Calculate the amount of energy evolved when 8 droplets of water ( surface tension of water =0.072 `Nm^(-1)` ) of radius `1//2`mm each combine into one.

Explanation

To calculate the amount of energy evolved when 8 droplets of water combine into one, we can follow these steps:

### Step 1: Understand the Given Data
- Surface tension of water, \( S = 0.072 \, \text{N/m} \)
- Radius of each droplet, \( r = \frac{1}{2} \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} \)
- Number of droplets, \( n = 8 \)

### Step 2: Calculate the Volume of the Small Droplets
The volume \( V \) of a single droplet is given by the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
For 8 droplets, the total volume \( V_{\text{total}} \) is:
\[ V_{\text{total}} = 8 \times \frac{4}{3} \pi r^3 = \frac{32}{3} \pi r^3 \]

### Step 3: Calculate the Radius of the Larger Droplet
Let the radius of the larger droplet be \( R \). Since the volume of the larger droplet is equal to the total volume of the 8 smaller droplets:
\[ \frac{4}{3} \pi R^3 = \frac{32}{3} \pi r^3 \]
Cancelling \( \frac{4}{3} \pi \) from both sides, we have:
\[ R^3 = 8r^3 \]
Taking the cube root:
\[ R = 2r \]
Substituting \( r = 0.5 \times 10^{-3} \, \text{m} \):
\[ R = 2 \times 0.5 \times 10^{-3} = 10^{-3} \, \text{m} \]

### Step 4: Calculate the Surface Area of the Larger Droplet
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi R^2 \]
Substituting \( R = 10^{-3} \, \text{m} \):
\[ A_{\text{large}} = 4 \pi (10^{-3})^2 = 4 \pi \times 10^{-6} \, \text{m}^2 \]

### Step 5: Calculate the Surface Area of the 8 Smaller Droplets
The surface area of one small droplet is:
\[ A_{\text{small}} = 4 \pi r^2 \]
For 8 droplets:
\[ A_{\text{total small}} = 8 \times 4 \pi r^2 = 32 \pi r^2 \]
Substituting \( r = 0.5 \times 10^{-3} \):
\[ A_{\text{total small}} = 32 \pi (0.5 \times 10^{-3})^2 = 32 \pi \times 0.25 \times 10^{-6} = 8 \pi \times 10^{-6} \, \text{m}^2 \]

### Step 6: Calculate the Decrease in Surface Area
The decrease in surface area \( \Delta A \) when the droplets combine is:
\[ \Delta A = A_{\text{total small}} - A_{\text{large}} = 8 \pi \times 10^{-6} - 4 \pi \times 10^{-6} = 4 \pi \times 10^{-6} \, \text{m}^2 \]

### Step 7: Calculate the Energy Evolved
The energy \( E \) evolved due to the change in surface area is given by:
\[ E = S \times \Delta A \]
Substituting the values:
\[ E = 0.072 \times 4 \pi \times 10^{-6} \]
Calculating this gives:
\[ E \approx 0.072 \times 4 \times 3.14 \times 10^{-6} \approx 9.05 \times 10^{-7} \, \text{J} \]

### Final Answer
The amount of energy evolved when 8 droplets of water combine into one is approximately:
\[ E \approx 9.05 \times 10^{-7} \, \text{J} \]