Answer to Question #144026 in Physics for etroks

Question

The density of gold is 19.30 g/cm³ at 20.0°C, and the coefficient of linear expansion is 14.3 x 10^-6 °C^-1. Compute the density of gold at 90°C.

Answer 1

Let's assume that gols is an isotropic material. Then the volumetric thermal expansion coefficient is three times the linear coefficient:

$$\alpha_V = 3\alpha_L = 3\cdot 14.3\times 10^{-6}\degree C^{-1} = 4.29\times 10^{-5}\degree C^{-1}$$

Let's take a piece of gold of mass $m$. Then its volume at $T_0 = 20.0°C$ is:

$$V_0 = \dfrac{m}{\rho_0}$$

According to the ermal expansion law, the volume of mass 1 g at $T = 90\degree C$ is:

$$V = V_0 + \alpha_VV_0(T-T_0) = \dfrac{m}{\rho} \\$$

where $\rho$ is the new density. Obtain

$$\rho = \dfrac{m}{V_0 + \alpha_VV_0(T-T_0)} = \dfrac{m}{\dfrac{m}{\rho_0} + \alpha_V\dfrac{m}{\rho_0}(T-T_0)}=\\ =\dfrac{\rho_0}{1 + \alpha_V(T-T_0)}$$

Substituting the numbers, obtain:

$$\rho = \dfrac{\rho_0}{1 + \alpha_V(T-T_0)} = \dfrac{19.3}{1 + 4.29\times 10^{-5}\cdot 70} \approx 19.24\space g/cm^3$$

Answer. $19.24\space g/cm^3$.