Thermal Expansion of two Metals

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SUMMARY

The discussion centers on calculating the common temperature at which an aluminum disk fits precisely into a steel ring, given their thermal expansion coefficients: alpha(Steel) = 1.3e-5 and alpha(Aluminum) = 1.9e-5. The user attempted to solve the problem using the formula deltaL/Lo = alpha*deltaT but encountered an impossible result of a temperature below absolute zero. This indicates a misunderstanding in the application of thermal expansion principles, particularly in how the coefficients affect the dimensions of the materials as temperature changes.

PREREQUISITES
  • Understanding of thermal expansion coefficients
  • Familiarity with the formula deltaL/Lo = alpha*deltaT
  • Basic knowledge of temperature scales, including absolute zero
  • Ability to set up and solve equations involving linear dimensions
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  • Review the principles of thermal expansion and how they apply to different materials
  • Learn about the implications of coefficients of thermal expansion in engineering applications
  • Explore the concept of absolute zero and its significance in thermodynamics
  • Practice solving thermal expansion problems with varying coefficients and initial conditions
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Students in physics or engineering, particularly those studying material properties and thermal dynamics, as well as educators looking for examples of thermal expansion applications.

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Homework Statement



The inner diameter of a steel ring is 2.0000 cm, and the diameter of an aluminum
disk is 2.0100 cm. Both are at 430 degrees C. At what common temperature will the disk t
precisely into the hole in the steel ring?

alpha(Steel) = 1.3e-5
alpha(Al) = 1.9e-5

Homework Equations


deltaL/Lo = alpha*deltaT


The Attempt at a Solution



What I told myself is this: If these the steel is to fall through the aluminum ring, then the length of its diameter at the temperature of interest must equal the length of the ring's diameter. Therefore, we must solve for the length of each diameter at temperature T and set them equal to each other. Therefore:

deltaL(steel) = alpha(steel)*Lo(steel)*deltaT
Lfinal Steel = Lo(steel) + alpha(steel)*Lo(steel)*(T-430) <----Eq.1

deltaL(Al) = alpha(Al) * Lo(Al) * deltaT
Lfinal Al = Lo(Al) +alpha(steel)*Lo(Al)*(T-430)<------Eq.2

Setting Eq.1 and Eq.2 equal to each other, I solved for T.

The T I got was below absolute Zero... Now I intuitvtely know that the T must be less than 430 since the coefficient of expansion of Al is higher than Steel, and so it contracts faster. However, I do not see how I could get a value less than 0K...

Can someone please tell me where I am going wrong? Thank you very much
 
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I got delta T = 820 degrees, same conclusion.
Wikipedia has alpha values more divergent - 2.3 for Al and 1.1 to 1.3 for steel (x 10^-5).
 

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