Exercise 6 | Liquids and Solids

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 = $\frac{{\mathrm{V}}_{\mathrm{initial} \mathrm{drop}}}{{\mathrm{V}}_{\mathrm{final} \mathrm{drop}}}$ = $\frac{{\displaystyle \frac{4}{3}}{\mathrm{\pi R}}^3}{{\displaystyle \frac{4}{3}}{\mathrm{\pi r}}^3}$ = ${\left(\frac{\mathrm{R}}{\mathrm{r}}\right)}^3$ = 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.