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 iron 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 will 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 given values:**
- Mass of iron, \( m_{Fe} = 10 \, \text{g} \)
- Specific heat capacity of iron, \( C_{Fe} = 0.45 \, \text{J/g°C} \)
- Initial temperature of iron, \( T_{i,Fe} = 100 \, \text{°C} \)
- Mass of water, \( m_{H2O} = 25 \, \text{g} \)
- Specific heat capacity of water, \( C_{H2O} = 4.2 \, \text{J/g°C} \)
- Initial temperature of water, \( T_{i,H2O} = 27 \, \text{°C} \)

2. **Set up the heat transfer equations:**
- Heat lost by iron:
\[ Q_{Fe} = m_{Fe} \cdot C_{Fe} \cdot (T_{i,Fe} - T_f) \]
- Heat gained by water:
\[ Q_{H2O} = m_{H2O} \cdot C_{H2O} \cdot (T_f - T_{i,H2O}) \]

3. **Equate the heat lost by iron to the heat gained by water:**
\[ m_{Fe} \cdot C_{Fe} \cdot (T_{i,Fe} - T_f) = m_{H2O} \cdot C_{H2O} \cdot (T_f - T_{i,H2O}) \]

4. **Substitute the known values into the equation:**
\[ 10 \cdot 0.45 \cdot (100 - T_f) = 25 \cdot 4.2 \cdot (T_f - 27) \]

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

6. **Rearrange the equation to isolate \( T_f \):**
\[ 450 + 2835 = 105 T_f + 4.5 T_f \]
\[ 3285 = 109.5 T_f \]

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

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

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