# Homework Help: Find distance to which the center rises

1. Dec 6, 2012

### omc1

1. The problem statement, all variables and given/known data As a result of a temperature rise of 22 C, a bar with a crack at its center buckles upward. If the fixed distance is L = 2.79m and the coefficient of linear expansion is α = 2.79×10-4K^-1, find the distance x to which the center rises.

2. Relevant equations i used DeltaL=alpha*L1*DeltaT

3. The attempt at a solution i converted T to K by adding 273, then multiplied all the numbers and got .229 m ....not right....thanks for help!

2. Dec 6, 2012

### Staff: Mentor

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?

3. Dec 6, 2012

### omc1

the change in T=+295, i drew a picture we are looking for the height the the bar rises...

4. Dec 6, 2012

### Staff: Mentor

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. Dec 6, 2012

### omc1

i thought that for C to K conversion you add 273?

6. Dec 6, 2012

### Staff: Mentor

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. Dec 6, 2012

### omc1

um the K degree is the same as the C degree...so when is the appropriate time to use 273.15?

8. Dec 6, 2012

### Staff: Mentor

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. Dec 6, 2012

### omc1

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...

10. Dec 6, 2012

### Staff: Mentor

Right. That's the temperature change (and it's the same in °K or °C).
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. Dec 6, 2012

### omc1

ok, when I did that I got .0171 m then I added 2.79 and got 2.81 m but thats not working...it doesnt make sense to do that because it was the distance it rises...

12. Dec 6, 2012

### Staff: Mentor

You said that you drew a diagram. What does the new length of 2.81 m represent there?

13. Dec 6, 2012

### omc1

does it represent the new length of the bar?

14. Dec 6, 2012

### Staff: Mentor

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. Dec 6, 2012

### omc1

right, it split then rose, so i dont understand how to find that height...

16. Dec 6, 2012

### Staff: Mentor

Take a look at the geometry of the situation. What lengths do you know?

17. Dec 6, 2012

### omc1

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 isnt right...

18. Dec 6, 2012

### omc1

so i need to use the Pythagorean theorem
(2.81/2)^2-(2.79/2)^2=b^2 i got 0.167 m which isnt right..

19. Dec 6, 2012

### Staff: Mentor

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. Dec 6, 2012

### omc1

ok thank you!!!! thank worked finally :)