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 can follow these steps:

### Step 1: Write down the given data
- Mass of water, \( M_w = 2 \, \text{kg} \)
- Initial temperature, \( T_i = 0 \, \text{°C} \)
- Final temperature, \( T_f = 4 \, \text{°C} \)
- Specific heat capacity of water, \( C_w = 4200 \, \text{J kg}^{-1} \text{K}^{-1} \)
- Density at 0°C, \( \rho_i = 999.9 \, \text{kg m}^{-3} \)
- Density at 4°C, \( \rho_f = 1000 \, \text{kg m}^{-3} \)
- Atmospheric pressure, \( P = 10^5 \, \text{Pa} \)

### Step 2: Calculate the change in temperature
\[ \Delta T = T_f - T_i = 4 \, \text{°C} - 0 \, \text{°C} = 4 \, \text{K} \]

### Step 3: Apply the first law of thermodynamics
According to the first law of thermodynamics:
\[ Q = \Delta U + W \]
Where:
- \( Q \) is the heat added to the system
- \( \Delta U \) is the change in internal energy
- \( W \) is the work done by the system

### Step 4: Calculate the heat added to the water
The heat added to the water can be calculated using the formula:
\[ Q = M_w \cdot C_w \cdot \Delta T \]
Substituting the values:
\[ Q = 2 \, \text{kg} \cdot 4200 \, \text{J kg}^{-1} \text{K}^{-1} \cdot 4 \, \text{K} = 33600 \, \text{J} \]

### Step 5: Calculate the work done on/by the system
The work done can be expressed as:
\[ W = P \Delta V \]
Where \( \Delta V \) is the change in volume. The volume of water can be calculated using:
\[ V = \frac{M_w}{\rho} \]
Thus, the change in volume is:
\[ \Delta V = V_f - V_i = \frac{M_w}{\rho_f} - \frac{M_w}{\rho_i} \]
Substituting the values:
\[ \Delta V = \frac{2 \, \text{kg}}{1000 \, \text{kg m}^{-3}} - \frac{2 \, \text{kg}}{999.9 \, \text{kg m}^{-3}} = 0.002 - 0.0020002 \approx -0.0000002 \, \text{m}^3 \]
Now, substituting this into the work done equation:
\[ W = 10^5 \, \text{Pa} \cdot (-0.0000002 \, \text{m}^3) = -0.02 \, \text{J} \]

### Step 6: Solve for the change in internal energy
Now we can substitute \( Q \) and \( W \) back into the first law equation:
\[ Q = \Delta U + W \]
Rearranging gives:
\[ \Delta U = Q - W \]
Substituting the values:
\[ \Delta U = 33600 \, \text{J} - (-0.02 \, \text{J}) = 33600.02 \, \text{J} \]

### Final Answer
The change in internal energy of the water is approximately:
\[ \Delta U \approx 33600 \, \text{J} \]