Consider an ice cube of edge 1.0 cm kept in a gravity-free hall. Find the surface area of the water when the ice melts. Neglect the difference in densities of ice and water.

Explanation

To solve the problem of finding the surface area of water when an ice cube melts in a gravity-free environment, we can follow these steps:

### Step 1: Calculate the Volume of the Ice Cube
The volume \( V \) of a cube is given by the formula:
\[ V = a^3 \]
where \( a \) is the length of an edge of the cube. Given that the edge of the ice cube is \( 1.0 \, \text{cm} \):
\[ V = (1.0 \, \text{cm})^3 = 1.0 \, \text{cm}^3 \]

### Step 2: Understand the Phase Change
When the ice melts, it turns into water. Since we are neglecting the difference in densities between ice and water, the volume of water formed will be equal to the volume of the ice cube:
\[ V_{\text{water}} = V_{\text{ice}} = 1.0 \, \text{cm}^3 \]

### Step 3: Assume the Shape of the Water
In a gravity-free environment, the water will take a spherical shape due to surface tension. Let the radius of the sphere be \( r \).

### Step 4: Write the Volume of the Sphere
The volume \( V \) of a sphere is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Setting the volume of the sphere equal to the volume of the water:
\[ \frac{4}{3} \pi r^3 = 1.0 \, \text{cm}^3 \]

### Step 5: Solve for the Radius \( r \)
Rearranging the equation to solve for \( r^3 \):
\[ r^3 = \frac{3}{4\pi} \]
Taking the cube root:
\[ r = \left(\frac{3}{4\pi}\right)^{1/3} \]

### Step 6: Calculate the Surface Area of the Sphere
The surface area \( A \) of a sphere is given by:
\[ A = 4 \pi r^2 \]
Substituting the value of \( r \):
\[ A = 4 \pi \left(\left(\frac{3}{4\pi}\right)^{1/3}\right)^2 \]
This simplifies to:
\[ A = 4 \pi \cdot \left(\frac{3}{4\pi}\right)^{2/3} \]

### Step 7: Simplify the Surface Area Expression
Now simplifying further:
\[ A = 4 \pi \cdot \frac{3^{2/3}}{(4\pi)^{2/3}} = 4 \pi \cdot \frac{3^{2/3}}{4^{2/3} \cdot \pi^{2/3}} = 4 \cdot \frac{3^{2/3}}{4^{2/3}} \cdot \pi^{1 - \frac{2}{3}} = 4 \cdot \frac{3^{2/3}}{4^{2/3}} \cdot \pi^{1/3} \]
This further simplifies to:
\[ A = 4^{1 - \frac{2}{3}} \cdot 3^{2/3} \cdot \pi^{1/3} = 4^{1/3} \cdot 3^{2/3} \cdot \pi^{1/3} \]

### Final Result
Thus, the surface area of the water when the ice melts is:
\[ A = 36 \pi^{1/3} \, \text{cm}^2 \]