How Do You Solve for Temperature in Thermal Expansion Calculations?

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SUMMARY

The discussion focuses on solving for temperature in thermal expansion calculations using the differential equation α(T)dT = ∂L/L. Participants detail the integration process to derive the relationship between length and temperature, ultimately expressing l in terms of T. The correct approach involves substituting l for l_1 and solving for T_1 as a function of l, rather than isolating l directly. The equation α(K) = α₀ + α₁T is crucial for this transformation.

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temaire
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Homework Statement



2ebx3is.png


Homework Equations



\alpha(T)dT = \frac{\partial L}{L} <--- Differential Equation given on formula sheet

The Attempt at a Solution



\alpha(T)dT = \frac{\partial L}{L}

\int_{l_0}^{l_1} \frac{\partial L}{L} = \int_{T_0}^{T_1} \alpha(T)dT

ln(\frac{l_1}{l_0}) = \int_{T_0}^{T_1} [ \alpha_{0} + \alpha_{1} T]dT

ln(\frac{l_1}{l_0}) = \alpha_{0} (T_1 - T_0) + \frac{\alpha_{1} (T_1 - T_0)^{2}}{2}

\frac{l_1}{l_0} = e^{\alpha_{0} (T_1 - T_0) + \frac{\alpha_{1} (T_1 - T_0)^{2}}{2}}

I know my answer is not correct, since the question asked to have the equation in terms of l. I'm also confused as to whether or not I was even supposed to use the equation that was given in the preamble, \alpha(K) = \alpha_{0} + \alpha_{1} T. I would appreciate any help.
 
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temaire said:

Homework Statement



2ebx3is.png


Homework Equations



\alpha(T)dT = \frac{\partial L}{L} <--- Differential Equation given on formula sheet

The Attempt at a Solution



\alpha(T)dT = \frac{\partial L}{L}

\int_{l_0}^{l_1} \frac{\partial L}{L} = \int_{T_0}^{T_1} \alpha(T)dT

ln(\frac{l_1}{l_0}) = \int_{T_0}^{T_1} [ \alpha_{0} + \alpha_{1} T]dT

ln(\frac{l_1}{l_0}) = \alpha_{0} (T_1 - T_0) + \frac{\alpha_{1} (T_1 - T_0)^{2}}{2}
You are good up to here. Instead of solving for l as you do below, solve for T_1 as a function l_1. Taking l_0 and T_0 as given initial values, replace T_1 with T and l_1 with l to get T as a function of l.

\frac{l_1}{l_0} = e^{\alpha_{0} (T_1 - T_0) + \frac{\alpha_{1} (T_1 - T_0)^{2}}{2}}

I know my answer is not correct, since the question asked to have the equation in terms of l. I'm also confused as to whether or not I was even supposed to use the equation that was given in the preamble, \alpha(K) = \alpha_{0} + \alpha_{1} T. I would appreciate any help.
 
HallsofIvy said:
You are good up to here. Instead of solving for l as you do below, solve for T_1 as a function l_1. Taking l_0 and T_0 as given initial values, replace T_1 with T and l_1 with l to get T as a function of l.

ln(l_1) - ln(l_0) = \alpha_0 T_1 - \alpha T_0 + \frac{\alpha_1 (T_{1}^2 - 2T_1 T_0 + T_{0}^2)}{2}

ln(l_1) - ln(l_0) = \alpha_0 T_1 - \alpha T_0 + \frac{\alpha_1 T_{1}^2}{2} - \alpha_1 T_1 T_0 + \frac{\alpha_1 T_{0}^2}{2}

It's getting pretty messy. I'm not really sure how I can isolate for T_1
 

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