Thermodynamics Linear Expansion Problem

Click For Summary
SUMMARY

The problem involves calculating the common temperature required for a brass plug with a diameter of 8.749 cm to fit into an iron ring with an inside diameter of 8.739 cm. The coefficients of linear expansion are α(brass) = 19E-6 and α(iron) = 12E-6. By setting the linear expansion equations equal, the solution reveals that the temperature must be raised to approximately 183.05°C for the plug to fit into the ring. The calculations confirm that the initial approach and methodology used are correct.

PREREQUISITES
  • Understanding of linear expansion concepts
  • Familiarity with coefficients of linear expansion
  • Basic algebra for solving equations
  • Knowledge of temperature scales (Celsius and Kelvin)
NEXT STEPS
  • Study the principles of thermal expansion in solids
  • Learn how to apply linear expansion equations in practical scenarios
  • Explore the effects of temperature on different materials
  • Investigate advanced thermodynamic concepts related to material properties
USEFUL FOR

Students in physics or engineering courses, particularly those focusing on thermodynamics and material science, will benefit from this discussion.

Ling_Ling
Messages
7
Reaction score
0

Homework Statement



A brass plug is to be placed in a ring made of iron. At room temperature (20°C), the diameter of the plug is 8.749 cm and that of the inside of the ring is 8.739 cm. They must both be brought to what common temperature in order to fit?

It's a linear expansion problem. As a a ring is heated, the diameter of the hole expands.
d-plug = 8.749 cm
d-ring = 8.739 cm
⌂diameter = .01cm
α(brass) = 19E-6
α(iron) = 12E-6
⌂T = 20-T or 293-T (⌂T is constant)
To = 20°C = 293°K (Which one should I use?)

T = ?

Homework Equations


L = Lo*(1+α⌂T)


The Attempt at a Solution


Set the L's of the brass and iron rings equal to each other.
Lo*(1+α⌂T)=Lo*(1+α⌂T)
8.749(1+19E-6*⌂T) = 8.739(1+12E-6*⌂T)
8.749 + .000166⌂T = 8.739 + .000105⌂T
.01 = -.000061⌂T
⌂T = -163.05 °C (I think it's Celsius)
20-(-163.05) = 183.05°C

I attempted a few different methods to solving this, but they proved to be wrong. Am I correct in my thinking?
 
Physics news on Phys.org
You work looks good to me.
 
Okay, thank you. I just needed to check that I had the right work and thinking.
 

Similar threads

Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
9K
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
11
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K