# Thermal Expansion of a brass plug

1. Jul 16, 2008

### 0338jw

1. The problem statement, all variables and given/known data
A brass plug is to be placed in a ring made of iron. At 20 C the diameter of the plug is 8.753 cm and that of the inside ring is 8.743 cm. They must both be brought to what common temperature in order to fit? What if the plug were iron and the ring brass?

2. Relevant equations

$$\Delta T$$ = $$\frac{\Delta L}{\alpha * L_o}$$
$$\Delta L$$= $$\alpha L_o * \Delta T$$
3. The attempt at a solution
I've attempted for the solution by setting the second equation with all the information equal to each other and solving for $$\Delta T$$ but I got 162 C fr the first part. I checked the other threads for this problem but I still can;t get the right answer using this way: https://www.physicsforums.com/archive/index.php/t-72993.html . Is the algebra in the previous thread correct? This is the only problem I'm stuck on and the final is tomorrow. All help is greatly appreciated!!!

2. Jul 16, 2008

### LowlyPion

Is this the "second" equation you were using because if you set just the $$\Delta$$'s equal you wouldn't get the right answer now would you?

$$L + \Delta L$$= $$L + \alpha L_o * \Delta T$$

Then they match when

$$L1 + \Delta L1$$= $$L2 + \Delta L2$$

This you solve for $$\Delta T$$ correct?

3. Jul 16, 2008

### 0338jw

I did use the first equation. Is the algebra correct from the other thread, because i tried that method as well and i still did not get the right answer. Was I correct in setting the second equation equal to the other side? I did my algebra differently from the original thread and got closer to the answer but still not there.

4. Jul 16, 2008

### LowlyPion

I think the equations I gave you in my last post were equivalent to the method of the archive thread. Sometimes it's easier to solve for the variable in terms of the other variables and then plug the numbers in.

5. Jul 16, 2008

### 0338jw

I'm sorry I'm still a little confused. With the equations thatyou gave me, do I solve for the second change then plug that in and solve for T? Again, sorry if im not picking up the hints immediately. I'm not one to complain but i do have quite a bit on my plate for the next two weeks..

Last edited: Jul 16, 2008
6. Jul 16, 2008

### LowlyPion

They are asking at what temp (which you find from change in temp + original temp) will the diameters of the 2 pieces be equal. The piece in the hole touches uniformly the edge of the hole.

Writing 2 equations

$$L_o_1 + \Delta L1$$ = $$L_o_1 + \alpha_1 L_o_1 * \Delta T$$
$$L_o_2 + \Delta L2$$ = $$L_o_2 + \alpha_2 L_o_2 * \Delta T$$

To satisfy the condition of the problem, the two LH sides must be equal to each other. Hence aren't the 2 RH sides equal?

$$L_o_1 + \alpha_1 L_o_1 * \Delta T$$ = $$L_o_2 + \alpha_2 L_o_2 * \Delta T$$

I would solve for $$\Delta T$$ as variables and then plug in the variable values.
For part 2 swap the expansion coefficients. (Presumably these values are the same as the archive.)

7. Jul 16, 2008

### 0338jw

I solved for T, and I got -189.97. Do I need to plug this back into anything for the change in temperature? Is my T correct?

8. Jul 16, 2008

### LowlyPion

Sorry I didn't do the math, I don't have the coefficients, but think about it. You should get a hotter number if the ring is smaller but expands faster and a colder number if it shrinks more slowly than the rod.

Remember your number from the equation is $$\Delta T$$ and you need to add it to the initial temp.

Good luck.

9. Jul 17, 2008

### 0338jw

Thank you VERY much for your help! Turns out I did have the right answer (-162.89) and i just needed to add the initial temp. again, thanks for the speedy reply and consistent help!