Find the increase in mass of water when `1.0 kg` of water absorbs `4.2 xx 10^3` J of energy to produce a temperature rise of `1K`.

Explanation

To find the increase in mass of water when it absorbs energy, we will use Einstein's mass-energy equivalence principle, which states that energy (E) is related to mass (Δm) by the equation:

\[ E = \Delta m \cdot c^2 \]

Where:
- \( E \) is the energy absorbed,
- \( \Delta m \) is the increase in mass,
- \( c \) is the speed of light in a vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

### Step-by-step Solution:

1. **Identify the given values**:
- Energy absorbed, \( E = 4.2 \times 10^3 \, \text{J} \)
- Speed of light, \( c = 3 \times 10^8 \, \text{m/s} \)

2. **Write the mass-energy equivalence formula**:
\[ E = \Delta m \cdot c^2 \]

3. **Rearrange the formula to solve for Δm**:
\[ \Delta m = \frac{E}{c^2} \]

4. **Calculate \( c^2 \)**:
\[ c^2 = (3 \times 10^8 \, \text{m/s})^2 = 9 \times 10^{16} \, \text{m}^2/\text{s}^2 \]

5. **Substitute the values into the equation**:
\[ \Delta m = \frac{4.2 \times 10^3 \, \text{J}}{9 \times 10^{16} \, \text{m}^2/\text{s}^2} \]

6. **Perform the division**:
\[ \Delta m = \frac{4.2}{9} \times 10^{3 - 16} = \frac{4.2}{9} \times 10^{-13} \]

7. **Calculate \( \frac{4.2}{9} \)**:
\[ \frac{4.2}{9} \approx 0.4667 \]

8. **Final calculation of Δm**:
\[ \Delta m \approx 0.4667 \times 10^{-13} \, \text{kg} = 4.67 \times 10^{-14} \, \text{kg} \]

### Final Answer:
The increase in mass of the water is approximately \( 4.7 \times 10^{-14} \, \text{kg} \).