When water droplets merge to form a bigger drop

Options

A. energy is liberated
B. energy is absorbed
C. energy is neither liberated nor absorbed
D. energy is sometimes liberated and sometimes absorbed

Answer

To solve the question of whether energy is liberated or absorbed when water droplets merge to form a bigger drop, we can follow these steps:

### Step 1: Understand the Initial and Final States
We start with several small water droplets, each with a radius denoted as \( r \). When these droplets merge, they form a larger droplet with a radius denoted as \( R \).

### Step 2: Calculate Initial Surface Energy
The surface energy of a droplet is given by the formula:
\[ \text{Surface Energy} = S \times \text{Surface Area} \]
For a single droplet, the surface area is \( 4\pi r^2 \). Therefore, for \( n \) droplets, the total initial surface energy \( E_I \) is:
\[ E_I = n \times S \times 4\pi r^2 \]

### Step 3: Calculate Final Surface Energy
After merging, the surface area of the larger droplet with radius \( R \) is:
\[ \text{Surface Area} = 4\pi R^2 \]
Thus, the final surface energy \( E_F \) is:
\[ E_F = S \times 4\pi R^2 \]

### Step 4: Compare Initial and Final Energies
To determine whether energy is liberated or absorbed, we need to compare \( E_F \) and \( E_I \).

### Step 5: Conservation of Volume
The total volume before and after merging must be conserved. The initial volume \( V_I \) of \( n \) droplets is:
\[ V_I = n \times \frac{4}{3}\pi r^3 \]
The final volume \( V_F \) of the larger droplet is:
\[ V_F = \frac{4}{3}\pi R^3 \]
Setting these equal gives:
\[ n \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \]
From this, we can solve for \( n \):
\[ n = \frac{R^3}{r^3} \]

### Step 6: Substitute \( n \) into the Energy Ratio
Now we substitute \( n \) into the expression for \( E_I \):
\[ E_I = \left(\frac{R^3}{r^3}\right) \times S \times 4\pi r^2 = S \times 4\pi \frac{R^3}{r} \]
Now, we can find the ratio \( \frac{E_F}{E_I} \):
\[ \frac{E_F}{E_I} = \frac{S \times 4\pi R^2}{S \times 4\pi \frac{R^3}{r}} = \frac{R^2 \cdot r}{R^3} = \frac{r}{R} \]

### Step 7: Analyze the Ratio
Since \( r < R \) (the radius of the smaller droplets is less than that of the larger droplet), we have:
\[ \frac{r}{R} < 1 \]
This implies that:
\[ E_F < E_I \]

### Step 8: Conclusion
Since the final energy \( E_F \) is less than the initial energy \( E_I \), it indicates that energy is liberated when the droplets merge.

### Final Answer
**Energy is liberated when water droplets merge to form a bigger drop.**

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