#$&* PHY 202
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Given Solution: ** The coefficient of volume expansion for quartz is 1 x 10^(-6) C^(-1). The sphere has diameter 8.75 cm, so its volume is 4/3 pi r^3 = 4/3 pi ( 4.38 cm)^3 = 352 cm^3, approx.. The coefficient of volume expansion is the proportionality constant beta in the relationship `dV = beta * V0 * `dT (completely analogous to the concept of a coefficient of linear expansion). We therefore have `dV = beta* V0*dT = 3 x 10^(-6) C^ (-1) * 352 cm^3 * (200C - 30 C) = 0.06 cm^3 ** Self-Critique: I am at a total loss as to where the sphere diameter in your solution came from. I found myself unable to answer the second part of the question because I know where to find the information for initial volume or temperature change. ------------------------------------------------ Self-Critique Rating:
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: @& The full statement of the problem is in the text. I can't completely quote the text in these documents, due to copyright laws. So the statements you see here are abbreviated statements, meant for reference only. ** ** I looked in my textbook and my #13.12 is entirely different than the one you are referencing. I have the sixth edition of Physics by Giancoli, which you had stated would be okay for the course. Am I looking in the wrong place or are you referencing the newer edition? ** ** @& I'm referencing the sixth edition, which I have in my hand. Problem 13.12 is at the bottom of the first column of problems at the end of the chapter, and it does reference the quartz sphere. What is the nature of problem 13.12 in your text? *@