A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be `75^(@)C` . T is given by : (Given : room temperature = `30^(@)C`, specific heat of copper `=0.1 cal//gm^(@)C`)

Options

A. `1250^(@)C`
B. `825^(@)C`
C. `800^(@)C`
D. `885^(@)C`

Explanation

To solve the problem, we need to apply the principle of conservation of energy, which states that the heat gained by the water and the calorimeter will be equal to the heat lost by the copper ball.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Mass of the copper ball, \( m_{cu} = 100 \, \text{gm} \)
- Mass of the calorimeter, \( m_{cal} = 100 \, \text{gm} \)
- Mass of the water, \( m_{w} = 170 \, \text{gm} \)
- Final temperature of the system, \( T_f = 75 \, ^\circ C \)
- Initial temperature (room temperature), \( T_0 = 30 \, ^\circ C \)
- Specific heat of copper, \( s_{cu} = 0.1 \, \text{cal/gm} \, ^\circ C \)
- Specific heat of water, \( s_{w} = 1 \, \text{cal/gm} \, ^\circ C \)

2. **Write the Heat Gain and Heat Loss Equations:**
- Heat gained by the calorimeter and water:
\[ Q_{gain} = m_{cal} \cdot s_{cu} \cdot (T_f - T_0) + m_{w} \cdot s_{w} \cdot (T_f - T_0) \]
- Heat lost by the copper ball:
\[ Q_{loss} = m_{cu} \cdot s_{cu} \cdot (T - T_f) \]

3. **Set the Heat Gain Equal to Heat Loss:**
\[ Q_{gain} = Q_{loss} \]

4. **Substituting Values into the Equations:**
- For the heat gained:
\[ Q_{gain} = 100 \cdot 0.1 \cdot (75 - 30) + 170 \cdot 1 \cdot (75 - 30) \]
\[ Q_{gain} = 100 \cdot 0.1 \cdot 45 + 170 \cdot 1 \cdot 45 \]
\[ Q_{gain} = 450 + 7650 = 8100 \, \text{cal} \]

- For the heat lost:
\[ Q_{loss} = 100 \cdot 0.1 \cdot (T - 75) \]

5. **Equate and Solve for T:**
\[ 8100 = 100 \cdot 0.1 \cdot (T - 75) \]
\[ 8100 = 10(T - 75) \]
\[ 810 = T - 75 \]
\[ T = 810 + 75 = 885 \, ^\circ C \]

### Final Answer:
The initial temperature \( T \) of the copper ball is \( 885 \, ^\circ C \).