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 \(0^\circ C\) to \(4^\circ C\), we can follow these steps:

### Step 1: Identify the Given Values
- Mass of water, \(m = 2 \, \text{kg}\)
- Specific heat capacity of water, \(c = 4200 \, \text{J kg}^{-1} \text{K}^{-1}\)
- Initial temperature, \(T_1 = 0^\circ C = 273.15 \, \text{K}\)
- Final temperature, \(T_2 = 4^\circ C = 277.15 \, \text{K}\)
- Density at \(0^\circ C\), \(\rho_1 = 999.9 \, \text{kg m}^{-3}\)
- Density at \(4^\circ C\), \(\rho_2 = 1000 \, \text{kg m}^{-3}\)
- Atmospheric pressure, \(P = 10^5 \, \text{Pa}\)

### Step 2: Calculate the Change in Temperature
The change in temperature, \(\Delta T\), is given by:
\[ \Delta T = T_2 - T_1 = 4^\circ C - 0^\circ C = 4 \, \text{K} \]

### Step 3: Calculate the Heat Added to the Water
Using the formula for heat added:
\[ Q = m \cdot c \cdot \Delta T \]
Substituting the values:
\[ Q = 2 \, \text{kg} \cdot 4200 \, \text{J kg}^{-1} \text{K}^{-1} \cdot 4 \, \text{K} = 33600 \, \text{J} \]

### Step 4: Calculate the Change in Volume
The volume of water can be calculated using the formula:
\[ V = \frac{m}{\rho} \]
Calculating the initial and final volumes:
- Initial volume at \(0^\circ C\):
\[ V_1 = \frac{m}{\rho_1} = \frac{2 \, \text{kg}}{999.9 \, \text{kg m}^{-3}} \approx 0.0020002 \, \text{m}^3 \]
- Final volume at \(4^\circ C\):
\[ V_2 = \frac{m}{\rho_2} = \frac{2 \, \text{kg}}{1000 \, \text{kg m}^{-3}} = 0.002 \, \text{m}^3 \]
Thus, the change in volume, \(\Delta V\), is:
\[ \Delta V = V_2 - V_1 = 0.002 \, \text{m}^3 - 0.0020002 \, \text{m}^3 \approx -0.0000002 \, \text{m}^3 \]

### Step 5: Calculate the Work Done
Using the formula for work done at constant pressure:
\[ W = P \cdot \Delta V \]
Substituting the values:
\[ W = 10^5 \, \text{Pa} \cdot (-0.0000002 \, \text{m}^3) = -0.02 \, \text{J} \]

### Step 6: Apply the First Law of Thermodynamics
According to the first law of thermodynamics:
\[ Q = W + \Delta U \]
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:
\[ \Delta U \approx 33600 \, \text{J} \]

---