Question
(a) Suppose a meter stick made of steel and one made of invar (an alloy of iron and nickel) are the same length at 0C0^\circ\textrm{C}. What is their difference in length at 22.0C22.0^\circ\textrm{C}? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes.
Question by OpenStax is licensed under CC BY 4.0
Final Answer
  1. 2.4×104 m2.4\times 10^{-4}\textrm{ m}
  2. 7.4 mm7.4\textrm{ mm}

Solution video

OpenStax College Physics, Chapter 13, Problem 16 (Problems & Exercises)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We considered two meter sticks which will have an initial length of one meter because they are meter sticks. And one is made out of steel which has a coefficient of linear expansion of 12 times 10 to the minus 6, and the other one is made out of invar an alloy of nickel and iron which has a coefficient of linear expansion of 0.9 times 10 to the minus 6. And the initial temperature is 0 degrees Celsius and the final temperature is 22. And the question is: what will their length differences be after the temperature increases from zero to 22? So they both have an initial length of one meter at 0 degrees. So this is the before scenario and then after the temperature increases to 22 the steel will increase more and the invar will increase less. And the question is: what will their length difference be? And so we have L 0 plus Delta L subscript S for steel is the total length of the steel bar minus the total length of the invar meter stick L 0 plus change in length of invar. And then if you distribute the minus into the brackets there you get L 0 plus delta L S minus L 0 minus Delta L invar. And so these L-naughts make zero and we're left with the difference in lengths will be the difference in the length changes of steel versus invar. And so, the change in length of one of the bars will be coefficient of linear expansion multiplied by its initial length times the change in temperature. So we make substitutions for each of these here. They have common factors L 0 and delta T. This delta T does not get a subscript s and this one does not get a subscript inv because the change in temperatures are the same for both of them. So there's no need to distinguish them. And same with L 0 , they both have the same initial length. So those common factors can be factored out of the bracket. And the change in temperature is the final minus the initial temperatures. And so we have 1 meter initial length multiplied by 22 degrees Celsius minus 0 degrees Celsius. And I'm taking this to have some extra digits of precision here. The question doesn't explicitly say that but why not? All the other numbers have three significant figures. And we multiply that by 12 times 10 to the minus 6 per Celsius degree minus 0.9 times 10 to the minus 6 per Celsius degree. And we get 2.4 times 10 to the minus 4 meters will be their differences in their lengths at 22 degrees Celsius. And that is 0.2 millimeters. Now what if you had a 30-meter surveyors measuring tape? Then the difference in length will be all the same as before except we're using the number 30 for the initial length instead of 1 meter for the initial length. So we have 30 meters times change in temperature times the difference in the coefficients of linear expansion and that gives 7.4 millimeters.