To solve this problem, we can use the principle of conservation of energy, which states that the heat lost by the hot mixture will be equal to the heat gained by the water and the aluminum bowl.
Given Data
- Mass of the mixture, \( m_m = 300 , \text{g} = 0.3 , \text{kg} \)
- Initial temperature of the mixture, \( T_{m} = 85^\circ C \)
- Final temperature of the mixture and water, \( T_f = 40^\circ C \)
- Mass of water, \( m_w = 500 , \text{g} = 0.5 , \text{kg} \)
- Initial temperature of water, \( T_{w} = 26^\circ C \)
- Heat capacity of the aluminum bowl, \( C_b = 90 , \text{J/K} \)
- Specific heat capacity of water, \( c_w = 4200 , \text{J/(kg K)} \)
Calculations
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Heat lost by the mixture:
\[ Q_{m} = m_m \cdot c_m \cdot (T_{m} - T_f) \]
where \( c_m \) is the specific heat capacity of the mixture.
-
Heat gained by the water:
\[ Q_{w} = m_w \cdot c_w \cdot (T_f - T_{w}) \]
-
Heat gained by the aluminum bowl:
\[ Q_{b} = C_b \cdot (T_f - T_{w}) \]
-
Setting up the conservation of energy equation:
According to conservation of energy:
\[ Q_{m} = Q_{w} + Q_{b} \]
Substituting for \( Q_{m} \), \( Q_{w} \), and \( Q_{b} \):
\[ m_m \cdot c_m \cdot (T_{m} - T_f) = m_w \cdot c_w \cdot (T_f - T_{w}) + C_b \cdot (T_f - T_{w}) \]
Now plug in the known values:
1. Calculate \( Q_{w} \):
\[ Q_{w} = 0.5 , \text{kg} \cdot 4200 , \text{J/(kg K)} \cdot (40^\circ C - 26^\circ C) \] \[ Q_{w} = 0.5 \cdot 4200 \cdot 14 = 29400 , \text{J} \]
2. Calculate \( Q_{b} \):
\[ Q_{b} = 90 , \text{J/K} \cdot (40^\circ C - 26^\circ C) \] \[ Q_{b} = 90 \cdot 14 = 1260 , \text{J} \]
3. Calculate total heat gained:
\[ Q_{w} + Q_{b} = 29400 + 1260 = 30660 , \text{J} \]
4. Solve for \( c_m \):
Setting \( Q_{m} = Q_{w} + Q_{b} \):
\[ 0.3 , \text{kg} \cdot c_m \cdot (85^\circ C - 40^\circ C) = 30660 \] \[ 0.3 \cdot c_m \cdot 45 = 30660 \]
Now solve for \( c_m \):
\[ c_m = \frac{30660}{0.3 \cdot 45} = \frac{30660}{13.5} \approx 2274.44 , \text{J/(kg K)} \]
Thus, the specific heat capacity of the mixture is approximately 2274.44 J/(kg K).