Twenty seven droplets of water, each of radius 0.1 mm coalesce into a single drop. Find the change in surface energy. Surface tension of water is 0.072 N/m.

Answer

To find the change in surface energy when 27 droplets of water, each with a radius of 0.1 mm, coalesce into a single drop, we can follow these steps:

### Step 1: Calculate the volume of one small droplet
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
For one droplet with radius \( r = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} \):
\[ V_{\text{small}} = \frac{4}{3} \pi (0.1 \times 10^{-3})^3 \]

### Step 2: Calculate the total volume of 27 small droplets
The total volume \( V_{\text{total}} \) of 27 small droplets is:
\[ V_{\text{total}} = 27 \times V_{\text{small}} = 27 \times \frac{4}{3} \pi (0.1 \times 10^{-3})^3 \]

### Step 3: Calculate the radius of the larger droplet
Let the radius of the larger droplet be \( R \). The volume of the larger droplet is equal to the total volume of the smaller droplets:
\[ V_{\text{large}} = \frac{4}{3} \pi R^3 \]
Setting the two volumes equal:
\[ 27 \times \frac{4}{3} \pi (0.1 \times 10^{-3})^3 = \frac{4}{3} \pi R^3 \]
Canceling \( \frac{4}{3} \pi \) from both sides:
\[ 27 (0.1 \times 10^{-3})^3 = R^3 \]
Taking the cube root:
\[ R = (27)^{1/3} \times (0.1 \times 10^{-3}) = 3 \times (0.1 \times 10^{-3}) = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m} \]

### Step 4: Calculate the surface area of the small droplets and the large droplet
The surface area \( S \) of a sphere is given by:
\[ S = 4 \pi r^2 \]
For 27 small droplets:
\[ S_{\text{small}} = 27 \times 4 \pi (0.1 \times 10^{-3})^2 \]
For the larger droplet:
\[ S_{\text{large}} = 4 \pi (0.3 \times 10^{-3})^2 \]

### Step 5: Calculate the change in surface area
The change in surface area \( \Delta S \) is:
\[ \Delta S = S_{\text{small}} - S_{\text{large}} \]
Substituting the values:
\[ \Delta S = 27 \times 4 \pi (0.1 \times 10^{-3})^2 - 4 \pi (0.3 \times 10^{-3})^2 \]
Factoring out \( 4 \pi \):
\[ \Delta S = 4 \pi \left( 27 (0.1 \times 10^{-3})^2 - (0.3 \times 10^{-3})^2 \right) \]

### Step 6: Calculate the change in surface energy
The change in surface energy \( \Delta E \) is given by:
\[ \Delta E = \text{Surface Tension} \times \Delta S \]
Substituting the surface tension \( \gamma = 0.072 \, \text{N/m} \):
\[ \Delta E = 0.072 \times \Delta S \]

### Step 7: Final Calculation
Now substituting the value of \( \Delta S \):
\[ \Delta S = 4 \pi \left( 27 (0.1 \times 10^{-3})^2 - (0.3 \times 10^{-3})^2 \right) \]
Calculating \( \Delta S \) and then substituting it into the formula for \( \Delta E \) will give us the final answer.