Thermal Expansion of a brass plug

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SUMMARY

The discussion centers on calculating the common temperature required for a brass plug to fit into an iron ring, given their initial diameters at 20°C. The equations used include ΔT = ΔL / (α * L₀) and ΔL = αL₀ * ΔT. The user initially calculated a temperature of 162°C but later confirmed the correct answer to be -162.89°C after adding the initial temperature. The discussion also explores the scenario where the plug is iron and the ring is brass, emphasizing the importance of thermal expansion coefficients.

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  • Understanding of thermal expansion concepts
  • Familiarity with the equations of linear expansion
  • Knowledge of the coefficients of linear expansion for brass and iron
  • Basic algebra skills for solving equations
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0338jw
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Homework Statement


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?


Homework Equations



\Delta T = \frac{\Delta L}{\alpha * L_o}
\Delta L= \alpha L_o * \Delta T

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!
 
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0338jw said:

Homework Statement


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?


Homework Equations



\Delta T = \frac{\Delta L}{\alpha * L_o}
\Delta L= \alpha L_o * \Delta T

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!

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?
 
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.
 
0338jw said:
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.

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.
 
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 I am 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:
0338jw said:
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 I am 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..

What are they asking?

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.)
 
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?
 
0338jw said:
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?

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.
 
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!
 

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