In an experiment a sphere of aluminium of mass 0.20 kg is heated upto `150^(@)`C. Immediately, it is put into water of volume 150 cc at `27^(@)C` kept in a calorimeter of water equivalent to 0.025 kg. Final temperature of the system is `40^(@)C`. The specific heat of aluminium is (take 4.2 Joule = 1 calorie)

Options

A. `434J/kg-""^(@)C`
B. `378J/kg-""^(@)C`
C. `315J/kg-""^(@)C`
D. `476J/kg-""^(@)C`

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Mass of aluminum sphere, \( m_{Al} = 0.20 \, \text{kg} = 200 \, \text{g} \)
- Initial temperature of aluminum, \( T_{i, Al} = 150^\circ C \)
- Volume of water, \( V_{water} = 150 \, \text{cc} = 150 \, \text{g} \) (since the density of water is \( 1 \, \text{g/cc} \))
- Initial temperature of water, \( T_{i, water} = 27^\circ C \)
- Water equivalent of calorimeter, \( W_{cal} = 0.025 \, \text{kg} = 25 \, \text{g} \)
- Final temperature of the system, \( T_f = 40^\circ C \)

2. **Calculate the Heat Lost by Aluminum:**
The heat lost by the aluminum sphere can be calculated using the formula:
\[ Q_{lost} = m_{Al} \cdot c_{Al} \cdot (T_{i, Al} - T_f) \]
where \( c_{Al} \) is the specific heat of aluminum.

3. **Calculate the Heat Gained by Water and Calorimeter:**
The total heat gained by the water and the calorimeter is given by:
\[ Q_{gained} = (m_{water} + W_{cal}) \cdot c_{water} \cdot (T_f - T_{i, water}) \]
where \( c_{water} = 4.2 \, \text{J/g}^\circ C \).

4. **Set Heat Lost Equal to Heat Gained:**
According to the principle of conservation of energy:
\[ Q_{lost} = Q_{gained} \]
Substituting the equations from steps 2 and 3:
\[ m_{Al} \cdot c_{Al} \cdot (T_{i, Al} - T_f) = (m_{water} + W_{cal}) \cdot c_{water} \cdot (T_f - T_{i, water}) \]

5. **Substitute the Values:**
\[ 200 \cdot c_{Al} \cdot (150 - 40) = (150 + 25) \cdot 4.2 \cdot (40 - 27) \]
Simplifying:
\[ 200 \cdot c_{Al} \cdot 110 = 175 \cdot 4.2 \cdot 13 \]

6. **Calculate the Right Side:**
\[ 175 \cdot 4.2 \cdot 13 = 175 \cdot 54.6 = 9565 \, \text{J} \]

7. **Solve for \( c_{Al} \):**
\[ 200 \cdot c_{Al} \cdot 110 = 9565 \]
\[ c_{Al} = \frac{9565}{200 \cdot 110} \]
\[ c_{Al} = \frac{9565}{22000} \approx 0.434 \, \text{J/g}^\circ C \]

### Final Answer:
The specific heat of aluminum is approximately \( 0.434 \, \text{J/g}^\circ C \).