A metal ball of mass 0.1 Kg is heated upto `500^(@)C` and drooped into a vessel of heat capacity `800 JK^(-1)` and containing 0.5 kg wate. The initial temperature of water and vessel is `30^(@)C`. What is the approximate percentage increment in the temperature of the water ? [Specific Heat Capacities of wate and metal are, repectively,`4200 Jkg^(-1) and 400 Jkg^(-1) K^(-1)`

Options

A. `15%`
B. `30%`
C. `25%`
D. `20%`

Answer

To solve the problem, we will use the principle of conservation of energy, which states that the heat lost by the hot object (the metal ball) will be equal to the heat gained by the cooler objects (the water and the vessel).

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of the metal ball, \( m_m = 0.1 \, \text{kg} \)
- Initial temperature of the metal ball, \( T_m = 500 \, \text{°C} \)
- Mass of water, \( m_w = 0.5 \, \text{kg} \)
- Initial temperature of water and vessel, \( T_i = 30 \, \text{°C} \)
- Heat capacity of the vessel, \( C_v = 800 \, \text{J/K} \)
- Specific heat capacity of water, \( c_w = 4200 \, \text{J/(kg·K)} \)
- Specific heat capacity of the metal, \( c_m = 400 \, \text{J/(kg·K)} \)

2. **Set Up the Heat Transfer Equation:**
The heat lost by the metal ball is equal to the heat gained by the water and the vessel.
\[ m_m c_m (T_m - T_f) = m_w c_w (T_f - T_i) + C_v (T_f - T_i) \]
Where \( T_f \) is the final equilibrium temperature.

3. **Substituting the Values:**
\[ 0.1 \times 400 \times (500 - T_f) = 0.5 \times 4200 \times (T_f - 30) + 800 \times (T_f - 30) \]
Simplifying the left side:
\[ 40 (500 - T_f) = 0.5 \times 4200 (T_f - 30) + 800 (T_f - 30) \]
\[ 20000 - 40 T_f = 2100 (T_f - 30) + 800 (T_f - 30) \]

4. **Combine the Right Side:**
\[ 20000 - 40 T_f = (2100 + 800)(T_f - 30) \]
\[ 20000 - 40 T_f = 2900 (T_f - 30) \]

5. **Distributing on the Right Side:**
\[ 20000 - 40 T_f = 2900 T_f - 87000 \]

6. **Rearranging the Equation:**
\[ 20000 + 87000 = 2900 T_f + 40 T_f \]
\[ 107000 = 2940 T_f \]

7. **Solving for \( T_f \):**
\[ T_f = \frac{107000}{2940} \approx 36.4 \, \text{°C} \]

8. **Calculating the Temperature Increase:**
The increase in temperature of the water is:
\[ \Delta T = T_f - T_i = 36.4 - 30 = 6.4 \, \text{°C} \]

9. **Calculating the Percentage Increase:**
\[ \text{Percentage Increase} = \left(\frac{\Delta T}{T_i}\right) \times 100 = \left(\frac{6.4}{30}\right) \times 100 \approx 21.33\% \]

### Final Answer:
The approximate percentage increment in the temperature of the water is **21.33%**.