# Relationship between coefficients of linear and volume expansion

by madmartigano
Tags: coefficients, expansion, linear, relationship, volume
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 P: 2 1. The problem statement, all variables and given/known data If a solid material is in the form of a block rather than a rod, its volume will grow larger when it is heated, and a coefficient of volume expansion beta defined by $$\beta = \frac{{{V_2} - {V_1}}}{{{V_1}\left( {{t_2} - {t_1}} \right)}}$$ may be quoted. Here $${V_1}$$ and $${V_2}$$ are the initial and final volumes of the block, and $${t_1}$$ and $${t_2}$$ are the initial and final temperatures. Find the relation between the coefficients $$\alpha$$ and $$\beta$$. 2. Relevant equations $$\alpha = \frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$$ 3. The attempt at a solution I'm assuming I need to set $${V_1} = {L_1}{W_1}{H_1}$$ and $${V_2} = {L_2}{W_2}{H_2}$$ and attempt to extract $$\frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$$ from $$\frac{{{L_2}{W_2}{H_2} - {L_1}{W_1}{H_1}}}{{{L_1}{W_1}{H_1}\left( {{t_2} - {t_1}} \right)}}$$ I've only gotten so far: $${W_1}{H_1}B = \frac{{{L_2}{W_2}{H_2} - {L_1}{W_1}{H_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$$ but I can't figure out the rest of the algebraic manipulation. Is this possible, or am I going about the problem incorrectly?
HW Helper
P: 2,322
 Quote by madmartigano I'm assuming I need to set $${V_1} = {L_1}{W_1}{H_1}$$ and $${V_2} = {L_2}{W_2}{H_2}$$ and attempt to extract $$\frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$$ from $$\frac{{{L_2}{W_2}{H_2} - {L_1}{W_1}{H_1}}}{{{L_1}{W_1}{H_1}\left( {{t_2} - {t_1}} \right)}}$$
Now, express L2, W2, and H2 in terms of L1, W1, and H1. Remember that the linear expansion equation, $$\alpha = \frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$$, applies for the width and height too.

A less messy way to do this problem is to write the linear expansion equation as Lf=Li(1+alpha*delta-T). Then LWH=Li(1+alpha*delta-T)*W*(1+alpha*delta-T)...you get the idea.

 I've only gotten so far: $${W_1}{H_1}B = \frac{{{L_2}{W_2}{H_2} - {L_1}{W_1}{H_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}$$
That step is correct algebraically, but it gets you farther from the solution.
 P: 2 You helped me see that I was just over-thinking the problem--I got it figured out. Thank you.

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