Relationship between coefficients of linear and volume expansion

AI Thread Summary
The discussion focuses on deriving the relationship between the coefficients of linear expansion (α) and volume expansion (β) for a solid block. The volume expansion is defined by the formula β = (V2 - V1) / (V1(t2 - t1)), while linear expansion is described by α = (L2 - L1) / (L1(t2 - t1)). Participants suggest expressing the volume in terms of its dimensions (length, width, height) and applying the linear expansion formula to each dimension. A simplified approach is recommended, using the equation Lf = Li(1 + α * ΔT) to derive the volume expansion relation. Ultimately, the discussion concludes with a participant successfully solving the problem after clarifying their approach.
madmartigano
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Homework Statement



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.


Homework Equations



\alpha = \frac{{{L_2} - {L_1}}}{{{L_1}\left( {{t_2} - {t_1}} \right)}}


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?
 
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madmartigano said:
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.
 
You helped me see that I was just over-thinking the problem--I got it figured out. Thank you.
 
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