# Exercise 6 | Liquids and Solids

The surface tension of water is 72 mJ.m-2

Calculate the amount of energy required to disperse one spherical drop of radius R = 3.0 mm into spherical drops of radius r = 3.0 x 10-3 mm.

Data: surface area of a sphere = 4πr2

volume of a sphere = πr3

Energy of the surface E of the initial drop (in J):

E (in J) = surface tension (in J.m-2) x surface area (in m2)

Surface tension = 72 mJ.m-2 = 72 x 10-3 J.m-2

Surface area = 4πR2 = 4π x (3.0 x 10-3)2 = 1.13 x 10-4 m2

E = 72 x 10-3 x 1.13 x 10-4 = 8.14 x 10-6 J

Energy of the surface E’ of the final drops:

E’ = surface tension (in J.m-2) x surface area of 1 drop (in m2) x number of drops

Surface tension = 72 mJ.m-2 = 72 x 10-3 J.m-2

Surface area = 4πr2 = 4π x (3.0 x 10-6)2 = 1.13 x 10-10 m2

Number of drops =  = $\frac{\frac{4}{3}{\mathrm{\pi R}}^{3}}{\frac{4}{3}{\mathrm{\pi r}}^{3}}$ = ${\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^{3}$ = 109 drops

E’ = 72 x 10-3 x 1.13 x 10-10 x 109 = 8.14 x 10-3 J

Amount of energy required to create the new area is 8.14 x 10-3 J.