Chapter 14, Problem 14.85CP

An ice cube whose edges measure 20.0 mm is floating in a glass of ice-cold water, and one of the ice cube's faces is parallel to the water’s surface, (a) How far below the water surface is the bottom face of the block? (b) Ice-cold ethyl alcohol is gently poured onto the water surface to form a layer 5.00 mm thick above the water. The alcohol does not mix with the water. When the ice cube again attains hydrostatic equilibrium, what is the distance from the top of the water to the bottom face of the block? (c) Additional cold ethyl alcohol is poured onto the water’s surface until the top surface of the alcohol coincides with the top surface of the ice cube (in hydrostatic equilibrium). How thick is the required layer of ethyl alcohol?

To determine

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

Answer to Problem 14.85CP

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

Explanation of Solution

Given info: The edge of the ice cube is $20.0\text{mm}$.

Formula to calculate the weight of the ice is,

$W_{\text{c}}={\rho }_{\text{i}}g{s}^3$

Here,

$W_{\text{c}}$ is the weight of the ice cube.

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

$s$ is the side of the edge of ice.

$g$ is the acceleration due to gravity.

Formula to calculate the buoyant force is,

$F_{\text{B}}={\rho }_{\text{w}}gh{s}^2$

Here,

$F_{\text{B}}$ is the buoyant force.

${\rho }_{\text{w}}$ is the density of water.

$h$ is the distance between the water surface and the bottom face of the ice block.

By the Archimedes’s principle the buoyant force is equal to the weight of the ice cube.

$F_{\text{B}}=W_{\text{c}}$

Substitute ${\rho }_{\text{w}}gh{s}^2$ for $F_{\text{B}}$ and ${\rho }_{\text{i}}g{s}^3$ for $W_{\text{c}}$ in the above equation.

${\rho }_{\text{w}}gh{s}^2={\rho }_{\text{i}}g{s}^3$

Rearrange the above expression for $h$.

$h=\frac{{\rho }_{\text{i}}}{{\rho }_{\text{w}}}s$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $917\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{i}}$ and $20.0\text{mm}$ for $s$ in the above equation to find the value of $h$.

$\begin{array}{c}h=\frac{917\text{kg}/{\text{m}}^{\text{3}}}{1000\text{kg}/{\text{m}}^{\text{3}}}\left(20.0\text{mm}\right)\\ =18.3\text{mm}\end{array}$

Conclusion:

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

To determine

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

Answer to Problem 14.85CP

The distance between the top of the water and the bottom face of the block is $14.3\text{mm}$.

Explanation of Solution

Given info: The edge of the ice cube is $20.0\text{mm}$ and the thickness of alcohol layer is $5.00\text{mm}$.

Formula to calculate the buoyant force due to water is,

${\left(F_{\text{B}}\right)}_1={\rho }_{\text{w}}g{h}_{\text{w}}{s}^2$

Here,

${\left(F_{\text{B}}\right)}_1$ is the buoyant force due to the water.

${\rho }_{\text{w}}$ is the density of water.

$h$ is the distance between the top of the water and the bottom face of the block.

Formula to calculate the buoyant force due to alcohol is,

${\left(F_{\text{B}}\right)}_2={\rho }_{\text{a}}g{h}_{\text{a}}{s}^2$

Here,

${\left(F_{\text{B}}\right)}_2$ is the buoyant force due to the alcohol.

${\rho }_{\text{a}}$ is the density of alcohol.

$h_{\text{a}}$ is the thickness of alcohol.

By the Archimedes’s principle the buoyant force is equal to the weight of the ice cube.

${\left(F_{\text{B}}\right)}_1+{\left(F_{\text{B}}\right)}_2=W_{\text{c}}$

Substitute ${\rho }_{\text{w}}g{h}_{\text{w}}{s}^2$ for ${\left(F_{\text{B}}\right)}_1$, ${\rho }_{\text{a}}g{h}_{\text{a}}{s}^2$ for ${\left(F_{\text{B}}\right)}_2$ and ${\rho }_{\text{i}}g{s}^3$ for $W_{\text{c}}$ in the above equation.

