A piece of metal of mass `112 g` is heated to `100^@ C` and dropped into a copper calorimeter of mass `40 g` containing `200 g` of water at `16^@ C`. Neglecting heat loss, the specific heat of the metal is nearly, if the equilibrium temperature reached is `24^@ C`.
`(S_(cu) = 0.1 cal//g-.^(@) C)`.

Options

A. `0.292 cal//g -.^(@) C`
B. `0.392 cal//g -.^(@) C`
C. `0.192 cal//g -.^(@) C`
D. `0.492 cal//g -.^(@) C`

Explanation

To solve the problem, we will apply the principle of calorimetry, which states that the heat lost by the hot object (the metal) is equal to the heat gained by the cold objects (the water and the calorimeter).

### Step-by-Step Solution:

1. **Identify the given data:**
- Mass of the metal, \( m_m = 112 \, \text{g} \)
- Initial temperature of the metal, \( T_{m_i} = 100^\circ C \)
- Mass of the copper calorimeter, \( m_{cu} = 40 \, \text{g} \)
- Mass of the water, \( m_w = 200 \, \text{g} \)
- Initial temperature of the water, \( T_{w_i} = 16^\circ C \)
- Final equilibrium temperature, \( T_f = 24^\circ C \)
- Specific heat of copper, \( S_{cu} = 0.1 \, \text{cal/g} \cdot ^\circ C \)
- Specific heat of water, \( S_w = 1 \, \text{cal/g} \cdot ^\circ C \)

2. **Calculate the heat lost by the metal:**
The heat lost by the metal can be calculated using the formula:
\[ Q_{lost} = m_m \cdot S_m \cdot (T_{m_i} - T_f) \]
Substituting the values:
\[ Q_{lost} = 112 \cdot S_m \cdot (100 - 24) = 112 \cdot S_m \cdot 76 \]

3. **Calculate the heat gained by the water:**
The heat gained by the water is given by:
\[ Q_{gained, water} = m_w \cdot S_w \cdot (T_f - T_{w_i}) \]
Substituting the values:
\[ Q_{gained, water} = 200 \cdot 1 \cdot (24 - 16) = 200 \cdot 1 \cdot 8 = 1600 \, \text{cal} \]

4. **Calculate the heat gained by the calorimeter:**
The heat gained by the copper calorimeter is:
\[ Q_{gained, cu} = m_{cu} \cdot S_{cu} \cdot (T_f - T_{w_i}) \]
Substituting the values:
\[ Q_{gained, cu} = 40 \cdot 0.1 \cdot (24 - 16) = 40 \cdot 0.1 \cdot 8 = 32 \, \text{cal} \]

5. **Total heat gained by the water and calorimeter:**
\[ Q_{gained, total} = Q_{gained, water} + Q_{gained, cu} = 1600 + 32 = 1632 \, \text{cal} \]

6. **Set heat lost equal to heat gained:**
According to the principle of calorimetry:
\[ Q_{lost} = Q_{gained, total} \]
Therefore:
\[ 112 \cdot S_m \cdot 76 = 1632 \]

7. **Solve for the specific heat of the metal \( S_m \):**
\[ S_m = \frac{1632}{112 \cdot 76} \]
\[ S_m = \frac{1632}{8512} \approx 0.1917 \, \text{cal/g} \cdot ^\circ C \]

8. **Final Result:**
The specific heat of the metal is approximately:
\[ S_m \approx 0.192 \, \text{cal/g} \cdot ^\circ C \]