Question

A cube of ice on the edge $4{\text{ }}cm$ is placed in an empty cylindrical glass of inner diameter $6{\text{ }}cm$ . Assume that the ice melts uniformly from each side so that it always retains its cubical shape. Remembering that ice is lighter than water, find the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass.

Answer 1

Solid water, sometimes known as ice, has a lower density than water.Water has a higher density than ice because the orientation of hydrogen bonds allows molecules to push farther apart, resulting in a lower density. As a result, water is heavier than ice. Ice's lattice architecture stops water molecules from moving around.

Complete step by step answer:
In order to answer the given question, let us first consider the density of water is $\rho$ and the density of ice is $\rho '$.Now, let the area of one of the cube's surfaces be ${L^2}$.Assuming the ice cube to float up to a height of $h$ , the mass of water expelled will be $\rho h{L^2}$ .
Therefore, buoyant force $= \rho h{L^2}$
The ice cube's weight $= \rho 'g{L^3}$
But, as we know that
$\rho '\left( {64{\text{ }}-{\text{ }}{L^3}} \right){\text{ }} =$Buoyant force = Weight of the ice cube
$\rho h{L^2} = \rho 'g{L^3}$
Hence, from here we will deduce the value of $h$ and name it as equation $\left( i \right)$
$\Rightarrow h = \dfrac{{\rho 'L}}{\rho }\,\,\,\,.......\left( i \right)$

Now, in order for the cube to just float in water, the height of the water must be equal to the height of the cube.The volume of water $\left( V \right)$ in the cylinder that corresponds to the height $h$ will be:
Cylinder volume at height $(h)$ $-$ volume occupied by ice
$\Rightarrow V = \pi \left( {\dfrac{6}{2}} \right)2h - h{L^2}\,\,\,........\left( {ii} \right)$
However, this water came from the ice cube melting, therefore
$\rho '\left( {64 - {L^3}} \right) =$ Mass of water obtained from ice melting (remembering density of ice is lesser than that of water)
The volume of this water mass $= \dfrac{{\rho '\left( {64 - {L^3}} \right)}}{\rho }\,\,\,\,..........\left( {iii} \right)$

By equations $\left( {ii} \right)$ and $\left( {iii} \right)$ we can deduce that
$\left[ {\pi \left( {\dfrac{6}{2}} \right)} \right]2h - h{L^2} = \rho '\dfrac{{\left( {64 - {L^3}} \right)}}{\rho }$
By equation $\left( i \right)$

\Rightarrow 9\pi L = 64 \\\ \therefore L = 2.26cm \\\ $$ **Therefore, the length of the edge of the ice cube at the instant it just leaves contact with the bottom of the glass is $$2.26\,cm$$.** **Note:** It's important to keep in mind that at $$4^\circ C$$ , ice has a higher density. It also begins to freeze at this point. When the temperature is reduced further to $${0^o}C$$ , intermolecular hydrogen bonding forms a three-dimensional cage structure, with half of the space remaining empty. This isn't to say that the structures of water and ice aren't similar. It's just that the amount of empty space occupied by liquid water is greater.