${\rho }_{\text{w}}g{h}_{\text{w}}{s}^2+{\rho }_{\text{a}}g{h}_{\text{a}}{s}^2={\rho }_{\text{i}}g{s}^3$

Rearrange the above expression for $h_{\text{w}}$.

$h_{\text{w}}=\frac{s{\rho }_{\text{i}}-h_{\text{a}}{\rho }_{\text{a}}}{{\rho }_{\text{w}}}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $917\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{i}}$, $806\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{a}}$, $5.00\text{mm}$ for $h_{\text{a}}$ and $20.0\text{mm}$ for $s$  in the above equation to find the value of $h_{\text{w}}$.

$\begin{array}{c}{h}_{\text{w}}=\frac{\left(20.0\text{mm}\right)\left(917\text{kg}/{\text{m}}^{\text{3}}\right)-\left(5.00\text{mm}\right)\left(806\text{kg}/{\text{m}}^{\text{3}}\right)}{1000\text{kg}/{\text{m}}^{\text{3}}}\\ =14.3\text{mm}\end{array}$

Conclusion:

Therefore, the distance between the top of the water and the bottom face of the block is $14.3\text{mm}$.

To determine

The thickness of the layer of ethyl alcohol.

Answer to Problem 14.85CP

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

Explanation of Solution

Given info: The edge of the ice cube is $20.0\text{mm}$.

Formula to calculate the buoyant force due to alcohol is,

${\left(F_{\text{B}}\right)}_3={\rho }_{\text{a}}g{h^{\prime }}_{\text{a}}{s}^2$

Here,

${\left(F_{\text{B}}\right)}_3$ is the buoyant force due to the alcohol.

${\rho }_{\text{a}}$ is the density of alcohol.

${h^{\prime }}_{\text{a}}$ is the thickness of alcohol.

Formula to calculate the buoyant force due to water is,

${\left(F_{\text{B}}\right)}_4={\rho }_{\text{w}}g\left(s-{h^{\prime }}_{\text{a}}\right)s^2$

Here,

${\left(F_{\text{B}}\right)}_4$ is the buoyant force due to the water.

${\rho }_{\text{w}}$ is the density of water.

By the Archimedes’s principle the buoyant force is equal to the weight of the ice cube.

${\left(F_{\text{B}}\right)}_3+{\left(F_{\text{B}}\right)}_4=W_{\text{c}}$

Substitute ${\rho }_{\text{a}}g{h^{\prime }}_{\text{a}}{s}^2$ for ${\left(F_{\text{B}}\right)}_3$, ${\rho }_{\text{w}}g\left(s-{h^{\prime }}_{\text{a}}\right)s^2$ for ${\left(F_{\text{B}}\right)}_4$ and ${\rho }_{\text{i}}g{s}^3$ for $W_{\text{c}}$ in the above equation.

${\rho }_{\text{a}}g{h^{\prime }}_{\text{a}}{s}^2+{\rho }_{\text{w}}g\left(s-{h^{\prime }}_{\text{a}}\right)s^2={\rho }_{\text{i}}g{s}^3$

Rearrange the above expression for ${h^{\prime }}_{\text{a}}$.

${h^{\prime }}_{\text{a}}=s\frac{\left({\rho }_{\text{w}}-{\rho }_{\text{i}}\right)}{\left({\rho }_{\text{w}}-{\rho }_{\text{a}}\right)}$

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $917\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{i}}$, $806\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{a}}$, and $20.0\text{mm}$ for $s$ in the above equation to find the value of ${h^{\prime }}_{\text{a}}$.

$\begin{array}{c}{h^{\prime }}_{\text{a}}=\left(20.0\text{mm}\right)\frac{\left[\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)-\left(917\text{kg}/{\text{m}}^{\text{3}}\right)\right]}{\left[\left(1000\text{kg}/{\text{m}}^{\text{3}}\right)-\left(806\text{kg}/{\text{m}}^{\text{3}}\right)\right]}\\ =8.56\text{mm}\end{array}$

Conclusion:

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