Calculate the increase in the internal energy of `10 g` of water when it is heated from `0^(0)C to 100^(0)C` and converted into steam at `100 kPa`. The density of steam `=0.6 kg m^(-3)`, specific heat capacity of water `=4200 J//kgC` ,latent heat of vaporization of water `=2.25xx10^6 J kg^(-1)`

Explanation

To calculate the increase in internal energy of 10 g of water when it is heated from 0°C to 100°C and converted into steam at 100 kPa, we can follow these steps:

### Step 1: Convert mass from grams to kilograms
The mass of water given is 10 g. To convert this to kilograms:
\[ \text{mass} = 10 \, \text{g} = \frac{10}{1000} \, \text{kg} = 0.01 \, \text{kg} \]

### Step 2: Calculate the initial volume of water
Using the density of water, which is approximately 1000 kg/m³:
\[ \text{Initial Volume} = \frac{\text{mass}}{\text{density}} = \frac{0.01 \, \text{kg}}{1000 \, \text{kg/m}^3} = 10^{-5} \, \text{m}^3 \]

### Step 3: Calculate the final volume of steam
Using the density of steam, which is given as 0.6 kg/m³:
\[ \text{Final Volume} = \frac{\text{mass}}{\text{density}} = \frac{0.01 \, \text{kg}}{0.6 \, \text{kg/m}^3} = 0.01667 \, \text{m}^3 \]

### Step 4: Calculate the change in internal energy
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Where:
- \( Q \) is the heat supplied,
- \( W \) is the work done.

The heat supplied \( Q \) consists of two parts:
1. Heating the water from 0°C to 100°C:
\[ Q_1 = m \cdot s \cdot \Delta \theta = 0.01 \, \text{kg} \cdot 4200 \, \text{J/(kg°C)} \cdot (100 - 0) \, \text{°C} = 4200 \, \text{J} \]

2. The latent heat for the phase change from water to steam:
\[ Q_2 = m \cdot L = 0.01 \, \text{kg} \cdot 2.25 \times 10^6 \, \text{J/kg} = 22500 \, \text{J} \]

Thus, the total heat supplied \( Q \) is:
\[ Q = Q_1 + Q_2 = 4200 \, \text{J} + 22500 \, \text{J} = 26700 \, \text{J} \]

### Step 5: Calculate the work done
The work done \( W \) is given by:
\[ W = P \cdot \Delta V \]
Where:
- \( P \) is the pressure (100 kPa = \( 10^5 \, \text{Pa} \)),
- \( \Delta V = V_{\text{final}} - V_{\text{initial}} = 0.01667 \, \text{m}^3 - 10^{-5} \, \text{m}^3 = 0.01667 - 0.00001 = 0.01666 \, \text{m}^3 \).

Calculating the work done:
\[ W = 10^5 \, \text{Pa} \cdot 0.01666 \, \text{m}^3 = 1666 \, \text{J} \]

### Step 6: Calculate the change in internal energy
Now substituting the values into the equation for change in internal energy:
\[ \Delta U = Q - W = 26700 \, \text{J} - 1666 \, \text{J} = 25034 \, \text{J} \]

### Final Answer
The increase in internal energy of the water is:
\[ \Delta U = 25034 \, \text{J} \]