Question

At room temperature, the surface tension of water is 72.0 mJ·m–2. What is the energy required to change a spherical drop of water with a diameter of 1.20 mm to five smaller spherical drops of equal size? The surface area of a sphere of radius r is 4π r2 and the volume is 4π r3/3.

Answer 1

It is given that the surface area of sphere is 4 π r² and its volume is (4/3 π r³)
With a diameter of 1.2 mm you have a radius of 0.6 mm so the surface area about 4.5 mm² and the volume is about 0.9 mm³ 
The total surface energy of the original droplet is (4.5 x 10⁻⁶ m x 72) = 3.24 x 10⁻⁴mJ
The five smaller droplets need to have the same volume as the original so:
5 V = 0.9 mm³ so the volume of smaller sphere will equal 0.18 mm³
Since this smaller volume still have volume (4/3 π r³) so r = 0.35 mm
Each of the smaller droplets has a surface are = 1.54 mm² 
The surface energy of the 5 smaller droplet is then (5 x 1.54 x 10⁻⁶ m x 72) = 5.54 x 10⁻⁴ mJ
From this radius the surface energy of all smaller droplets is 5.54 x 10⁻⁴ and the difference in energy is (5.54 x 10⁻⁴) - (3.24 x 10⁻⁴) = 2.3 x 10⁻⁴ mJ
Therefore we need about 2.3 x 10⁻⁴ mJ of energy to change a spherical droplet of water of diameter 1.2 mm into 5 identical smaller droplets