The increase in internal energy of 1 kg of water at `100^(@)` C when it is converted into steam at the same temperature and 1 atm (100 kPa) will be :
[The density of water and steam are `1000 kg//m^(3) "and" 0.6 kg//m^(3)` respectively. The latent heat of vapourisation of water is `2.25xx10^(6) J//kg`.]

Options

A. `2.08xx10^(6) J`
B. `4xx10^(7) J`
C. `3.27xx10^(8) J`
D. `5xx10^(9) J`

Explanation

To find the increase in internal energy of 1 kg of water at 100°C when it is converted into steam at the same temperature and 1 atm, we can follow these steps:

### Step 1: Write down the given data
- Mass of water (m) = 1 kg
- Latent heat of vaporization (L_v) = 2.25 × 10^6 J/kg
- Density of water (ρ_water) = 1000 kg/m³
- Density of steam (ρ_steam) = 0.6 kg/m³
- Pressure (P) = 100 kPa = 10^5 Pa

### Step 2: Calculate the heat (Q) absorbed during the phase change
The heat absorbed during the conversion from water to steam can be calculated using the formula:
\[ Q = m \times L_v \]

Substituting the values:
\[ Q = 1 \, \text{kg} \times 2.25 \times 10^6 \, \text{J/kg} = 2.25 \times 10^6 \, \text{J} \]

### Step 3: Calculate the change in volume (ΔV)
To find the work done (W), we need to calculate the change in volume when 1 kg of water is converted into steam.

1. **Calculate the volume of 1 kg of water:**
\[ V_{water} = \frac{m}{\rho_{water}} = \frac{1 \, \text{kg}}{1000 \, \text{kg/m}^3} = 0.001 \, \text{m}^3 \]

2. **Calculate the volume of 1 kg of steam:**
\[ V_{steam} = \frac{m}{\rho_{steam}} = \frac{1 \, \text{kg}}{0.6 \, \text{kg/m}^3} \approx 1.6667 \, \text{m}^3 \]

3. **Calculate the change in volume (ΔV):**
\[ \Delta V = V_{steam} - V_{water} = 1.6667 \, \text{m}^3 - 0.001 \, \text{m}^3 \approx 1.6657 \, \text{m}^3 \]

### Step 4: Calculate the work done (W)
The work done during the phase change can be calculated using the formula:
\[ W = P \times \Delta V \]

Substituting the values:
\[ W = 10^5 \, \text{Pa} \times 1.6657 \, \text{m}^3 \approx 166570 \, \text{J} \]

### Step 5: Calculate the change in internal energy (ΔU)
Using the first law of thermodynamics:
\[ \Delta U = Q - W \]

Substituting the values:
\[ \Delta U = 2.25 \times 10^6 \, \text{J} - 166570 \, \text{J} \]
\[ \Delta U \approx 2.08343 \times 10^6 \, \text{J} \]

### Final Result
The increase in internal energy of 1 kg of water at 100°C when it is converted into steam at the same temperature and 1 atm is approximately:
\[ \Delta U \approx 2.08 \times 10^6 \, \text{J} \]

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