Sure, let's solve this step-by-step.
### Step 1: Understanding the problem
We need to find the specific heat capacity of iron given:
- Mass of iron, [tex]\( m_{\text{iron}} = 2.15 \, \text{kg} \)[/tex]
- Initial temperature of iron, [tex]\( T_{\text{iron, initial}} = 100^{\circ} \text{C} \)[/tex]
- Final temperature of the mixture, [tex]\( T_{\text{final}} = 24.4^{\circ} \text{C} \)[/tex]
- Mass of water, [tex]\( m_{\text{water}} = 2.3 \, \text{kg} \)[/tex]
- Initial temperature of water, [tex]\( T_{\text{water, initial}} = 17^{\circ} \text{C} \)[/tex]
- Water equivalent of the vessel, [tex]\( m_{\text{vessel}} = 0.18 \, \text{kg} \)[/tex]
- Specific heat capacity of water, [tex]\( c_{\text{water}} = 4186 \, \text{J/(kg}\cdot\text{°C)} \)[/tex]
### Step 2: Calculating the effective mass of water
The water equivalent of the vessel is considered as additional mass of water:
[tex]\[ m_{\text{effective, water}} = m_{\text{water}} + m_{\text{vessel}} = 2.3 \, \text{kg} + 0.18 \, \text{kg} = 2.48 \, \text{kg} \][/tex]
### Step 3: Heat gained by water and vessel
The heat gained by the water and vessel can be calculated using:
[tex]\[ Q_{\text{gained, water}} = m_{\text{effective, water}} \cdot c_{\text{water}} \cdot (T_{\text{final}} - T_{\text{water, initial}}) \][/tex]
Substitute the values:
[tex]\[ Q_{\text{gained, water}} = 2.48 \, \text{kg} \cdot 4186 \, \text{J/(kg}\cdot\text{°C)} \cdot (24.4^{\circ} \text{C} - 17^{\circ} \text{C}) \][/tex]
[tex]\[ Q_{\text{gained, water}} = 2.48 \cdot 4186 \cdot 7.4 \][/tex]
[tex]\[ Q_{\text{gained, water}} = 2.48 \cdot 4186 \cdot 7.4 = 76914.88 \, \text{J} \][/tex]
### Step 4: Heat lost by iron
The heat lost by the iron must be equal to the heat gained by the water and vessel, as no heat is lost to the surroundings. The heat lost by iron can be represented as:
[tex]\[ Q_{\text{lost, iron}} = m_{\text{iron}} \cdot c_{\text{iron}} \cdot (T_{\text{iron, initial}} - T_{\text{final}}) \][/tex]
Given the heat lost by iron is equal to the heat gained by water:
[tex]\[ Q_{\text{lost, iron}} = Q_{\text{gained, water}} \][/tex]
### Step 5: Set up the equation for specific heat capacity of iron
[tex]\[ m_{\text{iron}} \cdot c_{\text{iron}} \cdot (T_{\text{iron, initial}} - T_{\text{final}}) = Q_{\text{gained, water}} \][/tex]
Substitute the known values:
[tex]\[ 2.15 \, \text{kg} \cdot c_{\text{iron}} \cdot (100^{\circ} \text{C} - 24.4^{\circ} \text{C}) = 76914.88 \, \text{J} \][/tex]
[tex]\[ 2.15 \cdot c_{\text{iron}} \cdot 75.6 = 76914.88 \][/tex]
### Step 6: Solving for [tex]\( c_{\text{iron}} \)[/tex]
[tex]\[ c_{\text{iron}} = \frac{76914.88}{2.15 \cdot 75.6} \][/tex]
[tex]\[ c_{\text{iron}} = \frac{76914.88}{162.54} \][/tex]
[tex]\[ c_{\text{iron}} \approx 473.27 \, \text{J/(kg}\cdot\text{°C)} \][/tex]
### Conclusion
The specific heat capacity of iron is approximately [tex]\( 473.27 \, \text{J/(kg}\cdot\text{°C)} \)[/tex].
