When `1g` of water at `0^(@)C` and `1 xx 10^(4)Nm^(-2)` pressure is converted into ice of volume `1.091 cm^(3)`, find the work done by water? `(rho_(w) = 1 gm//cm^(3))`

Explanation

To solve the problem of finding the work done by water when it is converted into ice, we can follow these steps:

### Step 1: Identify the given values
- Mass of water, \( m = 1 \, \text{g} \)
- Initial temperature, \( T = 0^\circ C \)
- Initial pressure, \( P = 1 \times 10^4 \, \text{N/m}^2 \)
- Volume of ice, \( V_f = 1.091 \, \text{cm}^3 \)
- Density of water, \( \rho_w = 1 \, \text{g/cm}^3 \)

### Step 2: Calculate the initial volume of water
Using the formula for volume based on mass and density:
\[ V_i = \frac{m}{\rho_w} \]
Substituting the values:
\[ V_i = \frac{1 \, \text{g}}{1 \, \text{g/cm}^3} = 1 \, \text{cm}^3 \]

### Step 3: Calculate the change in volume
The change in volume, \( \Delta V \), is given by:
\[ \Delta V = V_f - V_i \]
Substituting the values:
\[ \Delta V = 1.091 \, \text{cm}^3 - 1 \, \text{cm}^3 = 0.091 \, \text{cm}^3 \]

### Step 4: Convert the change in volume to cubic meters
Since the pressure is given in \( \text{N/m}^2 \), we need to convert the volume from \( \text{cm}^3 \) to \( \text{m}^3 \):
\[ 0.091 \, \text{cm}^3 = 0.091 \times 10^{-6} \, \text{m}^3 = 9.1 \times 10^{-8} \, \text{m}^3 \]

### Step 5: Calculate the work done
The work done by the water during the conversion is given by:
\[ W = P \Delta V \]
Substituting the values:
\[ W = (1 \times 10^4 \, \text{N/m}^2) \times (9.1 \times 10^{-8} \, \text{m}^3) \]
Calculating this gives:
\[ W = 1 \times 10^4 \times 9.1 \times 10^{-8} = 9.1 \times 10^{-4} \, \text{J} \]

### Final Answer
The work done by the water is \( 9.1 \times 10^{-4} \, \text{J} \).

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