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 kg^(-1 ^(0)C^(-3)` latent heat of vaporization of water `=2.25xx10^(6) J kg^(-1)`

Explanation

To calculate the increase in the internal energy of 10 g of water when it is heated from \(0^{\circ}C\) to \(100^{\circ}C\) and converted into steam at 100 kPa, we will follow these steps:

### Step 1: Convert mass from grams to kilograms
The mass of water is given as 10 g. We need to convert this to kilograms for our calculations.

\[ m = 10 \, \text{g} = \frac{10}{1000} \, \text{kg} = 0.01 \, \text{kg} \]

### Step 2: Calculate the heat required to heat water from \(0^{\circ}C\) to \(100^{\circ}C\)
We can use the formula for heat \(Q_1\) required to raise the temperature:

\[ Q_1 = m \cdot c \cdot \Delta T \]

Where:
- \(m = 0.01 \, \text{kg}\) (mass of water)
- \(c = 4200 \, \text{J/kg}^{\circ}C\) (specific heat capacity of water)
- \(\Delta T = 100^{\circ}C - 0^{\circ}C = 100^{\circ}C\)

Substituting the values:

\[ Q_1 = 0.01 \, \text{kg} \cdot 4200 \, \text{J/kg}^{\circ}C \cdot 100^{\circ}C = 42000 \, \text{J} \]

### Step 3: Calculate the heat required to convert water at \(100^{\circ}C\) to steam
The heat \(Q_2\) required for the phase change from water to steam is given by:

\[ Q_2 = m \cdot L \]

Where:
- \(L = 2.25 \times 10^6 \, \text{J/kg}\) (latent heat of vaporization)

Substituting the values:

\[ Q_2 = 0.01 \, \text{kg} \cdot 2.25 \times 10^6 \, \text{J/kg} = 22500 \, \text{J} \]

### Step 4: Calculate the total heat supplied
The total heat \(Q\) supplied to the system is the sum of \(Q_1\) and \(Q_2\):

\[ Q = Q_1 + Q_2 = 42000 \, \text{J} + 22500 \, \text{J} = 64500 \, \text{J} \]

### Step 5: Calculate the work done during the phase change
The work done \(W\) can be calculated using the formula:

\[ W = P \cdot \Delta V \]

Where:
- \(P = 100 \, \text{kPa} = 10^5 \, \text{Pa}\)
- \(\Delta V = V_{\text{final}} - V_{\text{initial}}\)

The volume of steam can be calculated using its density:

\[ V_{\text{steam}} = \frac{m}{\text{density of steam}} = \frac{0.01 \, \text{kg}}{0.6 \, \text{kg/m}^3} = 0.01667 \, \text{m}^3 \]

The volume of water at \(0^{\circ}C\):

\[ V_{\text{water}} = \frac{m}{\text{density of water}} = \frac{0.01 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.00001 \, \text{m}^3 \]

Now, calculate \(\Delta V\):

\[ \Delta V = V_{\text{steam}} - V_{\text{water}} = 0.01667 \, \text{m}^3 - 0.00001 \, \text{m}^3 = 0.01666 \, \text{m}^3 \]

Now calculate the work done:

\[ W = P \cdot \Delta V = 10^5 \, \text{Pa} \cdot 0.01666 \, \text{m}^3 = 1666 \, \text{J} \]

### Step 6: Apply the first law of thermodynamics
According to the first law of thermodynamics:

\[ \Delta U = Q - W \]

Substituting the values we calculated:

\[ \Delta U = 64500 \, \text{J} - 1666 \, \text{J} = 62834 \, \text{J} \]

### Final Answer
The increase in the internal energy of the water when it is heated and converted to steam is:

\[ \Delta U = 62834 \, \text{J} \]