How Do You Solve for Temperature in Thermal Expansion Calculations?

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The discussion focuses on solving for temperature in thermal expansion calculations using the differential equation α(T)dT = ∂L/L. The user attempts to integrate this equation but realizes their solution is incorrect as it does not express the equation in terms of length (l). Guidance is provided to instead solve for T_1 as a function of l_1, while keeping initial values l_0 and T_0 constant. The user expresses confusion about using the provided equation for α(K) and struggles with isolating T_1 in their calculations. The conversation emphasizes the need for clarity in rearranging the equation to find temperature based on length.
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
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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