The density of mercury at exactly 0^∘ C is 13600 kg / m^3,...

The density of mercury at exactly $0^{\circ} \mathrm{C}$ is $13600 \mathrm{~kg} / \mathrm{m}^{3}$, and its volume expansion coefficient is $1.82 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. Calculate the density of mercury at $50.0^{\circ} \mathrm{C}$. Let $$ \begin{array}{l} \rho_{0}=\text { Density of mercury at } 0{ }^{\circ} \mathrm{C} \\ \rho_{1}=\text { Density of mercury at } 50^{\circ} \mathrm{C} \\ V_{0}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 0{ }^{\circ} \mathrm{C} \\ V_{1}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 50^{\circ} \mathrm{C} \end{array} $$ Since the mass does not change, $m=\rho_{0} V_{0}=\rho_{1} V_{1}$, from which it follows that $$ \rho_{1}=\rho_{0} \frac{V_{0}}{V_{1}}=\rho_{0} \frac{V_{0}}{V_{0}+\Delta V}=\rho_{0} \frac{1}{1+\left(\Delta V / V_{0}\right)} $$ But $$ \frac{\Delta V}{V_{0}}=\beta \Delta T=\left(1.82 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\right)\left(50.0^{\circ} \mathrm{C}\right)=0.00910 $$ Substitution into the first equation yields $$ \rho_{1}=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right) \frac{1}{1+0.00910}=13.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3} $$

The density of mercury at exactly $0^{\circ} \mathrm{C}$ is $13600 \mathrm{~kg} / \mathrm{m}^{3}$, and its volume expansion coefficient is $1.82 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}$. Calculate the density of mercury at $50.0^{\circ} \mathrm{C}$.
Let
$$
\begin{array}{l}
\rho_{0}=\text { Density of mercury at } 0{ }^{\circ} \mathrm{C} \\
\rho_{1}=\text { Density of mercury at } 50^{\circ} \mathrm{C} \\
V_{0}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 0{ }^{\circ} \mathrm{C} \\
V_{1}=\text { Volume of } m \mathrm{~kg} \text { of mercury at } 50^{\circ} \mathrm{C}
\end{array}
$$
Since the mass does not change, $m=\rho_{0} V_{0}=\rho_{1} V_{1}$, from which it follows that
$$
\rho_{1}=\rho_{0} \frac{V_{0}}{V_{1}}=\rho_{0} \frac{V_{0}}{V_{0}+\Delta V}=\rho_{0} \frac{1}{1+\left(\Delta V / V_{0}\right)}
$$
But
$$
\frac{\Delta V}{V_{0}}=\beta \Delta T=\left(1.82 \times 10^{-4}{ }^{\circ} \mathrm{C}^{-1}\right)\left(50.0^{\circ} \mathrm{C}\right)=0.00910
$$
Substitution into the first equation yields
$$
\rho_{1}=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right) \frac{1}{1+0.00910}=13.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}
$$

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Key Concepts

Conservation of Mass

Conservation of mass is a principle stating that mass remains constant in a closed system, regardless of changes in the state of matter or physical conditions. This concept is essential when linking changes in volume to changes in density, as it ensures that, despite thermal expansion, the total mass of the material does not change.

Density

Density is a fundamental physical property defined as mass per unit volume. It plays a crucial role in understanding how substances behave under various conditions, such as changes in temperature, because as volume increases or decreases while mass remains constant, density inversely adjusts accordingly.

Thermal Expansion

Thermal expansion refers to the tendency of matter to change in volume in response to a change in temperature. It is an important concept in thermodynamics because it explains how objects expand upon heating and contract when cooled, affecting properties like density and material stability.

Volume Expansion Coefficient

The volume expansion coefficient is a material-specific parameter that quantifies the fractional change in volume per unit change in temperature. It is critical in calculations involving thermal expansion, as it enables predicting how much a substance's volume will increase with temperature, thereby impacting its density.

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