Question

An ice cube whose edges measure $\mathbf{20}.\mathbf{0}\mathbf{mm}$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?

Short answer

(a)The water surface in the bottom face of the block is below.

(b) The distance from the top of the water to the bottom face of the block is .

(c) The required layer thickness of ethyl alcohol is .

Step-by-step solution

Bernoulli's equation:

The sum of pressure, kinetic energy per unit volume, and gravitational potential energy per unit volume have the same value for an ideal fluid at all points along the streamline. This result is summarized in Bernoulli's equation:

$P+\frac{1}{2}\rho v^2+\rho gy=cons\tan t$

Here, $P$is the pressure, $\rho$is the density, $g$is the gravity, and $y$is the distance.

(a) The water surface in the bottom face of the block:

Let s stand for the edge of the cube, h for the depth of immersion, ${\rho }_{ice}$for the density of the ice, ${\rho }_w$for the density of water, and ${\rho }_{al}$for the density of the alcohol.

According to Archimedes’s principle, at equilibrium you have

$$\begin{gathered} B=F_g \\ {\rho }_{w}gh{s}^2={\rho }_{ice}g{s}^3 \end{gathered}$$

$h=s\frac{{\rho }_{ice}}{{\rho }_w}$ ….. (1)

Here,

Density of ice, ${\rho }_{ice}=917kg/m^3$

Density of water,${\rho }_{water}=1000kg/m^3$${\rho }_{water}=1000kg/m^3$

Distance,$s=20.0mm$

Substitute these values into equation (1), and you have

$$\begin{gathered} h=20.0\left(\frac{917kg/m^3}{1000kg/m^3}\right) \\ =20.0\left(0.917\right) \\ =18.34 \\ =18.3mm \end{gathered}$$

(b) The distance from the top of the water to the bottom face of the block:

Assume that the top of the cube is still above the alcohol surface. Letting $h_{al}$stand for the thickness of the alcohol layer, you have

${\rho }_{al}g{s}^2{h}_{al}+{\rho }_{w}g{s}^2{h}_w={\rho }_{ice}g{s}^3$

So,

$h_w=\left(\frac{{\rho }_{ice}}{{\rho }_w}\right)s-\left(\frac{{\rho }_{al}}{{\rho }_w}\right)h_{al}$ ….. (2)

With

${\rho }_{al}=0.806\times {10}^{3}kg/m^3$

And

$h_{al}=5.00mm$,

Substitute these into equation (2).

$$\begin{gathered} h_w=\left(\frac{917kg/m^3}{1000kg/m^3}\right)\left(20.0mm\right)-\left(\frac{0.806\times {10}^{3}kg/m^3}{1000kg/m^3}\right)\left(5.00mm\right) \\ =18.34-0.806\left(5.00\right) \\ =14.3mm \end{gathered}$$

To check our assumption above, the bottom of the cube is below the top surface of the alcohol, that is,

$14.4mm+5.00mm=19.3mm$

So, the top of the cube is above the surface of the alcohol is,

$20.0mm+19.3mm=0.7mm$

The assumption was valid.

(c) The layer thickness of ethyl alcohol:

Here,$h{\text{'}}_w=s-h{\text{'}}_{al}$

So, Archimedes’s principle gives

$$\begin{gathered} {\rho }_{al}g{s}^{2}h{\text{'}}_{al}+{\rho }_{w}g{s}^2\left(s-h{\text{'}}_{al}\right)={\rho }_{ice}g{s}^3 \\ {\rho }_{al}h{\text{'}}_{al}+{\rho }_w\left(s-h{\text{'}}_{al}\right)={\rho }_{ice}s \end{gathered}$$

$$\begin{gathered} h{\text{'}}_{al}=s\left(\frac{{\rho }_w-{\rho }_{ice}}{{\rho }_w-{\rho }_{al}}\right) \\ =\left(20.0mm\right)\left(\frac{1.000-0.917}{1.000-0.806}\right) \\ =8.557⩽mm \\ \approx 8.56mm \end{gathered}$$

Hence, the required layer thickness of ethyl alcohol is $8.56mm$.