Chapter 15, Problem 34PQ

(a)

To determine

The density of the salt water.

Answer to Problem 34PQ

The density of the salt water is $1.01\times {10}^3{\text{kg/m}}^3$.

Explanation of Solution

Write the expression for density of salt water.

    ${\rho }_{sw}=\frac{m_{\text{salt}}+m_{\text{water}}}{V_{\text{tot}}}$                                                                                   (I)

Here, ${\rho }_{sw}$ is the density of salt water, $m_{\text{salt}}$ is the mass of the salt water, $m_{\text{water}}$ is the mass of the water, and $V_{\text{tot}}$ is the total volume.

Write the expression for mass of the salt water.

    $m_{\text{salt}}={\rho }_{\text{salt}}{V}_{\text{salt}}$                                                                                          (II)

Here, ${\rho }_{\text{salt}}$ is the density of the salt and $V_{\text{salt}}$ is the volume of the salt.

Write the expression for mass of the water.

    $m_{\text{water}}={\rho }_{\text{water}}{V}_{\text{water}}$                                                                                   (III)

Here, ${\rho }_{\text{water}}$ is the density of the water and $V_{\text{water}}$ is the volume of the water.

Write the equation for total volume of the salt and water.

    $V_{\text{tot}}=V_{\text{salt}}+V_{\text{water}}$                                                                                     (IV)

Conclusion:

Substitute the equation (II), (IIII, and (IV) in the equation (I).

  ${\rho }_{sw}=\frac{{\rho }_{\text{salt}}{V}_{\text{salt}}+{\rho }_{\text{water}}{V}_{\text{water}}}{V_{\text{salt}}+V_{\text{water}}}$

Substitute $1233{\text{kg/m}}^3$ for ${\rho }_{\text{salt}}$, $12.5\text{ml}$ for $V_{\text{salt}}$, $998.2{\text{kg/m}}^3$ for ${\rho }_{\text{water}}$, and $225\text{ml}$ for $V_{\text{water}}$ in the above equation to find ${\rho }_{sw}$.

  $\begin{array}{c}{\rho }_{sw}=\frac{\left(1233{\text{kg/m}}^3\right)\left(12.5\text{ml}\right)\left(\frac{1\times {10}^{-6}{\text{m}}^3}{1\text{ml}}\right)+\left(998.2{\text{kg/m}}^3\right)\left(225\text{ml}\right)\left(\frac{1\times {10}^{-6}{\text{m}}^3}{1\text{ml}}\right)}{\left(12.5+225\right)\text{ml}\left(\frac{1\times {10}^{-6}{\text{m}}^3}{1\text{ml}}\right)}\\ =\frac{\left(1233{\text{kg/m}}^3\right)\left(1.25\times {10}^{-5}{\text{m}}^3\right)+\left(998.2{\text{kg/m}}^3\right)\left(2.25\times {10}^{-4}{\text{m}}^3\right)}{\left(237.5\times {10}^{-6}{\text{m}}^3\right)}\\ =\frac{1541.25\times {10}^{-5}\text{kg}+22459.5\times {10}^{-5}\text{kg}}{237.5\times {10}^{-6}{\text{m}}^3}\\ =1.01\times {10}^3{\text{kg/m}}^3\end{array}$

Thus, the density of the salt water is $1.01\times {10}^3{\text{kg/m}}^3$.

(b)

To determine

The density and mass of a grape.

Answer to Problem 34PQ

The density of a grape is $1.01\times {10}^3{\text{kg/m}}^3$ and mass of a grape is $2.53\text{g}$.

Explanation of Solution

Since the grapes are floating, so their acceleration is zero, and the buoyant force equals to the weight of the grape.

Write the expression for the buoyant force equals to the weight of the grape.

  $F_B=F_g$                                                                                               (V)

Write the expression for force due to gravity of the grape.

  $F_g={\rho }_{\text{grape}}{V}_{\text{grape}}g$                                                                                 (VI)

Here, ${\rho }_{\text{grape}}$ is the density of the grape, $V_{\text{grape}}$ is the volume of the grape, and $g$ is the acceleration due to gravity.

Write the expression for buoyant force.

  $F_B={\rho }_{\text{sw}}{V}_{\text{grape}}g$                                                                                    (VII)

Write the expression for mass of a grape.

    $m_{\text{grape}}={\rho }_{\text{grape}}\frac{V_{\text{grape}}}{5}$                                                                              (VIII)

Here, $m_{\text{grape}}$ is the mass of a grape and $V_{\text{grape}}$ is the volume of the grape is one fifth of the total volume.

Conclusion:

Substitute the equation (VI) and (VII) in the equation (V).

  $\begin{array}{c}{\rho }_{\text{sw}}{V}_{\text{grape}}g={\rho }_{\text{grape}}{V}_{\text{grape}}g\\ {\rho }_{\text{sw}}={\rho }_{\text{grape}}\end{array}$

Substitute $1.01\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{sw}}$ in the above equation to find ${\rho }_{\text{grape}}$.

  ${\rho }_{\text{grape}}=1.01\times {10}^3{\text{kg/m}}^3$

Substitute $1.01\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{grape}}$ and $12.5\text{ml}$ for $V_{\text{grape}}$ in the equation (VIII) to find $m_{\text{grape}}$.

  $\begin{array}{c}{m}_{\text{grape}}=\left(1.01\times {10}^3{\text{kg/m}}^3\right)\frac{\left(12.5\text{ml}\right)\left(\frac{1\times {10}^{-6}{\text{m}}^3}{1\text{ml}}\right)}{5}\\ =\left(1.01\times {10}^3{\text{kg/m}}^3\right)\frac{\left(1.25\times {10}^{-5}{\text{m}}^3\right)}{5}\\ =2.53\times {10}^{-3}\text{kg}\left(\frac{{10}^3\text{g}}{1\text{kg}}\right)\\ =2.53\text{g}\end{array}$

Therefore, The density of a grape is $1.01\times {10}^3{\text{kg/m}}^3$ and mass of a grape is $2.53\text{g}$.

(c)

To determine

If the raisin is denser than a grape.

Answer to Problem 34PQ

Yes, the raisin is denser than a grape.

Explanation of Solution

Raisins have a complicated, wrinkly surface and even though they are dried out grapes can contain a substantial amount of water inside.

Since the raisin remains in the bottom of the glass, it must be denser than salt water and thus, it is denser than a grape.

Conclusion:

Thus, the raisin is denser than a grape.