Chapter 15, Problem 65P

(a)

To determine

The distance below the water surface at which is the bottom face of the block.

Answer to Problem 65P

The distance below the water surface at which is the bottom face of the block is $18.3\text{ mm}$.

Explanation of Solution

Write the condition for equilibrium.

  $B=F_g$        (I)

Here, $B$ is the buoyant force and $F_g$ is the weight of the ice cube.

Write the equation for the buoyant force.

  $B={\rho }_{w}g{V}_i$        (II)

Here, ${\rho }_w$ is the density of water, $g$ is the acceleration due to gravity and $V_i$ is volume of ice cube immersed in water.

Write the equation for $V_i$ .

  $V_i=h{s}^2$        (III)

Here, $h$ is the depth of immersion and $s$ is the edge of the cube.

Put equation (III) in equation (II).

  $B={\rho }_{w}gh{s}^2$        (IV)

Write the equation for the force of gravity on the ice cube.

  $F_g=mg$        (V)

Here, $m$ is the mass of the cube.

Write the equation for density of ice.

  ${\rho }_{\text{ice}}=\frac{m}{V}$

Here, ${\rho }_{\text{ice}}$ is the density of ice and $V$ is the volume of cube.

Rewrite the above equation for $m$ .

  $m={\rho }_{\text{ice}}V$        (VI)

Write the equation for $V$ .

  $V=s^3$

Put the above equation in equation (VI).

  $m={\rho }_{\text{ice}}{s}^3$

Put the above equation in equation (V).

  $F_g={\rho }_{\text{ice}}{s}^{3}g$        (VII)

Put equations (IV) and (VII) in equation (I) and rearrange it for $h$ .

  $\begin{array}{c}{\rho }_{w}gh{s}^2={\rho }_{\text{ice}}{s}^{3}g\\ {\rho }_{w}h={\rho }_{\text{ice}}s\\ h=\frac{{\rho }_{\text{ice}}s}{{\rho }_w}\end{array}$        (VIII)

Conclusion:

The density of ice is $917{\text{ kg/m}}^3$ and the density of water is $1000{\text{ kg/m}}^3$.

Substitute $917{\text{ kg/m}}^3$ for ${\rho }_{\text{ice}}$ , $20.0\text{ mm}$ for $s$ and $1000{\text{ kg/m}}^3$ for ${\rho }_w$ in equation (VIII) to find $h$.

    $\begin{array}{c}h=\frac{\left(917{\text{ kg/m}}^3\right)\left(20.0\text{ mm}\right)}{1000{\text{ kg/m}}^3}\\ =18.34\text{ mm}\\ \approx \text{18}\text{.3 mm}\end{array}$

Therefore, the distance below the water surface at which is the bottom face of the block is $18.3\text{ mm}$.

(b)

To determine

The distance below the water surface at which is the bottom face of the block after the alcohol is poured into water surface.

Answer to Problem 65P

The distance below the water surface at which is the bottom face of the block after the alcohol is poured into water surface is $14.3\text{ mm}$.

Explanation of Solution

Assume that the top of the cube is still above the alcohol surface.

Write the equation for the buoyant force.

  $B={\rho }_{\text{al}}g{V}_{\text{al}}+{\rho }_{w}g{V}_i$        (IX)

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

Write the equation for $V_{\text{al}}$.

  $V_{\text{al}}=h_{\text{al}}{s}^2$        (X)

Here, $h_{\text{al}}$ is the thickness of the alcohol layer.

Put equations (III) and (X) in equation (IX).

  $B={\rho }_{\text{al}}g{h}_{\text{al}}{s}^2+{\rho }_{w}gh{s}^2$        (XI)

Put equations (VII) and (XI) in equation (I) and rearrange it for $h$ .

  $\begin{array}{c}{\rho }_{\text{al}}g{h}_{\text{al}}{s}^2+{\rho }_{w}gh{s}^2={\rho }_{\text{ice}}{s}^{3}g\\ {\rho }_{\text{al}}{h}_{\text{al}}+{\rho }_{w}h={\rho }_{\text{ice}}s\\ {\rho }_{w}h={\rho }_{\text{ice}}s-{\rho }_{\text{al}}{h}_{\text{al}}\\ h=\frac{{\rho }_{\text{ice}}}{{\rho }_w}s-\frac{{\rho }_{\text{al}}}{{\rho }_w}{h}_{\text{al}}\end{array}$        (XII)

Conclusion:

The density of alcohol is $806{\text{ kg/m}}^3$.

