Calculating Surface Energy Released During Coalescence

When two smaller water drops coalesce into a single larger drop, the surface area changes, leading to a release or absorption of surface energy. Surface tension acts to minimize the surface area.

Step 1: Determine the radius of the large drop

Let the radius of the initial two drops be '$r$'. Their individual volumes are:

$V_{small} = \frac{4}{3}\pi r^3$

The total volume of the two drops is:

$V_{total} = 2 \times V_{small} = 2 \times \frac{4}{3}\pi r^3$

Let the radius of the new, single large drop be '$R$'. Its volume is:

$V_{large} = \frac{4}{3}\pi R^3$

Since volume is conserved during coalescence:

$V_{large} = V_{total}$

$\frac{4}{3}\pi R^3 = 2 \times \frac{4}{3}\pi r^3$

$R^3 = 2 r^3$

$R = \sqrt[3]{2} r = 2^{\frac{1}{3}} r$

Step 2: Calculate the initial and final surface areas

The surface area of one small drop is:

$A_{small} = 4\pi r^2$

The total initial surface area of the two small drops is:

$A_{initial} = 2 \times A_{small} = 2 \times (4\pi r^2) = 8\pi r^2$

The surface area of the single large drop is:

$A_{large} = 4\pi R^2$

Substitute the expression for R:

$A_{large} = 4\pi (2^{\frac{1}{3}} r)^2 = 4\pi (2^{\frac{2}{3}} r^2) = 4\pi r^2 2^{\frac{2}{3}}$

Step 3: Calculate the surface energy released

The surface energy is given by the product of surface tension ('T') and surface area ('A'). The energy released is the difference between the initial surface energy and the final surface energy.

$E_{released} = E_{initial} - E_{final}$

$E_{released} = T \times A_{initial} - T \times A_{large}$

$E_{released} = T (A_{initial} - A_{large})$

Substitute the calculated areas:

$E_{released} = T (8\pi r^2 - 4\pi r^2 2^{\frac{2}{3}})$

Factor out common terms:

$E_{released} = 4\pi r^2 T (2 - 2^{\frac{2}{3}})$

This matches the surface energy released.