How Do You Solve for Temperature in Thermal Expansion Calculations?

[temaire]

Homework Statement

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.

 

[HallsofIvy]

Homework Statement

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.

 

[temaire]

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 $T_1$