Question

What amount of energy is required to change a spherical drop of water with a diameter of 1.80 mm to three smaller spherical drops of equal size? The surface tension, γ, of water at room temperature is 72.0 mJ/m2.

Answer 1

This is a straightforward question related to the surface energy of the droplet. 

You know the surface area of a sphere is 4π r² and its volume is (4/3) π r³. 

With a diameter of 1.4 mm you have an original droplet with a radius of 0.7 mm so the surface area is roughly 6.16 mm² (0.00000616 m²) and the volume is roughly 1.438 mm³. 

The total surface energy of the original droplet is 0.00000616 * 72 ~ 0.00044 mJ 

The five smaller droplets need to have the same volume as the original. Therefore 

5 V = 1.438 mm³ so the volume of one of the smaller spheres is 1.438/5 = 0.287 mm³. 

Since this smaller volume still has the volume (4/3) π r³ then r = cube_root(0.287/(4/3) π) = cube_root(4.39) = 0.4 mm. 

Each of the smaller droplets has a surface area of 4π r² = 2 mm² or 0.0000002 m². 

The surface energy of the 5 smaller droplets is then 5 * 0.000002 * 72.0 = 0.00072 mJ 
From this radius the surface energy of all smaller droplets is 0.00072 and the difference in energy is 0.00072- 0.00044 mJ = 0.00028 mJ. 

Therefore you need roughly 0.00028 mJ or 0.28 µJ of energy to change a spherical droplet of water of diameter 1.4 mm into 5 identical smaller droplets.