Giancoli's Physics: Principles with Applications, 7th Edition

13

Temperature and Kinetic Theory

Change chapter13-1: Atomic Theory

13-2: Temperature and Thermometers

13-4: Thermal Expansion

13-5: Gas Laws; Absolute Temperature

13-6 and 13-7: Ideal Gas Law

13-8: Ideal Gas Law in Terms of Molecules; Avogadro's Number

13-9: Molecular Interpretation of Temperature

13-11: Real Gases; Phase Changes

13-12: Vapor Pressure and Humidity

13-13: Diffusion

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.

Problem 11

Q

A

$\Delta l_{inv} = 1.7\% \Delta l_s$

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VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. The amount that the 1.8 meter table made of super invar will expand equals the coefficient of linear expansion for a super invar of 0.2 times 10 to the minus 6 times 1.8 meters times the 6 Celsius degrees temperature change, and that's 2.2 times 10 to the minus 6 meters which is very small, 2.2 micrometers. The ratio of change in length of the super invar table versus the change in length of a steel table is coefficient of linear expansion for invar times *l naught* times *Δt* divided by coefficient of linear expansion for steel times the same *l naught* times *Δt*. So, these things cancel meaning ratio and length changes will be the ratio in the coefficients of linear expansion. So, that's 0.2 times 10 to the minus 6 divided by 12 times 10 to the minus 6 which is 0.01667 and multiply that by 100% and you get this. So, the change in length for the super invar table will be about 1.7% the change in length that would happen with a steel table.

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