To find the final temperature of the mixture, we can use the principle of conservation of energy, which states that the heat lost by the hot water will be equal to the heat gained by the cold water and the heat lost to the vessel.
Let:
- \( m_1 = 0.05 \) kg (mass of hot water)
- \( T_1 = 80 \) °C (initial temperature of hot water)
- \( m_2 = 0.06 \) kg (mass of cold water)
- \( T_2 = 10 \) °C (initial temperature of cold water)
- \( C_v = 28 \) J/K (heat capacity of the vessel)
Let \( T_f \) be the final temperature of the mixture.
The specific heat capacity of water is approximately \( c = 4184 \) J/(kg·°C).
The heat lost by the hot water is given by:
\[ Q_{\text{hot}} = m_1 \cdot c \cdot (T_1 - T_f) \]
The heat gained by the cold water is:
\[ Q_{\text{cold}} = m_2 \cdot c \cdot (T_f - T_2) \]
The heat absorbed by the vessel is:
\[ Q_{\text{vessel}} = C_v \cdot (T_f - T_2) \]
Since energy is conserved:
\[ Q_{\text{hot}} = Q_{\text{cold}} + Q_{\text{vessel}} \]
Substituting in the energy expressions gives:
\[ m_1 \cdot c \cdot (T_1 - T_f) = m_2 \cdot c \cdot (T_f - T_2) + C_v \cdot (T_f - T_2) \]
Substituting the given values:
\[ 0.05 \cdot 4184 \cdot (80 - T_f) = 0.06 \cdot 4184 \cdot (T_f - 10) + 28 \cdot (T_f - 10) \]
Now, simplifying the equation:
\[ 209.2 (80 - T_f) = 251.04 (T_f - 10) + 28 (T_f - 10) \]
This expands as follows:
\[ 209.2 (80 - T_f) = 251.04 T_f - 2510.4 + 28 T_f - 280 \]
Combining terms:
\[ 209.2 (80 - T_f) = (251.04 + 28) T_f - 2510.4 - 280 \] \[ 209.2 (80 - T_f) = 279.04 T_f - 2790.4 \]
Now expand the left side:
\[ 16736 - 209.2 T_f = 279.04 T_f - 2790.4 \]
Rearranging gives:
\[ 16736 + 2790.4 = 279.04 T_f + 209.2 T_f \] \[ 19526.4 = 488.24 T_f \]
Now, solving for \( T_f \):
\[ T_f = \frac{19526.4}{488.24} \approx 40.00 °C \]
Thus, the final temperature of the mixture is approximately \( 40.00 °C \).
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