Linear expansion coefficient of glass

AI Thread Summary
The linear expansion coefficient of glass is 9 × 10−6 (°C)−1, and the discussion revolves around calculating the minimum spacing needed around a windshield measuring 60 cm by 400 cm to prevent breakage due to a temperature change of 150°F. Participants emphasize the importance of correctly converting the temperature change to Celsius, noting that the conversion should be done as (delta t)*(5/9) without subtracting 32. The correct formula to use is delta L = alpha * Lo * delta t, where delta L represents the total expansion. One contributor clarifies that the temperature change is not a fixed temperature, which is crucial for accurate calculations. The final spacing required at each end is derived from the total expansion value calculated.
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Given: The linear expansion coefficient of glass is 9 × 10−6 (◦C)−1. An automobile windshield has dimensions of 60 cm by 400 cm. What minimum spacing around the wind-shield is needed to prevent the windshield from breaking if the temperature changes by 150◦F? Answer in units of mm.

I understand I use the equation alpha*Length*delta(t) but it won't pop out the right answer and I think the dimensions are what is throwing me off. Any advice is appreciated!
 
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Show what you did. I would have chosen the 400 cm dimension to work with, as it will change the most. Did you convert the temp change to degrees C?
 
I did not change to degrees Celsius. The conversion is (5/9)*(150-32) so my new temp is 65.556. Also, I have to change cm -> mm. so the dimension would be 4000 mm. So, my final equation would be (9e-6)(4000)(65.556) ??
 
nslinker said:
So, my final equation would be (9e-6)(4000)(65.556) ??
Yes, but don't forget that that's the total increase in that dimension. So what minimum spacing would you need at each end?
 
My answer from that comes out to be 2.36 but I don't know how to find the minimum spacing piece that you're talking about.
 
If the entire glass expands by 2.36, how much space do you have to allow at each end?
 
the given dimensions + expansion value.
 
so i don't know if you allready figured out your problem... but i just did that exact problem. the error that you made was that the temp that is given is a (delta t). to convert from F to C it is just (delta t)*(5/9)... you don't subtract 32. that should fix it! after that it is just:

delta L = alpha*Lo*delta t and that should be your answer.
 
mochi12 said:
the error that you made was that the temp that is given is a (delta t). to convert from F to C it is just (delta t)*(5/9)... you don't subtract 32.
Good catch! (I wasn't paying attention when the OP did the conversion. :rolleyes:)

nslinker said:
I did not change to degrees Celsius. The conversion is (5/9)*(150-32) so my new temp is 65.556.
Fix that conversion. You're converting a temperature change, not a temperature.
 
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