The densities of a liquid at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ are respectively 1.0127 and…
The densities of a liquid at $0^{\circ} \mathrm{C}$ and $100^{\circ} \mathrm{C}$ are respectively 1.0127 and 1 . A specific gravity bottle is filled with $300 \mathrm{~g}$ of the liquid at $0^{\circ} \mathrm{C}$ upto the brim and it is heated to $100^{\circ} \mathrm{C}$. Then, the mass of the liquid expelled in grams is: (Coefficient of linear expansion of glass $\left.=9 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right)$
$\frac{3}{10.1}$
$\frac{3}{1.01}$
$\frac{3.81}{1.0127}$
$\frac{3.81}{0.0127}$
Solution
Density at $0^{\circ} \mathrm{C}, \rho_0=1.0127$
Density at $100^{\circ} \mathrm{C}, \rho_{100}=1$
Coefficient of real expansion of liquid,
$\gamma_{\text {real }}=\frac{\rho_0-\rho_{100}}{\rho_{100} \times \Delta t}$
$=\frac{1.0127-1}{1 \times 100}=0.0127 \times 10^{-2}$
$=1.27 \times 10^{-4}$
$\gamma_{\text {real }}=\gamma_{\text {app }}+\gamma_g$
$\gamma_g=$ coefficient of volume expansion of glass $=3 \alpha$
$\therefore \quad 1.27 \times 10^{-4}=\gamma_{\mathrm{app}}+3 \alpha$
$1.27 \times 10^{-4}=\gamma_{\text {app }}+3 \times 9 \times 10^{-6}$
$\gamma_{\mathrm{app}}=1.27 \times 10^{-4}-27 \times 10^{-6}$
$=1.27 \times 10^{-4}-0.27 \times 10^{-4}$
$=1 \times 10^{-4}$
$\therefore \quad \gamma_{\text {app }}=\frac{\text { mass expelled }}{\text { remaining mass } \times \Delta t}$
$1 \times 10^{-4}=\frac{m_1-m_2}{m_2 \times 100}$
$\left(\frac{m_1}{m_2}-1\right)=1 \times 10^{-4} \times 100=10^{-2}$
$\frac{m_1}{m_2}=1+10^{-2}=1.01$
$m_2=\frac{m_1}{1.01}=\frac{300}{1.01}$
Mass expelled $=m_1-m_2$
$=300-\frac{300}{1.01}$
$=\frac{3}{1.01}$