1 kg of water is heated from `40^(@) C` to `70^(@)C`, If its volume remains constant, then the change in internal energy is (specific heat of water = `4148 J kg^(-1) K^(-1)`)

Options

A. `2.44xx10^(5)J`
B. `1.62xx10^(5)J`
C. `1.24xx10^(5)J`
D. `2.62xx10^(5)J`

Explanation

To find the change in internal energy when 1 kg of water is heated from \(40^\circ C\) to \(70^\circ C\) at constant volume, we can follow these steps:

### Step 1: Identify the Given Values
- Mass of water, \(m = 1 \, \text{kg}\)
- Initial temperature, \(T_1 = 40^\circ C = 40 \, \text{K}\)
- Final temperature, \(T_2 = 70^\circ C = 70 \, \text{K}\)
- Specific heat capacity of water, \(c = 4148 \, \text{J kg}^{-1} \text{K}^{-1}\)

### Step 2: Calculate the Change in Temperature
The change in temperature (\(\Delta T\)) can be calculated as:
\[ \Delta T = T_2 - T_1 = 70^\circ C - 40^\circ C = 30 \, \text{K} \]

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

Since the volume remains constant, the work done (\(\Delta W\)) is zero:
\[ \Delta W = 0 \]
Thus, the equation simplifies to:
\[ \Delta Q = \Delta U \]

### Step 4: Calculate the Heat Added to the System
The heat added to the system can be calculated using the formula:
\[ \Delta Q = m \cdot c \cdot \Delta T \]
Substituting the known values:
\[ \Delta Q = 1 \, \text{kg} \cdot 4148 \, \text{J kg}^{-1} \text{K}^{-1} \cdot 30 \, \text{K} \]

### Step 5: Perform the Calculation
Calculating \(\Delta Q\):
\[ \Delta Q = 1 \cdot 4148 \cdot 30 = 124440 \, \text{J} \]

### Step 6: Conclusion
Since \(\Delta Q = \Delta U\), we find:
\[ \Delta U = 124440 \, \text{J} \approx 1.244 \times 10^5 \, \text{J} \]

### Final Answer
The change in internal energy is approximately \(1.24 \times 10^5 \, \text{J}\).

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