Chapter 19, Problem 19PQ

To determine

The coefficient of volume expansion of the liquid.

Answer to Problem 19PQ

The coefficient of volume expansion of the liquid is $\underset{\_}{8.3\times {10}^{-4}°{\text{C}}^{-1}}$.

Explanation of Solution

After the ball sinks, the buoyant force from the displaced liquid must be equal to the gravitational force of the sphere.

Write the expression for the gravitational force of the sphere.

  $F_g=mg$                                                                                                            (I)

Here, $F_g$ is the gravitational force on the sphere, $m$ is the mass of the sphere, and $g$ is the acceleration due to gravity.

Write the expression for the buoyant force due to the displaced liquid.

  $F_b=\frac{4}{3}\pi R^3\rho g$                                                                                                  (II)

Here, $F_b$ is the buoyant force due to displaced liquid, $R$ is the radius of the sphere, and $\rho$ is the density of the liquid.

Write the expression to find the radius of the sphere.

  $R=\frac{d}{2}$                                                                                                             (III)

Here, $d$ is the dimeter of the sphere.

Equate the right hand sides of equations (I) and (II) and solve for $\rho$.

  $\begin{array}{c}mg=\frac{4}{3}\pi R^3\rho g\\ \rho =\frac{3m}{4\pi R^3}\end{array}$                                                                                               (IV)

As the temperature of the liquid increases, the volume of the liquid expands.

Write the expression to find the final volume of the liquid after the expansion of liquid.

  $V=V_0+\beta \left(T_2-T_1\right)V_0$                                                                                       (V)

Here, $V$ is the volume of the liquid at temperature $T_2$, $V_0$ is the volume at temperature $T_1$, $\beta$ is the coefficient of volume expansion, $T_2$ is the final temperature, and $T_1$ is the initial temperature of the liquid.

The mass of the liquid always remain constant at all the temperature and is equal to the product of density and the volume of the liquid.

  $m=\rho V={\rho }_0{V}_0$                                                                                                (VI)

From equation (VI) it can be deduced that,

  $\frac{{\rho }_0}{\rho }=\frac{V}{V_0}$                                                                                                         (VII)

From equation (V), take the ratio between the volumes of liquid.

  $\frac{V}{V_0}=1+\beta \left(T_2-T_1\right)$                                                                                       (VIII)

Use equation (VIII) in (VII).

  $\frac{{\rho }_0}{\rho }=1+\beta \left(T_2-T_1\right)$                                                                                         (IX)

Solve equation (IX) for $\beta$.

$\beta =\frac{\left(\frac{{\rho }_0}{\rho }-1\right)}{T_2-T_1}$ (X)

Conclusion:

Given that the density of the liquid at $0°\text{C}$ is $1.527\text{g}/{\text{cm}}^{\text{3}}$. The sphere sinks at a temperature of $35°\text{C}$.

Substitute $7.00\text{cm}$ for $d$ in equation (III) to find $R$.

  $\begin{array}{c}R=\frac{7.00\text{cm}}{2}\\ =3.5\text{cm}\end{array}$

Substitute $266.5\text{g}$ for $m$, and $3.5\text{cm}$ for $R$ in equation (IV) to find $\rho$.

  $\begin{array}{c}\rho =\frac{3\left(266.5\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right)}{4\pi {\left(3.5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}^3}\\ =1.484\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Substitute $1.527\text{g}/{\text{cm}}^{\text{3}}$ for ${\rho }_0$, $1.484\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, $0°\text{C}$ for $T_1$, and $35°\text{C}$ for $T_2$ in equation (X) to find $\beta$.

  $\begin{array}{c}\beta =\frac{\left(\frac{1.527\text{g}/{\text{cm}}^{\text{3}}\times \frac{1\text{kg}}{1000\text{g}}\times \frac{1{\text{cm}}^3}{1\times {10}^{-6}{\text{m}}^3}}{1.484\times {10}^3\text{kg}/{\text{m}}^{\text{3}}}-1\right)}{\left(35°\text{C}-0°\text{C}\right)}\\ =8.3\times {10}^{-4}°{\text{C}}^{-1}\end{array}$

Therefore, the coefficient of volume expansion of the liquid is $\underset{\_}{8.3\times {10}^{-4}°{\text{C}}^{-1}}$.