A piece of zinc at a temperature of `20^(@)C` weighing `65.38 g` is dropped into `180 g` of boiling water `(T=100^(@)C)` The specific heat of zince is `0.4 J g^(-1)C^(-1)` and that of water is `4.2J g^(-1^(@))C^(-1)`. What is the final common temperature reached by both the zinc and water ?

Options

A. `97.3^(@)C`
B. `33.4^(@)C`
C. `80.1^(@)C`
D. `60.0^(@)C`

Explanation

To find the final common temperature reached by both the zinc and water, we can use the principle of conservation of energy, which states that the heat lost by the water will be equal to the heat gained by the zinc.

### Step-by-step Solution:

1. **Identify the known values:**
- Mass of zinc, \( m_z = 65.38 \, \text{g} \)
- Initial temperature of zinc, \( T_{z_i} = 20^\circ C \)
- Specific heat of zinc, \( s_z = 0.4 \, \text{J g}^{-1} \text{C}^{-1} \)
- Mass of water, \( m_w = 180 \, \text{g} \)
- Initial temperature of water, \( T_{w_i} = 100^\circ C \)
- Specific heat of water, \( s_w = 4.2 \, \text{J g}^{-1} \text{C}^{-1} \)

2. **Set up the heat transfer equations:**
- Heat gained by zinc:
\[ Q_z = m_z \cdot s_z \cdot (T_f - T_{z_i}) \]
- Heat lost by water:
\[ Q_w = m_w \cdot s_w \cdot (T_{w_i} - T_f) \]

3. **Equate the heat gained by zinc to the heat lost by water:**
\[ m_z \cdot s_z \cdot (T_f - T_{z_i}) = m_w \cdot s_w \cdot (T_{w_i} - T_f) \]

4. **Substitute the known values into the equation:**
\[ 65.38 \cdot 0.4 \cdot (T_f - 20) = 180 \cdot 4.2 \cdot (100 - T_f) \]

5. **Simplify the equation:**
- Left side:
\[ 65.38 \cdot 0.4 = 26.152 \]
So,
\[ 26.152 \cdot (T_f - 20) \]
- Right side:
\[ 180 \cdot 4.2 = 756 \]
So,
\[ 756 \cdot (100 - T_f) \]

6. **Expand both sides:**
\[ 26.152 T_f - 523.04 = 75600 - 756 T_f \]

7. **Combine like terms:**
\[ 26.152 T_f + 756 T_f = 75600 + 523.04 \]
\[ 782.152 T_f = 76123.04 \]

8. **Solve for \( T_f \):**
\[ T_f = \frac{76123.04}{782.152} \approx 97.3^\circ C \]

### Final Answer:
The final common temperature reached by both the zinc and water is approximately \( 97.3^\circ C \).

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