Question

Two smaller spherical raindrops collide and merge into one larger spherical raindrop. What is the radius and surface energy of this larger drop, compared to the two smaller ones?

Answer 1

Radius increases by 26% while the surface energy drops 21%

Explanation:

For a sphere, its volume and surface A area are:
`V=(4pi)/2R^3; A = 4pir^2`
`E_s` = surface energy = gamma

A) Volume consideration.

Let
`r_1 =`radius of small droplet 1
`r_2 =`radius of small droplet 2
`R =`radius of merged droplet

Because density of water and total mass of the droplets do not change, the volume before and after the merger must be the same.

Volume of two small droplets = Volume of merged droplet

`(4pi)/3r_1^3+(4pi)/3r_2^3 =(4pi)/3R^3`

`R^3 =r_1^3+r_2^3`
`R = (r_1^3+r_2^3)^(⅓)`

If the small droplets have the same size, then `r_1 =r_2=r`,

`R= (2r^3)^(⅓) = root(3)2 r = 1.26r`

`(Delta r)/r =`(1.26r-r)/r = 0.26 = 26%

The radius of the merged droplet is 26% larger.

(B) Surface area consideration , assuming small droplets have identical radius.

Let
`sigma =` surface tension
`E_s =` surface energy `= sigma A = 4pir^2sigma`
`E_(1)=E_(2) = 4pir^2sigma =` surface energy of small droplets
`E_(3)= 4piR^2sigma` = surface energy of the merger droplet

The percentage change in surface energy is:

`(Delta E)/(E_1+E_2)= (E_(3)- (E_(1)+E_(2)))/(E_1+E_2) =(E_3-2E_1)/(2E_1)= E_3/(2E_1)-1`

`because R = root(3)2r`
`therefore E_3/(2E_1)-1=(4piR^2sigma)/(8pir^2sigma)-1= (4pisigma(root(3)2r)^2)/(8pir^2sigma)-1=-0.21 =- 21%`

The surface energy has reduced by 21%.