- #1
omc1
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Find distance to which the center rises
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Homework Help Overview
The problem involves a bar that buckles upward due to a temperature rise of 22°C, with a fixed length of 2.79m and a coefficient of linear expansion. Participants are tasked with determining the distance the center of the bar rises as a result of this temperature change.
Discussion Character
- Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking
Approaches and Questions Raised
- Participants discuss the correct interpretation of temperature change and its conversion between Celsius and Kelvin. There is an exploration of the formula for linear expansion and how it applies to the problem. The geometry of the situation is also examined, particularly in relation to the Pythagorean theorem.
Discussion Status
The discussion has progressed with participants clarifying the temperature change and its implications for calculations. Some have provided guidance on maintaining precision in calculations and the importance of understanding the geometry involved in the problem. There is ongoing exploration of how to accurately determine the height the center of the bar rises.
Contextual Notes
Participants are working within the constraints of the problem statement, focusing on the implications of temperature change and the physical setup of the bar. There is an emphasis on avoiding inaccuracies in calculations due to rounding errors.
- #2
gneill
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DeltaT is not a temperature, it's a change in temperature. What's the change in temperature in K?
Now, draw a picture of the before and after scenarios. What distance are you looking for?
Now, draw a picture of the before and after scenarios. What distance are you looking for?
- #3
omc1
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the change in T=+295, i drew a picture we are looking for the height the the bar rises...
- #4
gneill
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omc1 said:
How can the change in temperature be 295K if the problem states that it changes by 22C? How does the size of a K degree compare to a C degree?
- #5
omc1
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i thought that for C to K conversion you add 273?
- #6
gneill
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omc1 said:
That's to convert a particular temperature on the C scale to one on the K scale. Here you are given a CHANGE in temperature. So what counts is the size of the degrees on each scale.
I'll ask again, how does the size of a K degree compare to a C degree?
You could also go through the exercise of picking an arbitrary starting temperature in C, along with a change in temperature ΔC, and then converting starting and ending temperatures from C to K and finding the difference in the resulting K temperatures...
- #7
omc1
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um the K degree is the same as the C degree...so when is the appropriate time to use 273.15?
- #8
gneill
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omc1 said:
Use the 273K offset when you need to determine a particular temperature on the absolute scale (for example, if you have a formula that needs an absolute temperature value, such as the Steffan-Boltzmann law for radiated power, or the junction temperature in the Diode Equation, or if you want to determine the total heat energy held by a given body compared to an absolute zero background). If you're just working with temperature differences, then they will be the same on both scales.
- #9
omc1
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ok, I think I see now...so I should use 22 instead of 295...
is the formula I used right, I am still confused about how to solve this...
is the formula I used right, I am still confused about how to solve this...
- #10
gneill
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Right. That's the temperature change (and it's the same in °K or °C).omc1 said:
The formula you've used will give you the change in length. Add the change in length to the original length to get the new overall length.
- #11
omc1
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ok, when I did that I got .0171 m then I added 2.79 and got 2.81 m but that's not working...it doesn't make sense to do that because it was the distance it rises...
- #12
gneill
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omc1 said:
You said that you drew a diagram. What does the new length of 2.81 m represent there?
- #13
omc1
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does it represent the new length of the bar?
- #14
gneill
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omc1 said:
Yes. Well, to be more specific, it represents the total length of the two halves of the bar... remember it split along a crack at its center...
- #15
omc1
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right, it split then rose, so i don't understand how to find that height...
- #16
gneill
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omc1 said:
Take a look at the geometry of the situation. What lengths do you know?
- #17
omc1
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ok so if i split the triangle in half we know the hypotenuse and the length of one side, so I would just need to use the Pythagorean theorem to find the height?
(2.81/2)^2-(2.79/2)^2=b^2 i got 0.167 m which isn't right...
(2.81/2)^2-(2.79/2)^2=b^2 i got 0.167 m which isn't right...
- #18
omc1
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so i need to use the Pythagorean theorem
(2.81/2)^2-(2.79/2)^2=b^2 i got 0.167 m which isn't right..
(2.81/2)^2-(2.79/2)^2=b^2 i got 0.167 m which isn't right..
- #19
gneill
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omc1 said:
When values are being squared and square-rooted, you should keep more decimal places in the intermediate results in order to avoid introducing inaccuracies due to roundoff errors. You'll find that the result is very sensitive to small differences in the values. In fact, it's always a good idea to keep several "extra" decimal places in all intermediate values, and only round the final result.
Recalculate your "new" length, keeping a few more decimal places in the result.
- #20
omc1
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ok thank you! thank worked finally :)
- #21
gneill
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omc1 said:
Excellent. Just remember to always retain full precision through intermediate calculations...
- #22
omc1
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ok thanks :)
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