The surface energy of a liquid drop is u. It is sprayed into 1000 identical drops. What will be the total surface energy?
If the droplet is sprayed into multiple droplets, the total volume is unchanged. Thus, we can write
n(4/3)πr³ = (4/3)πR³
where R is the radius of the initial drop and r is the radius of the droplets.
We can divide the equation by (4/3)π and we get
nr³ = R³
r = R(1/n)¹/³
The total surface area of the droplets is n4πr².
We rewrite it by using the last equation
n4πr² = n4πR²(1/n)²/³ = 4πR²n¹/³
4πR² is the surface area of the original drop, so the total area has increased by a factor of n¹/³.
The surface energy is the work required to increase the surface area of a substance by unit area. Then, the surface energy is proportional to surface area.
Hence, the surface energy has increased by a factor n¹/³ and is n¹/³u
The original surface energy was u, so the increase in surface energy is
n¹/³u – u = [n¹/³ – 1]u
If n = 1000 then n¹/³ = 10
and [n¹/³ – 1]u = [10 – 1]u = 9u
Finally, the total surface energy will be 9u.