When `0.2 kg` of brass at `100 .^(@) C` is dropped into `0.5 kg` of water at `20 .^(@) C`,the resulting temperature is `23 .^(@) C`. The specific heat of brass is.

Options

A. `(a)0.41 xx 10^3 Jkg^-1 .^(@) C^-1`.
B. `(b)0.41 xx 10^2 Jkg^-1 .^(@) C^-1`.
C. `(c)0.41 xx 10^4 Jkg^-1 .^(@) C^-1`.
D. `(d)0.41 Jkg^-1 .^(@) C^-1`.

Explanation

To find the specific heat of brass, we will use the principle of conservation of energy, which states that the heat lost by the brass will be equal to the heat gained by the water.

### Step-by-Step Solution:

1. **Identify the Known Values:**
- Mass of brass, \( m_1 = 0.2 \, \text{kg} \)
- Initial temperature of brass, \( T_{1i} = 100 \, ^\circ C \)
- Mass of water, \( m_2 = 0.5 \, \text{kg} \)
- Initial temperature of water, \( T_{2i} = 20 \, ^\circ C \)
- Final temperature of the system, \( T_f = 23 \, ^\circ C \)

2. **Calculate the Change in Temperature:**
- Change in temperature of brass, \( \Delta T_1 = T_{1i} - T_f = 100 - 23 = 77 \, ^\circ C \)
- Change in temperature of water, \( \Delta T_2 = T_f - T_{2i} = 23 - 20 = 3 \, ^\circ C \)

3. **Use the Heat Transfer Equation:**
- Heat lost by brass = Heat gained by water
- \( m_1 \cdot S_1 \cdot \Delta T_1 = m_2 \cdot S_2 \cdot \Delta T_2 \)
- Where \( S_1 \) is the specific heat of brass (unknown), and \( S_2 = 4.2 \times 10^3 \, \text{J/kg} \cdot ^\circ C \) (specific heat of water).

4. **Substitute the Known Values:**
- \( 0.2 \cdot S_1 \cdot 77 = 0.5 \cdot 4.2 \times 10^3 \cdot 3 \)

5. **Calculate the Right Side:**
- \( 0.5 \cdot 4.2 \times 10^3 \cdot 3 = 0.5 \cdot 12600 = 6300 \, \text{J} \)

6. **Set Up the Equation:**
- \( 0.2 \cdot S_1 \cdot 77 = 6300 \)

7. **Solve for \( S_1 \):**
- \( S_1 = \frac{6300}{0.2 \cdot 77} \)
- \( S_1 = \frac{6300}{15.4} \)
- \( S_1 \approx 409.09 \, \text{J/kg} \cdot ^\circ C \)

8. **Final Answer:**
- The specific heat of brass is approximately \( 0.41 \times 10^3 \, \text{J/kg} \cdot ^\circ C \).