Thermal Expansion — Question 5

Question 5

Expansion lowers density without removing material

A solid cube of fixed mass has density $\rho_0=8000\,\mathrm{kg/m^3}$ at the reference temperature. Each edge obeys the given law $L/L_0=1+\alpha\Delta T$, with $\alpha=2.00\times10^{-5}\,\mathrm{K^{-1}}$ and $\Delta T=200\,\mathrm K$.

1. Find the volume ratio and final density exactly within this edge law.

2. Compare the first-order density approximation $\rho\simeq\rho_0(1-3\alpha\Delta T)$ with the exact modeled result. Does lower density imply lower total mass?

Question 5 - Solution

Expansion lowers density without removing material

(a) Method and calculation

$$\frac V{V_0}=(1+.004)^3=1.012048064.$$$$\rho=\frac mV=\frac{8000}{1.012048064}\approx7904.763\,\mathrm{kg/m^3}.$$

(b) Interpretation and further calculation

$$\rho_{\mathrm{lin}}=8000(1-.012)=7904\,\mathrm{kg/m^3}.$$$$\rho V=\rho_0V_0=m\quad\text{unchanged}.$$The approximation is about $.763\,\mathrm{kg/m^3}$ low. Density falls because the denominator grows, not because material disappears.

Physical check

For positive expansion the density must decrease; taking the same fractional density change as the edge change would miss the three-dimensional effect.