linear expansion/contraction and it gives the same value, never knew I can use this to check.
:smile: Now you know!
 
  • Like
Likes paulie
  • #5
Thanks for you help! :)
 

1. What is thermal expansion?

Thermal expansion is the tendency of a material to expand or contract when exposed to changes in temperature. This is due to the atoms within the material vibrating at a higher frequency when heated, causing them to take up more space.

2. Why does glass expand when heated?

Glass expands when heated because it is an amorphous solid, meaning its atoms are arranged randomly and have more freedom to move compared to a crystalline solid. As the temperature increases, the atoms vibrate more and take up more space, causing the glass to expand.

3. How does thermal expansion affect a glass rectangle?

Thermal expansion can cause a glass rectangle to expand or contract depending on the temperature. This can lead to changes in the dimensions of the rectangle, such as the length, width, and thickness.

4. What is the coefficient of thermal expansion for glass?

The coefficient of thermal expansion for glass varies depending on the type of glass. On average, it is around 8.5 x 10^-6/°C, meaning that for every degree Celsius increase in temperature, the glass will expand by 0.0000085 times its original length.

5. How can thermal expansion be managed in glass rectangles?

To manage thermal expansion in glass rectangles, materials with similar coefficients of thermal expansion can be used for the frame or support of the glass. Additionally, the use of expansion joints or allowing for some flexibility in the structure can help accommodate the expansion and prevent damage to the glass.

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