Question

A cube of ice whose side is $4.0{\text{ cm}}$ is allowed to melt. The volume of water found to be $58.24{\text{ c}}{{\text{m}}^3}.$ Find the density of ice.

Answer 1

Hint We are provided with the value of the edge of the ice cube and the volume of the water which is formed when the given ice cube melts. We can solve this problem by equating the density and volume of ice and water.

Complete step by step answer
Density: density of an object or a liquid is defined as the thickness of the object. It is calculated by dividing the mass of the object or liquid by the volume of the object or the liquid.
The density of the ice cube is less than that of the water. Water has higher density compared to the ice cube formed out of it.
The ice melts and becomes water, so one will think that the volume of ice and the volume of water formed out of it should be equal. But it is not equal, the volume of ice cube is larger than the volume of water formed because the water expands to form ice so its volume also increases when the ice cube melts again the water molecules shrink and the volume decreases.
So here the volume of ice cube and the volume of water formed out of ice is not equal.
Given,
The length of a side of the cube is $a = 4{\text{ cm}}$
The volume of the cube is given by the formula ${a^3}$
The volume of the ice cube ${a^3} = {4^3} = 64{\text{ c}}{{\text{m}}^3}$
The volume of water formed is $58.24{\text{ c}}{{\text{m}}^3}$
We know the density of water ${\rho _w} = 1g$
We have to find the density of the ice ${\rho _i} = ?$
A cube of ice which is $4{\text{ cm}}$ melts and becomes water which has $58.24{\text{ c}}{{\text{m}}^3}$ density. The ice cube melts and becomes water.
Since the ice cube melts and becomes water the product of density and volume of the ice cube is equal to the product of density and volume of water formed.
Volume of ice × density of ice = volume of water × density of water
Substitute the known values in the formula
$\Rightarrow 64 \times {\rho _i}{\text{ = 58}}{\text{.24}} \times {\text{1}}$
$\Rightarrow {\rho _i}{\text{ = }}\dfrac{{{\text{58}}{\text{.24}}}}{{64}}$
$\Rightarrow {\rho _i}{\text{ = 0}}{\text{.91gc}}{{\text{m}}^{ - 3}}$

Note We can understand that water is higher density than ice cube by doing a simple experiment, if you take an ice cube and put it in water the ice cube floats because ice cubes are less dense than water.