A `10g` piece of iron `( C =0.45 J//g^(@)C )` at `100^(@)C` is dropped into `25g` of water `( C =4.2 J//g^(@)C)` at `27^(@)C`. Find temperature of the rion and water system at thermal equilibrium .

Options

A. `30^(@)C`
B. `33^(@)C`
C. `40^(@)C`
D. None of these

Explanation

To find the temperature of the iron and water system at thermal equilibrium, we can use the principle of conservation of energy, which states that the heat lost by the iron will be equal to the heat gained by the water.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Mass of iron, \( m_{Fe} = 10 \, g \)
- Specific heat of iron, \( C_{Fe} = 0.45 \, J/g^{\circ}C \)
- Initial temperature of iron, \( T_{i,Fe} = 100 \, ^{\circ}C \)
- Mass of water, \( m_{H_2O} = 25 \, g \)
- Specific heat of water, \( C_{H_2O} = 4.2 \, J/g^{\circ}C \)
- Initial temperature of water, \( T_{i,H_2O} = 27 \, ^{\circ}C \)
- Final equilibrium temperature, \( T_f \) (unknown)

2. **Write the Heat Transfer Equations:**
- Heat lost by iron:
\[ Q_{lost} = m_{Fe} \cdot C_{Fe} \cdot (T_{i,Fe} - T_f) \]
- Heat gained by water:
\[ Q_{gained} = m_{H_2O} \cdot C_{H_2O} \cdot (T_f - T_{i,H_2O}) \]

3. **Set Up the Equation Based on Conservation of Energy:**
Since the heat lost by iron is equal to the heat gained by water, we can write:
\[ m_{Fe} \cdot C_{Fe} \cdot (T_{i,Fe} - T_f) = m_{H_2O} \cdot C_{H_2O} \cdot (T_f - T_{i,H_2O}) \]

4. **Substitute the Known Values:**
\[ 10 \cdot 0.45 \cdot (100 - T_f) = 25 \cdot 4.2 \cdot (T_f - 27) \]

5. **Simplify the Equation:**
\[ 4.5 \cdot (100 - T_f) = 105 \cdot (T_f - 27) \]
Expanding both sides:
\[ 450 - 4.5T_f = 105T_f - 2835 \]

6. **Combine Like Terms:**
\[ 450 + 2835 = 105T_f + 4.5T_f \]
\[ 3285 = 109.5T_f \]

7. **Solve for \( T_f \):**
\[ T_f = \frac{3285}{109.5} \approx 30 \, ^{\circ}C \]

### Final Answer:
The temperature of the iron and water system at thermal equilibrium is approximately \( 30 \, ^{\circ}C \).