Substitute $917{\text{ kg/m}}^3$ for ${\rho }_{\text{ice}}$ , $1000{\text{ kg/m}}^3$ for ${\rho }_w$ , $806{\text{ kg/m}}^3$ for ${\rho }_{\text{al}}$ , $20.0\text{ mm}$ for $s$ and $5.00\text{ mm}$ for $h_{\text{al}}$ in equation (XII) to find $h$.

    $\begin{array}{c}h=\left(\frac{917{\text{ kg/m}}^3}{1000{\text{ kg/m}}^3}\right)\left(20.0\text{ mm}\right)-\left(\frac{806{\text{ kg/m}}^3}{1000{\text{ kg/m}}^3}\right)\left(5.00\text{ mm}\right)\\ =14.31\text{ mm}\\ \approx \text{14}\text{.3 mm}\end{array}$

Therefore, the distance below the water surface at which is the bottom face of the block after the alcohol is poured into water surface is $14.3\text{ mm}$.

(c)

To determine

The thickness of the layer of ethyl alcohol required.

Answer to Problem 65P

The thickness of the layer of ethyl alcohol required is $8.56\text{ mm}$.

Explanation of Solution

Write the equation of $V_{\text{al}}$ for the given situation.

  $V_{\text{al}}={h^{\prime }}_{\text{al}}{s}^2$        (XIII)

Here, ${h^{\prime }}_{\text{al}}$ is the thickness of the ethyl alcohol required.

Write the equation of $h$ for the given situation.

  $h=s-{h^{\prime }}_{\text{al}}$

Put the above equation in equation (III).

  $V_i=\left(s-{h^{\prime }}_{\text{al}}\right)s^2$        (XIV)

Put equations (XIII) and (XIV) in equation (IX).

  $B={\rho }_{\text{al}}g{h^{\prime }}_{\text{al}}{s}^2+{\rho }_{w}g\left(s-{h^{\prime }}_{\text{al}}\right)s^2$        (XV)

Put equations (VII) and (XV) in equation (I) and rearrange it for ${h^{\prime }}_{\text{al}}$ .

  $\begin{array}{c}{\rho }_{\text{al}}g{h^{\prime }}_{\text{al}}{s}^2+{\rho }_{w}g\left(s-{h^{\prime }}_{\text{al}}\right)s^2={\rho }_{\text{ice}}{s}^{3}g\\ {\rho }_{\text{al}}{h^{\prime }}_{\text{al}}+{\rho }_w\left(s-{h^{\prime }}_{\text{al}}\right)={\rho }_{\text{ice}}s\\ \left({\rho }_{\text{al}}-{\rho }_w\right){h^{\prime }}_{\text{al}}=\left({\rho }_{\text{ice}}-{\rho }_w\right)s\\ {h^{\prime }}_{\text{al}}=s\frac{\left({\rho }_{\text{ice}}-{\rho }_w\right)}{\left({\rho }_{\text{al}}-{\rho }_w\right)}\end{array}$        (XVI)

Conclusion:

Substitute $917{\text{ kg/m}}^3$ for ${\rho }_{\text{ice}}$ , $1000{\text{ kg/m}}^3$ for ${\rho }_w$ , $806{\text{ kg/m}}^3$ for ${\rho }_{\text{al}}$ and $20.0\text{ mm}$ for $s$ in equation (XVI) to find ${h^{\prime }}_{\text{al}}$.

  $\begin{array}{c}{h^{\prime }}_{\text{al}}=\left(20.0\text{ mm}\right)\frac{\left(917{\text{ kg/m}}^3-1000{\text{ kg/m}}^3\right)}{\left(806{\text{ kg/m}}^3-1000{\text{ kg/m}}^3\right)}\\ =8.557\text{ mm}\\ \approx \text{8}\text{.56 mm}\end{array}$

Therefore, the thickness of the layer of ethyl alcohol required is $8.56\text{ mm}$.