Question

The density of lead is 11g/cm 3 at 20 0 C. Find its density at 200 0 C

Answer 1

The volume at $T_0 = 20\degree C$ is:

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

where $m$ is the mass, and $\rho_0 = 11g/cm^3 = 11000\space kg/m^ 3$ is the density at $T_0$.

The volume $V$ at $T = 200\degree C$ is given as follows (see https://en.wikipedia.org/wiki/Thermal_expansion):

$$\Delta V = \alpha_V\Delta TV_0$$

were $\Delta V = V - V_0$ is the change in volume, $\Delta T = T - T_0$ is the change in temperature, and $\alpha_V = 87\times 10^{-6}\space \degree C^{-1}$ is the volumetric expansion coefficient of lead. Thus, obtain:

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

The density at $T = 200\degree C$ is then:

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

Substituting the values, obtain:

$$\rho = \dfrac{11g/cm^3}{1 + 87\times 10^{-6} \space \degree C^{-1}(200\degree C-20\degree C)} \approx 10.8\space g/cm^3$$

Answer. $10.8\space g/cm^3$.