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 step by step, we will use the principle of conservation of energy, which states that the heat lost by the metal ball will be equal to the heat gained by the water and the vessel.

### Step 1: Identify the known values
- Mass of the metal ball, \( m_m = 0.1 \, \text{kg} \)
- Initial temperature of the metal ball, \( T_{m_i} = 500 \, ^\circ C \)
- Heat capacity of the metal, \( c_m = 400 \, \text{J/kg/K} \)
- Mass of the water, \( m_w = 0.5 \, \text{kg} \)
- Initial temperature of the water, \( T_{w_i} = 30 \, ^\circ C \)
- Heat capacity of the vessel, \( C_v = 800 \, \text{J/K} \)

### Step 2: Set up the heat loss and gain equations
The heat lost by the metal ball can be expressed as:
\[ Q_m = m_m \cdot c_m \cdot (T_{m_i} - T_f) \]
where \( T_f \) is the final equilibrium temperature.

The heat gained by the water and the vessel can be expressed as:
\[ Q_{w+v} = m_w \cdot c_w \cdot (T_f - T_{w_i}) + C_v \cdot (T_f - T_{w_i}) \]
where \( c_w = 4200 \, \text{J/kg/K} \) is the specific heat capacity of water.

### Step 3: Set the heat lost equal to the heat gained
According to the conservation of energy:
\[ Q_m = Q_{w+v} \]
Substituting the equations from Step 2:
\[ m_m \cdot c_m \cdot (T_{m_i} - T_f) = m_w \cdot c_w \cdot (T_f - T_{w_i}) + C_v \cdot (T_f - T_{w_i}) \]

### Step 4: Substitute known values
Substituting the known values into the equation:
\[ 0.1 \cdot 400 \cdot (500 - T_f) = 0.5 \cdot 4200 \cdot (T_f - 30) + 800 \cdot (T_f - 30) \]

### Step 5: Simplify the equation
Calculating the left side:
\[ 40 \cdot (500 - T_f) = 20000 - 40 T_f \]
Calculating the right side:
\[ (0.5 \cdot 4200 + 800)(T_f - 30) = (2100 + 800)(T_f - 30) = 2900(T_f - 30) \]
Expanding:
\[ 2900T_f - 87000 \]

### Step 6: Set the equation
Now we have:
\[ 40 \cdot (500 - T_f) = 2900T_f - 87000 \]
Expanding the left side:
\[ 20000 - 40T_f = 2900T_f - 87000 \]

### Step 7: Rearranging the equation
Rearranging gives:
\[ 20000 + 87000 = 2900T_f + 40T_f \]
\[ 107000 = 2940T_f \]

### Step 8: Solve for \( T_f \)
\[ T_f = \frac{107000}{2940} \approx 36.4 \, ^\circ C \]

### Step 9: Calculate the percentage increase in temperature
The initial temperature of the water was \( 30 \, ^\circ C \), so the change in temperature \( \Delta T \) is:
\[ \Delta T = T_f - T_{w_i} = 36.4 - 30 = 6.4 \, ^\circ C \]
The percentage increase in temperature is:
\[ \text{Percentage Increase} = \left( \frac{\Delta T}{T_{w_i}} \right) \times 100 = \left( \frac{6.4}{30} \right) \times 100 \approx 21.33\% \]

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