Chapter 14, Problem 85CP

(a)

To determine

The distance between the water surface and the bottom face of the block.

Answer to Problem 85CP

The distance between the water surface and the bottom face of the block is $\underset{\_}{18.3\text{mm}}$.

Explanation of Solution

The ice cube is floating in a glass of ice cold water. According to Archimedes principle the upward buoyant force exerted on the ice cube by the ice cold water is equal to the weight of the ice cold water displaced.

    $B_w=F_g$                                                                                                          (I)

Here, $B_w$ is the buoyant force acting on the ice cube, and $F_g$ is the weight of the ice.

Write the expression for the buoyant force exerted by the ice cold water on ice.

    $B_w={\rho }_{w}gV$                                                                                                      (II)

Here, ${\rho }_w$ is the density of water, $g$ is acceleration due to gravity, and $V$ is the volume of the water.

One of the surface of the ice is parallel to the surface of the water. So dimension of ice cube and water are same.

Write the expression for volume of water in terms of the height and edge length.

    $V=hA$                                                                                                             (III)

Here, $h$ is the distance between ice and water surface, and $A$ is the surface area of the water.

Write the expression for the surface area of water surface.

    $A=s^2$                                                                                                            (IV)

Here, $s$ is the length of water surface.

Use expressions (IV) in (III).

    $V=h{s}^2$                                                                                                        (V)

Use expression (V) in (II) to find $B_w$.

    $B_w={\rho }_{w}gh{s}^2$                                                                                                            (VI)

Write the expression for the weight of ice.

    $F_g=mg$                                                                                                                (VII)

Here, $m$ is the mass of the ice, and $g$ is the acceleration due to gravity.

Write the expression for the mass of ice in terms of volume and density.

    $m={\rho }_{\text{ice}}{s}^3$                                                                                                             (VIII)

Here, ${\rho }_{\text{ice}}$ is the density of ice.

Use expression (VIII) in (VII).

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

Use expressions (VI) and (IX) in expression (I) to find $h$.

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

Conclusion:

Substitute $0.917\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{ice}}$, $1.00\times {10}^3{\text{kg/m}}^3$ for ${\rho }_w$, and $20.0\text{mm}$ for $s$ in equation (X) to find $h$.

    $\begin{array}{c}h=\left(20.0\text{mm}\right)\frac{\left(0.917\times {10}^3{\text{kg/m}}^3\right)}{\left(1.00\times {10}^3{\text{kg/m}}^3\right)}\\ =18.34\text{mm}\\ \approx \text{18}\text{.3}\text{mm}\end{array}$

Therefore, the distance between the water surface and the bottom face of the block is $\underset{\_}{18.3\text{mm}}$.

(b)

To determine

The distance from the top of the water to the bottom face of the block.

Answer to Problem 85CP

The distance from the top of the water to the bottom face of the block is $\underset{\_}{14.3\text{mm}}$.

Explanation of Solution

Alcohol is poured on top of the water surface so that it forms a layer above the water. After attaining hydrostatic equilibrium also the top of cube is still above the alcohol.

Again apply Archimedes principle. That is the sum of buoyant force exerted by the water surface and alcohol surface is equal to the weight of the ice.

    $B_w+B_{\text{al}}=F_g$                                                                                                           (XI)

Here, $B_{\text{al}}$ is the buoyant force exerted by alcohol.

Write the expression for buoyant force exerted by alcohol.

    $B_{\text{al}}={\rho }_{\text{al}}g{s}^2{h}_{\text{al}}$                                                                                                        (XII)

Here, $B_{\text{al}}$ is the buoyant force exerted by alcohol, ${\rho }_{\text{al}}$ is the density of alcohol, $h_{\text{al}}$ is the height of alcohol layer.

Use expressions (XII), (IX) and (VI) in expression (XI) and solve for $h_w$.

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

Conclusion:

Substitute $0.917\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{ice}}$, $1.00\times {10}^3{\text{kg/m}}^3$ for ${\rho }_w$, $0.806\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{al}}$, $20.0\text{mm}$ for $s$, and $5.00\text{mm}$ for $h_{\text{al}}$ in expression (XIII) to find $h_w$.

    $\begin{array}{c}{h}_w=\left(\frac{0.917\times {10}^3{\text{kg/m}}^3}{1.00\times {10}^3{\text{kg/m}}^3}\right)\left(20.0\text{mm}\right)-\left(\frac{0.806\times {10}^3{\text{kg/m}}^3}{1.00\times {10}^3{\text{kg/m}}^3}\right)\left(5.00\text{mm}\right)\\ =14.3\text{mm}\end{array}$

Therefore, the distance from the top of the water to the bottom face of the block is $\underset{\_}{14.3\text{mm}}$.

(c)

To determine

The thickness of ethyl alcohol required to cause complete coinciding of top surface of ice cube with alcohol layer.

Answer to Problem 85CP

A thickness of ethyl alcohol required to cause complete coinciding of top surface of ice cube with alcohol layer is $\underset{\_}{8.56\text{mm}}$.

Explanation of Solution

The ethyl alcohol is added to the surface of water. Thus height of the water layer will be modified. Apply Archimedes principle. The sum of buoyant force exerted by the water and alcohol layer after the addition of ethyl alcohol.

  ${B^{\prime }}_w+{B^{\prime }}_{\text{al}}=F_g$                                                                                                   (XIV)

Here, ${B^{\prime }}_{\text{al}}$ is the modified buoyant force exerted by alcohol, and ${B^{\prime }}_w$ is the buoyant force exerted by the water layer after addition of ethyl alcohol.

Write the expression for the new height of water layer.

    ${h^{\prime }}_w=s-{h^{\prime }}_{\text{al}}$                                                                                                           (XV)

Here, ${h^{\prime }}_w$ is the new height of water layer, ${h^{\prime }}_{\text{al}}$ is the new height of alcohol layer.

Write the expression for the buoyant force exerted by the alcohol layer.

    ${B^{\prime }}_{\text{al}}={\rho }_{\text{al}}g{s}^2{h^{\prime }}_{\text{al}}$                                                                                                      (XVI)

Write the expression for the buoyant force exerted by the water layer after addition of ethyl alcohol.

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

Use expressions (XVII), (XVI), and (IX) in expression (XIV) and solve for ${h^{\prime }}_{\text{al}}$.

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

Conclusion:

Substitute $20.0\text{mm}$ for $s$, $0.917\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{ice}}$, $1.00\times {10}^3{\text{kg/m}}^3$ for ${\rho }_w$, and $0.806\times {10}^3{\text{kg/m}}^3$ for ${\rho }_{\text{al}}$ in equation (XVIII) to find ${h^{\prime }}_{\text{al}}$.

    $\begin{array}{c}{h^{\prime }}_{\text{al}}=\left(20.0\text{mm}\right)\frac{\left(1.00\times {10}^3{\text{kg/m}}^3-0.917\times {10}^3{\text{kg/m}}^3\right)}{\left(1.00\times {10}^3{\text{kg/m}}^3-0.806\times {10}^3{\text{kg/m}}^3\right)}\\ =8.56\text{mm}\end{array}$

Therefore, the thickness of ethyl alcohol required to cause complete coinciding of top surface of ice cube with alcohol layer is $\underset{\_}{8.56\text{mm}}$.