To solve this problem, we need to apply the principle of conservation of energy. The heat lost by the hot mixture will be equal to the heat gained by the water and the aluminum bowl.
Given:
- Mass of the mixture (\(m_m\)) = 300 g = 0.3 kg
- Initial temperature of the mixture (\(T_{m_i}\)) = 85 °C
- Final temperature of the mixture and water/bowl (\(T_f\)) = 40 °C
- Heat capacity of the aluminum bowl (\(C_{bowl}\)) = 90 J/K
- Mass of water (\(m_w\)) = 500 g = 0.5 kg
- Specific heat capacity of water (\(c_w\)) = 4200 J/(kg·K)
- Initial temperature of the water (\(T_{w_i}\)) = 26 °C
Heat Loss by the Mixture:
The heat lost by the mixture can be calculated using:
\[ Q_m = m_m \cdot c_m \cdot (T_{m_i} - T_f) \]
Where \(c_m\) is the specific heat capacity of the mixture.
Heat Gained by the Water:
The heat gained by the water can be calculated using:
\[ Q_w = m_w \cdot c_w \cdot (T_f - T_{w_i}) \]
Heat Gained by the Aluminum Bowl:
The heat gained by the aluminum bowl can be calculated by:
\[ Q_b = C_{bowl} \cdot (T_f - T_{w_i}), \]
Setting the heat lost equal to the heat gained:
The total heat gained by the water and bowl is equal to the heat lost by the mixture.
\[ Q_m = Q_w + Q_b \]
Substituting the equations:
- For \(Q_w\):
\[ Q_w = 0.5 , \text{kg} \cdot 4200 , \text{J/(kg·K)} \cdot (40 - 26) = 0.5 \cdot 4200 \cdot 14 \]
\[ Q_w = 0.5 \cdot 4200 \cdot 14 = 29400 \text{ J} \]
- For \(Q_b\):
\[ Q_b = 90 , \text{J/K} \cdot (40 - 26) = 90 \cdot 14 \]
\[ Q_b = 90 \cdot 14 = 1260 \text{ J} \]
Total Heat Gained (water + bowl):
\[ Q_{total} = Q_w + Q_b = 29400 + 1260 = 30660 \text{ J} \]
- Now for the heat lost by the mixture:
So,
\[ m_m \cdot c_m \cdot (T_{m_i} - T_f) = Q_{total} \] \[ 0.3 , \text{kg} \cdot c_m \cdot (85 - 40) = 30660 \] \[ 0.3 \cdot c_m \cdot 45 = 30660 \] \[ c_m \cdot 1350 = 30660 \] \[ c_m = \frac{30660}{1350} = 22.73 \text{ J/(kg·K)} \]
Conclusion:
Thus, the specific heat capacity of the mixture is approximately \(22.73 \text{ J/(kg·K)}\).