Thermal Expansion — Question 6

Question 6

A temperature-dependent expansion coefficient must be integrated

A freely expanding rod obeys the differential definition $\alpha(T)=L^{-1}dL/dT=a+b(T-T_0)$, with $a=1.00\times10^{-5}\,\mathrm{K^{-1}}$, $b=2.00\times10^{-8}\,\mathrm{K^{-2}}$ and initial length $L_0=1.000\,\mathrm m$. Raise its temperature by $100\,\mathrm K$.

1. Integrate the differential relation and calculate the extension.

2. Compare with using only the initial coefficient in $\Delta L=\alpha(T_0)L_0\Delta T$. Find the error and explain why an arithmetic average of endpoint coefficients is useful here.

Question 6 - Solution

A temperature-dependent expansion coefficient must be integrated

(a) Method and calculation

$$\ln\frac L{L_0}=\int_{T_0}^{T_0+\Delta T}\alpha(T)dT=a\Delta T+\frac b2(\Delta T)^2=.0011.$$$$\Delta L=L_0(e^{.0011}-1)\approx1.100605\,\mathrm{mm}.$$

(b) Interpretation and further calculation

$$\Delta L_{\mathrm{initial}}=1(10^{-5})(100)=1.000\,\mathrm{mm}.$$$$\overline\alpha=\frac{a+(a+b\Delta T)}2=1.10\times10^{-5}\,\mathrm{K^{-1}}.$$The shortcut is low by $.100605$ mm. Because $\alpha$ is linear, the endpoint average gives the exact integral exponent; using $L_0\overline\alpha\Delta T$ still linearizes the exponential.

Physical check

The differential definition and the finite linear approximation are distinct; their difference is second order in the accumulated expansion.