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))` a) 2.44 x 10^5 J b) 1.62 x 10^5 J c) 1.24 x 10^5 J d) 2.62 x 10^5 J

Options

A. 2.44x`10^(5)J`
B. 1.62x`10^(5)J`
C. 1.24xx`10^(5)J`
D. 2.62x`10^(5)J`

Explanation

To find the change in internal energy when 1 kg of water is heated from 40°C to 70°C at constant volume, we can use the following steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Mass of water, \( m = 1 \, \text{kg} \)
- Initial temperature, \( T_i = 40^\circ C \)
- Final temperature, \( T_f = 70^\circ C \)
- Specific heat of water, \( c = 4148 \, \text{J kg}^{-1} \text{K}^{-1} \)

2. **Calculate the Change in Temperature:**
\[ \Delta T = T_f - T_i = 70^\circ C - 40^\circ C = 30 \, \text{K} \]

3. **Calculate the Heat Added (Q):**
Since the volume remains constant, the heat added to the system is given by:
\[ Q = m \cdot c \cdot \Delta T \]
Substituting the values:
\[ Q = 1 \, \text{kg} \cdot 4148 \, \text{J kg}^{-1} \text{K}^{-1} \cdot 30 \, \text{K} \]

4. **Perform the Calculation:**
\[ Q = 1 \cdot 4148 \cdot 30 = 124440 \, \text{J} \]

5. **Relate Heat Added to Change in Internal Energy:**
According to the first law of thermodynamics, since the volume is constant, the work done \( W = 0 \):
\[ \Delta U = Q - W = Q - 0 = Q \]
Therefore, the change in internal energy \( \Delta U \) is:
\[ \Delta U = 124440 \, \text{J} \]

6. **Convert to Scientific Notation:**
\[ \Delta U = 1.2444 \times 10^5 \, \text{J} \]

7. **Select the Correct Option:**
The closest answer to \( 1.2444 \times 10^5 \, \text{J} \) from the options provided is:
- **Option (c)**: \( 1.24 \times 10^5 \, \text{J} \)

### Final Answer:
The change in internal energy is \( \Delta U = 1.24 \times 10^5 \, \text{J} \) (Option c).