Assume that a drop of liquid evaporates by decreases in its surface energy, so that its temperature remains unchanged. What should be the minimum radius of the drop for this to be possible? The surface tension is T, density of liquid is `rho` and L is its latent heat of vaporization.

Explanation

To solve the problem, we need to find the minimum radius of a drop of liquid that can evaporate by decreasing its surface energy while keeping its temperature constant. The given parameters are surface tension \( T \), density of the liquid \( \rho \), and the latent heat of vaporization \( L \).

### Step-by-Step Solution:

1. **Understanding Surface Energy**:
The surface energy \( E \) of a drop is given by the formula:
\[ E = T \times A \]
where \( A \) is the surface area of the drop. For a spherical drop of radius \( r \), the surface area \( A \) is:
\[ A = 4\pi r^2 \]
Therefore, the surface energy becomes:
\[ E = T \times 4\pi r^2 \]

2. **Change in Surface Energy**:
When the drop evaporates, its radius decreases by a small amount \( dr \). The change in surface area \( dA \) due to this decrease is:
\[ dA = 4\pi r^2 - 4\pi (r-dr)^2 \]
Simplifying this gives:
\[ dA = 4\pi (r^2 - (r^2 - 2r \cdot dr + (dr)^2)) = 4\pi (2r \cdot dr - (dr)^2) \approx 8\pi r \cdot dr \quad \text{(for small } dr\text{)} \]

3. **Decrease in Surface Energy**:
The decrease in surface energy \( dE \) is then:
\[ dE = T \cdot dA = T \cdot 8\pi r \cdot dr \]

4. **Heat Required for Vaporization**:
The heat \( dQ \) required to vaporize the liquid is given by:
\[ dQ = \text{mass} \times L = \rho \cdot V \cdot L \]
The volume \( V \) of the drop is:
\[ V = \frac{4}{3}\pi r^3 \]
Thus, the mass \( m \) of the drop is:
\[ m = \rho \cdot \frac{4}{3}\pi r^3 \]
Therefore, the heat required is:
\[ dQ = \rho \cdot \frac{4}{3}\pi r^3 \cdot L \]

5. **Setting the Decrease in Surface Energy Equal to Heat Required**:
For the evaporation process to occur without a change in temperature, the decrease in surface energy must equal the heat required for vaporization:
\[ T \cdot 8\pi r \cdot dr = \rho \cdot \frac{4}{3}\pi r^3 \cdot L \]

6. **Cancelling Common Terms**:
We can cancel \( \pi \) from both sides and also \( dr \) (assuming \( dr \neq 0 \)):
\[ 8T = \frac{4}{3}\rho r^2 L \]

7. **Solving for Minimum Radius \( r \)**:
Rearranging the equation gives:
\[ r^2 = \frac{8T \cdot 3}{4\rho L} = \frac{6T}{\rho L} \]
Taking the square root:
\[ r = \sqrt{\frac{6T}{\rho L}} \]

### Final Answer:
The minimum radius \( r \) of the drop for it to evaporate without a change in temperature is:
\[ r = \sqrt{\frac{6T}{\rho L}} \]