Final answer:
The entropy change of the aluminum in the mixture with water can be calculated using the mass of aluminum, its specific heat, and the logarithmic ratio of the final and initial temperatures.
The entropy change (\( \Delta S_{\text{Al}} \)) for the aluminum is approximately \( 20.58 \, \text{J/K} \).
Step-by-step explanation:
To calculate the entropy change of the aluminum, we can use the formula:
\[ Q = mc\Delta T \]
where:
- \( Q \) is the heat transferred,
- \( m \) is the mass of the substance (in kg),
- \( c \) is the specific heat of the substance,
- \( \Delta T \) is the change in temperature.
The entropy change (\( \Delta S \)) can be obtained using the relation:
\[ \Delta S = \frac{Q}{T} \]
where:
- \( \Delta S \) is the entropy change,
- \( Q \) is the heat transferred,
- \( T \) is the temperature.
For aluminum, we need to calculate the heat transferred (\( Q_{\text{Al}} \)) using the formula:
\[ Q_{\text{Al}} = m_{\text{Al}}c_{\text{Al}}\Delta T_{\text{Al}} \]
And then use this value to calculate the entropy change (\( \Delta S_{\text{Al}} \)).
Let's break down the steps:
1. Calculate the heat transferred to aluminum (\( Q_{\text{Al}} \)):
\[ Q_{\text{Al}} = m_{\text{Al}}c_{\text{Al}}\Delta T_{\text{Al}} \]
\[ Q_{\text{Al}} = (0.170 \, \text{kg})(900 \, \text{J/kg} \cdot \text{K})(41.41 \, \text{K}) \]
2. Calculate the entropy change of aluminum (\( \Delta S_{\text{Al}} \)):
\[ \Delta S_{\text{Al}} = \frac{Q_{\text{Al}}}{T_{\text{Al}}} \]
\[ \Delta S_{\text{Al}} = \frac{(0.170 \, \text{kg})(900 \, \text{J/kg} \cdot \text{K})(41.41 \, \text{K})}{273.15 + 41.41 \, \text{K}} \]
Now, calculate \( \Delta S_{\text{Al}} \) to get the entropy change of aluminum.
Apologies for the interruption. Let's continue with the calculations:
\[ \Delta S_{\text{Al}} = \frac{(0.170 \, \text{kg})(900 \, \text{J/kg} \cdot \text{K})(41.41 \, \text{K})}{273.15 + 41.41 \, \text{K}} \]
\[ \Delta S_{\text{Al}} = \frac{(0.170 \, \text{kg})(900 \, \text{J/kg} \cdot \text{K})(41.41 \, \text{K})}{314.56 \, \text{K}} \]
\[ \Delta S_{\text{Al}} \approx \frac{(0.170)(900)(41.41)}{314.56} \]
\[ \Delta S_{\text{Al}} \approx \frac{6473.7}{314.56} \]
\[ \Delta S_{\text{Al}} \approx 20.58 \, \text{J/K} \]
So, the entropy change (\( \Delta S_{\text{Al}} \)) for the aluminum is approximately \( 20.58 \, \text{J/K} \).