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To solve this problem, we need to calculate the total heat energy \( Q \) absorbed by both the aluminum pot and the water when heated from \( 30^\circ \text{C} \) to \( 100^\circ \text{C} \). This is a combined heat transfer problem involving two components: the aluminum pot and the water.

Let's break it down step-by-step.

### Given Data
1. **Mass of aluminum pot** \( m_{\text{Al}} = 329 \, \text{g} = 0.329 \, \text{kg} \)
2. **Volume of water** \( V_{\text{H}_2\text{O}} = 1.7 \, \text{L} = 0.0017 \, \text{m}^3 \)
3. **Initial temperature** \( T_{\text{initial}} = 30^\circ \text{C} \)
4. **Final temperature** \( T_{\text{final}} = 100^\circ \text{C} \)
5. **Specific heat capacity of aluminum** \( c_{\text{Al}} = 920 \, \text{J/(kg} \cdot \text{°C)} \)
6. **Specific heat capacity of water** \( c_{\text{H}_2\text{O}} = 4200 \, \text{J/(kg} \cdot \text{°C)} \)
7. **Density of water** \( \rho_{\text{H}_2\text{O}} = 1000 \, \text{kg/m}^3 \)

### Steps
1. **Calculate the mass of water** using its density:
   \[
   m_{\text{H}_2\text{O}} = \rho_{\text{H}_2\text{O}} \times V_{\text{H}_2\text{O}} = 1000 \, \text{kg/m}^3 \times 0.0017 \, \text{m}^3 = 1.7 \, \text{kg}
   \]

2. **Calculate the heat absorbed by the aluminum pot**:
   \[
   Q_{\text{Al}} = m_{\text{Al}} \times c_{\text{Al}} \times \Delta T
   \]
   where \( \Delta T = T_{\text{final}} - T_{\text{initial}} = 100^\circ \text{C} - 30^\circ \text{C} = 70^\circ \text{C} \).
   
   Plugging in the values:
   \[
   Q_{\text{Al}} = 0.329 \, \text{kg} \times 920 \, \text{J/(kg} \cdot \text{°C)} \times 70^\circ \text{C} = 21228.6 \, \text{J}
   \]

3. **Calculate the heat absorbed by the water**:
   \[
   Q_{\text{H}_2\text{O}} = m_{\text{H}_2\text{O}} \times c_{\text{H}_2\text{O}} \times \Delta T
   \]
   Using the same \( \Delta T = 70^\circ \text{C} \),
   \[
   Q_{\text{H}_2\text{O}} = 1.7 \, \text{kg} \times 4200 \, \text{J/(kg} \cdot \text{°C)} \times 70^\circ \text{C} = 499800 \, \text{J}
   \]

4. **Calculate the total heat absorbed**:
   \[
   Q_{\text{total}} = Q_{\text{Al}} + Q_{\text{H}_2\text{O}} = 21228.6 \, \text{J} + 499800 \, \text{J} = 521028.6 \, \text{J}
   \]

### Answer
The total heat absorbed by the aluminum pot and the water is approximately:
\[
Q_{\text{total}} \approx 521029 \, \text{J}
\]

Would you like more details on any step or have further questions?

### Related Questions
1. How would the result change if the initial temperature was different?
2. What would happen if the pot were made of another material with a different specific heat?
3. How does the density of water affect the heat calculation in such problems?
4. Why is the density of water assumed to be \(1000 \, \text{kg/m}^3\) here?
5. How would this calculation change if we included the latent heat of vaporization?

### Tip
In heat capacity problems, always double-check units, especially when dealing with mass in grams and specific heat in J/(kg·°C).

Определи, какое количество теплоты получили алюминиевая кастрюля массой 329 г и находящаяся в ней вода объёмом 1,7 л при нагревании от 30°C до кипения при температуре 100°C. Удельная теплоёмкость алюминия равна 920 Дж/(кг·°C), удельная теплоёмкость воды — 4200 Дж/(кг·°C), плотность воды — 1000 кг/м³.

Calculate the total heat absorbed by an aluminum pot (329 g) and 1.7 L of water when heated from 30°C to 100°C. This problem involves key thermodynamics concepts, including specific heat capacities and density, suitable for students in Grades 10-12.

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