Question

A cube of ice with side 4.0 cm is allowed to melt. The volume of the water formed to be 58.24 cm³. Find the density of ice.​

Answer 1

Understanding the concept:

•We are given that a cube of ice with side of 4.0cm which is allowed to melt and volume of water formed is 58.24$\sf{cm}^{3}$ and we have to find the density of ice so first of all we have to change 1cm into 1g and then we will find the mass of 58.24g of water and after that we will find the volume of ice cube so as we know that formula of cube is side cubic and then we will find the density of ice so as we know the formula of density is mass upon volume so let's find it!

Given:-

•Side of. cube of ice = 4.0cm

•Volume of ice formed to be = 58.24$\sf{cm}^{3}$

To Find:-

•Find the density of ice =?

Solution:-

$\sf \:mass \: 1cm \: cubic \: of \: water = 1g \\ \\ \: \sf \: therefore \: mass \: of \: 58.24g \: of \: water = 58.24 \times 1 \\ \\ \: \sf \: = 58.24$

As we know

$\: \sf \large \: volume \: of \: cube = {side}^{3}$

Now put the values

$\: \sf \: volume \: of \: cube = {4}^{3} \\ \\ \: \sf \: volume \: of \: cube = 12 {cm}^{3}$

$\sf \: therefore$

As we know

$\: \sf \large \: density \: of \: ice = \frac{mass}{volume}$

Now put the values

$\: \sf \implies \: density = \frac{58.24}{12} \\ \\ \: \sf \implies \: density = 0.91g {cm}^{ - 3}$

$\sf \therefore \: hence \: density \: of \: ice \: is \: 0.91g {cm}^{ - 3}$