Explanation
To solve the problem of finding the specific heat capacity of the metal, we will use the principle of conservation of energy, which states that the heat lost by the metal will be equal to the heat gained by the water and the calorimeter.
### Step-by-Step Explanation
1. **Identify the Given Data:**
- Mass of the metal, \( m_m = 0.5 \, \text{kg} \)
- Initial temperature of the metal, \( T_{m_i} = 100 \, \text{°C} \)
- Final temperature of the mixture, \( T_f = 25 \, \text{°C} \)
- Heat capacity of the calorimeter, \( C_c = 80 \, \text{J/K} \)
- Heat capacity of the water, \( C_w = 420 \, \text{J/K} \)
- Initial temperature of the water, \( T_{w_i} = 15 \, \text{°C} \)
2. **Calculate the Heat Gained by the Water and Calorimeter:**
The heat gained by the water and the calorimeter can be calculated using the formula:
\[
Q = C \Delta T
\]
where \( \Delta T \) is the change in temperature.
- For the water:
\[
\Delta T_w = T_f - T_{w_i} = 25 \, \text{°C} - 15 \, \text{°C} = 10 \, \text{°C}
\]
\[
Q_w = C_w \Delta T_w = 420 \, \text{J/K} \times 10 \, \text{K} = 4200 \, \text{J}
\]
- For the calorimeter:
\[
\Delta T_c = T_f - T_{w_i} = 25 \, \text{°C} - 15 \, \text{°C} = 10 \, \text{°C}
\]
\[
Q_c = C_c \Delta T_c = 80 \, \text{J/K} \times 10 \, \text{K} = 800 \, \text{J}
\]
- Total heat gained by the water and calorimeter:
\[
Q_{total} = Q_w + Q_c = 4200 \, \text{J} + 800 \, \text{J} = 5000 \, \text{J}
\]
3. **Calculate the Heat Lost by the Metal:**
The heat lost by the metal can be expressed as:
\[
Q_m = m_m c_m (T_{m_i} - T_f)
\]
where \( c_m \) is the specific heat capacity of the metal.
The change in temperature for the metal is:
\[
\Delta T_m = T_{m_i} - T_f = 100 \, \text{°C} - 25 \, \text{°C} = 75 \, \text{°C}
\]
Therefore, the heat lost by the metal is:
\[
Q_m = 0.5 \, \text{kg} \cdot c_m \cdot 75 \, \text{°C}
\]
4. **Set Up the Equation Using Conservation of Energy:**
According to the conservation of energy:
\[
Q_m = Q_{total}
\]
Substituting the expressions we derived:
\[
0.5 \, \text{kg} \cdot c_m \cdot 75 \, \text{°C} = 5000 \, \text{J}
\]
5. **Solve for the Specific Heat Capacity \( c_m \):**
Rearranging the equation gives:
\[
c_m = \frac{5000 \, \text{J}}{0.5 \, \text{kg} \cdot 75 \, \text{°C}} = \frac{5000}{37.5} = 133.33 \, \text{J/kg/K}
\]
Rounding to two decimal places, we find:
\[
c_m \approx 133 \, \text{J/kg/K}
\]
### Conclusion
The specific heat capacity of the metal is approximately **133 J/kg/K**.
### Explanation of Options:
- **Option A (877 J/kg/K)**: This value is too high for a typical metal's specific heat capacity, which usually ranges from about 100 to 900 J/kg/K.
- **Option B (286 J/kg/K)**: This value is also higher than what we calculated and does not match the expected range for metals.
- **Option C (133 J/kg/K)**: This is the correct answer based on our calculations.
- **Option D (92 J/kg/K)**: This value is too low for most metals and does not fit our calculated result.
### Revision Summary:
- Use the principle of conservation of energy to equate heat lost and gained.
- Calculate heat gained using \( Q = C \Delta T \) for both water and calorimeter.
- Set the heat lost by the metal equal to the total heat gained.
- Solve for the specific heat capacity using the rearranged formula.