Consider an ice cube of edge 1.0 cm placed in a gravity-free hall. Determine the surface area of the water after the ice melts. Ignore the difference in densities between ice and water.

Step-by-step solution

Volume of the ice cube is V = a^3 = (1.0)^3 = 1.0 cm ^3 . Since the difference in densities is ignored, the volume of water formed is also V = 1.0 cm ^3 . In a gravity-free space, the liquid takes a spherical shape to minimize its surface area due to surface tension. Let the radius of the spherical water drop be R . Equating the volumes: 4 3 R^3 = 1 R = ( 3 4 )^ 1/3 The surface area of the spherical drop is given by A = 4 R^2 . A = 4 ( 3 4 )^ 2/3 A = (4 )^ 1 - 2/3 (3^2)^ 1/3 A = (4 )^ 1/3 (9)^ 1/3 = (36 )^ 1/3 cm ^