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To solve this problem, we need to find the initial temperature of a heated metal sample that was submerged in water. Here's a step-by-step breakdown:

### Given Data

1. **Mass of the metal** (\(m_{\text{metal}}\)) = 200 g = 0.2 kg
2. **Volume of water** = 1.3 L
   - Since the density of water is 1000 kg/m³, 1.3 L of water has a mass of 1.3 kg.
3. **Initial temperature of water** (\(T_{\text{initial water}}\)) = 17°C
4. **Final temperature of water** (\(T_{\text{final water}}\)) = 71°C
5. **Specific heat capacity of water** (\(c_{\text{water}}\)) = 4200 J/(kg·°C)
6. **Specific heat capacity of metal** (\(c_{\text{metal}}\)) = 500 J/(kg·°C)

### Objective
We need to find the initial temperature of the metal sample, denoted as \( T_{\text{initial metal}} \).

### Solution
1. **Heat gained by water**:
   Since the water's temperature increases, it absorbs heat from the metal. The heat \(Q_{\text{water}}\) absorbed by the water can be calculated by:
   \[
   Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final water}} - T_{\text{initial water}})
   \]

   Substituting values:
   \[
   Q_{\text{water}} = 1.3 \, \text{kg} \cdot 4200 \, \text{J/(kg·°C)} \cdot (71 - 17) \, \text{°C}
   \]
   \[
   Q_{\text{water}} = 1.3 \cdot 4200 \cdot 54
   \]
   \[
   Q_{\text{water}} = 294840 \, \text{J}
   \]

2. **Heat lost by metal**:
   The heat lost by the metal, \(Q_{\text{metal}}\), should be equal to the heat gained by the water (since no heat is lost to the surroundings).

   \[
   Q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{initial metal}} - T_{\text{final water}})
   \]

   Setting \(Q_{\text{metal}} = Q_{\text{water}}\), we get:
   \[
   m_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{initial metal}} - T_{\text{final water}}) = Q_{\text{water}}
   \]

   Substitute values:
   \[
   0.2 \cdot 500 \cdot (T_{\text{initial metal}} - 71) = 294840
   \]

3. **Solve for \(T_{\text{initial metal}}\)**:
   \[
   100 \cdot (T_{\text{initial metal}} - 71) = 294840
   \]
   \[
   T_{\text{initial metal}} - 71 = \frac{294840}{100}
   \]
   \[
   T_{\text{initial metal}} - 71 = 2948.4
   \]
   \[
   T_{\text{initial metal}} = 2948.4 + 71
   \]
   \[
   T_{\text{initial metal}} \approx 3019 \, \text{°C}
   \]

### Answer
The initial temperature of the metal sample is approximately **3019°C**.

Would you like a deeper explanation of any step, or do you have additional questions?

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### Additional Questions
1. How would the solution change if the metal had a different specific heat capacity?
2. What would happen if we included heat loss to the surroundings in the calculations?
3. How does the initial temperature of the metal affect the time it takes for the system to reach thermal equilibrium?
4. How would this calculation differ if the metal and water reached different final temperatures?
5. Why is it important to convert units (e.g., grams to kilograms) in heat transfer problems?

### Tip
Always ensure that units are consistent across all values when calculating physical quantities like heat. This helps avoid errors and ensures accurate results.

При обработке металлического образца массой 200 г он нагрелся до состояния ярко-красного каления. При опускании образца в воду объёмом 1,3 л и температурой 17°C температура воды увеличивается до 71°C. Определите температуру раскалённого образца. Справочные данные: удельная теплоёмкость воды — 4200 Дж/(кг·°C), удельная теплоёмкость металла — 500 Дж/(кг·°C), плотность воды — 1000 кг/м³.

Learn how to determine the initial temperature of a heated metal sample submerged in water, given specific data on mass, specific heat capacity, and temperature changes. This problem applies thermodynamic principles and the conservation of energy in an isolated system, suitable for high school physics students.

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