Question

Consider an ice cube of edge 1.0 cm kept in a gravity-free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.

Answer 1

Given:
Edge of the ice cube (a) = 1.0 cm
The water that is formed due to the melting of ice acquires a spherical surface.
In the absence of gravity, let the radius of the spherical surface be r.
Volume of ice cube = volume of spherical surface of water

$$\begin{gathered} \Rightarrow a^3=\frac{4}{3}\mathrm{\pi }{r}^3 \\ \Rightarrow r={\left[\frac{3{\mathrm{a}}^3}{4\mathrm{\pi }}\right]}^{1/3} \end{gathered}$$
Surface area of spherical water surface = 4πr²
$$\begin{gathered} =4\mathrm{\pi }{\left[\frac{3a^3}{4\mathrm{\pi }}\right]}^{2/3} \\ ={\left(36\mathrm{\pi }\right)}^{1/3}{\mathrm{cm}}^2 \end{gathered}$$