Find the change in the internal energy of `2 kg` of water as it heated from `0^(0)C to 4^(0)C`. The specific heat capacity of water is `4200 J kg^(-1) K^(-1)` and its densities at `0^(0)C` and `4^(0)C` are `999.9 kg m^(-3)` and `1000 kg m^(-3)` respectively. atmospheric pressure `=10^(5)` Pa.

Explanation

To find the change in the internal energy of 2 kg of water as it is heated from 0°C to 4°C, we will follow these steps:

### Step 1: Calculate the initial volume of water at 0°C
We can calculate the initial volume (V_initial) using the formula:
\[ V_{\text{initial}} = \frac{m}{\rho_{0°C}} \]
Where:
- \( m = 2 \, \text{kg} \) (mass of water)
- \( \rho_{0°C} = 999.9 \, \text{kg/m}^3 \) (density at 0°C)

Substituting the values:
\[ V_{\text{initial}} = \frac{2 \, \text{kg}}{999.9 \, \text{kg/m}^3} \approx 2.0002 \times 10^{-3} \, \text{m}^3 \]

### Step 2: Calculate the final volume of water at 4°C
Next, we calculate the final volume (V_final) using the density at 4°C:
\[ V_{\text{final}} = \frac{m}{\rho_{4°C}} \]
Where:
- \( \rho_{4°C} = 1000 \, \text{kg/m}^3 \)

Substituting the values:
\[ V_{\text{final}} = \frac{2 \, \text{kg}}{1000 \, \text{kg/m}^3} = 2 \times 10^{-3} \, \text{m}^3 \]

### Step 3: Calculate the change in volume
Now we find the change in volume (\(\Delta V\)):
\[ \Delta V = V_{\text{final}} - V_{\text{initial}} \]
Substituting the values:
\[ \Delta V = 2 \times 10^{-3} \, \text{m}^3 - 2.0002 \times 10^{-3} \, \text{m}^3 = -0.0002 \times 10^{-3} \, \text{m}^3 \]

### Step 4: Calculate the heat added (q)
The heat added (q) can be calculated using the formula:
\[ q = m \cdot c \cdot \Delta T \]
Where:
- \( c = 4200 \, \text{J/kg/K} \) (specific heat capacity)
- \( \Delta T = 4 - 0 = 4 \, \text{K} \)

Substituting the values:
\[ q = 2 \, \text{kg} \cdot 4200 \, \text{J/kg/K} \cdot 4 \, \text{K} = 33600 \, \text{J} \]

### Step 5: Calculate the work done (w)
The work done (w) due to the change in volume can be calculated using:
\[ w = P \cdot \Delta V \]
Where:
- \( P = 10^5 \, \text{Pa} \) (atmospheric pressure)

Substituting the values:
\[ w = 10^5 \, \text{Pa} \cdot (-0.0002 \times 10^{-3} \, \text{m}^3) = -20 \, \text{J} \]

### Step 6: Calculate the change in internal energy (\(\Delta U\))
Using the first law of thermodynamics:
\[ \Delta U = q - w \]
Substituting the values:
\[ \Delta U = 33600 \, \text{J} - (-20 \, \text{J}) = 33600 + 20 = 33620 \, \text{J} \]

### Final Answer:
The change in the internal energy of the water is approximately:
\[ \Delta U \approx 33620 \, \text{J} \]