Relationship between coefficients of linear and volume expansion


by madmartigano
Tags: coefficients, expansion, linear, relationship, volume
madmartigano
madmartigano is offline
#1
Jan24-10, 05:11 PM
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
[tex]\beta = \frac{{{V_2} - {V_1}}}{{{V_1}\left( {{t_2} - {t_1}} \right)}}[/tex]
may be quoted. Here [tex]{V_1}[/tex] and [tex]{V_2}[/tex] are the initial and final volumes of the block, and [tex]{t_1}[/tex] and [tex]{t_2}[/tex] are the initial and final temperatures. Find the relation between the coefficients [tex]\alpha[/tex] and [tex]\beta[/tex].


2. Relevant equations

[tex]\alpha = \frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}[/tex]


3. The attempt at a solution

I'm assuming I need to set [tex]{V_1} = {L_1}{W_1}{H_1}[/tex] and [tex]{V_2} = {L_2}{W_2}{H_2}[/tex]

and attempt to extract [tex]\frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}[/tex] from [tex]\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)}}[/tex]

I've only gotten so far:

[tex]{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)}}[/tex]

but I can't figure out the rest of the algebraic manipulation.

Is this possible, or am I going about the problem incorrectly?
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ideasrule
ideasrule is offline
#2
Jan24-10, 08:29 PM
HW Helper
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P: 2,324
Quote Quote by madmartigano View Post
I'm assuming I need to set [tex]{V_1} = {L_1}{W_1}{H_1}[/tex] and [tex]{V_2} = {L_2}{W_2}{H_2}[/tex]

and attempt to extract [tex]\frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}[/tex] from [tex]\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)}}[/tex]
Now, express L2, W2, and H2 in terms of L1, W1, and H1. Remember that the linear expansion equation, [tex]\alpha = \frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}[/tex], 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:

[tex]{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)}}[/tex]
That step is correct algebraically, but it gets you farther from the solution.
madmartigano
madmartigano is offline
#3
Jan24-10, 11:10 PM
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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