At `100^(@)C` the volume of `1 kg` of water is `10^(-3) m^(3)` and volume of `1 kg` of steam at normal pressure is `1.671 m^(3)`. The latent heat of steam is `2.3 xx 10^(6) J//kg` and the normal pressure is `5xx10^(5) N//m^(2)`. If `5 kg` of water at `100^(@)C` is converted into steam, the increase in the internal energy of water in this process will be

Options

A. `8.35 xx 10^(5) J`
B. `10.66 xx 10^(6) J`
C. `11.5 xx 10^(5) J`
D. zero

Explanation

To find the increase in the internal energy when 5 kg of water at 100°C is converted into steam, we can follow these steps:

### Step 1: Calculate the heat required to convert water into steam
The heat \( Q \) required to convert a mass \( m \) of water into steam is given by the formula:
\[ Q = m \times L \]
where \( L \) is the latent heat of steam.

Given:
- Mass of water, \( m = 5 \, \text{kg} \)
- Latent heat of steam, \( L = 2.3 \times 10^6 \, \text{J/kg} \)

Substituting the values:
\[ Q = 5 \, \text{kg} \times 2.3 \times 10^6 \, \text{J/kg} = 11.5 \times 10^6 \, \text{J} \]

### Step 2: Calculate the work done during the expansion
The work done \( W \) during the expansion of steam can be calculated using the formula:
\[ W = P \times \Delta V \]
where \( P \) is the pressure and \( \Delta V \) is the change in volume.

Given:
- Normal pressure, \( P = 5 \times 10^5 \, \text{N/m}^2 \)
- Volume of 1 kg of water at 100°C, \( V_{\text{water}} = 10^{-3} \, \text{m}^3 \)
- Volume of 1 kg of steam, \( V_{\text{steam}} = 1.671 \, \text{m}^3 \)

Calculating the change in volume for 5 kg of water:
\[ \Delta V = (5 \, \text{kg} \times 1.671 \, \text{m}^3/\text{kg}) - (5 \, \text{kg} \times 10^{-3} \, \text{m}^3/\text{kg}) \]
\[ \Delta V = (8.355 - 0.005) \, \text{m}^3 = 8.350 \, \text{m}^3 \]

Now substituting into the work formula:
\[ W = 5 \times 10^5 \, \text{N/m}^2 \times 8.350 \, \text{m}^3 = 4.175 \times 10^6 \, \text{J} \]

### Step 3: Calculate the change in internal energy
According to the first law of thermodynamics:
\[ \Delta U = Q - W \]
Substituting the values we calculated:
\[ \Delta U = 11.5 \times 10^6 \, \text{J} - 4.175 \times 10^6 \, \text{J} = 7.325 \times 10^6 \, \text{J} \]

### Final Answer
The increase in the internal energy of water during this process is:
\[ \Delta U = 7.325 \times 10^6 \, \text{J} \]