Thermal Expansion — Question 3

Question 3

Area expansion retains a small quadratic correction

A square plate has initial area $A_0=.500\,\mathrm{m^2}$. Every side follows $L=L_0(1+\alpha\Delta T)$, with $\alpha=2.00\times10^{-5}\,\mathrm{K^{-1}}$ and $\Delta T=100\,\mathrm K$. Treat this length law as the given model.

1. Find the area change exactly within that model and compare it with $\Delta A\simeq2\alpha A_0\Delta T$.

2. Find the largest positive temperature rise for which the omitted quadratic term is at most 1% of the retained linear area-change term.

Question 3 - Solution

Area expansion retains a small quadratic correction

(a) Method and calculation

$$A=A_0(1+\alpha\Delta T)^2,\quad\Delta A=A_0[2\alpha\Delta T+(\alpha\Delta T)^2].$$$$\Delta A=.500(.004+.000004)=.002002\,\mathrm{m^2}.$$The first-order result is $.002000\,\mathrm{m^2}$, low by $2.00\times10^{-6}\,\mathrm{m^2}$.

(b) Interpretation and further calculation

For a positive rise,$$\frac{A_0(\alpha\Delta T)^2}{2A_0\alpha\Delta T}=\frac{\alpha\Delta T}{2}\le.01.$$$$\Delta T\le\frac{.02}{2.00\times10^{-5}}=1000\,\mathrm K.$$This is a mathematical truncation limit; real material coefficients need not remain constant over so large a range.

Physical check

The correction is positive because expansion in both directions creates an additional corner area.