Math Problem Statement

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.

Solution

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 (mmetalm_{\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 (Tinitial waterT_{\text{initial water}}) = 17°C
  4. Final temperature of water (Tfinal waterT_{\text{final water}}) = 71°C
  5. Specific heat capacity of water (cwaterc_{\text{water}}) = 4200 J/(kg·°C)
  6. Specific heat capacity of metal (cmetalc_{\text{metal}}) = 500 J/(kg·°C)

Objective

We need to find the initial temperature of the metal sample, denoted as Tinitial metalT_{\text{initial metal}}.

Solution

  1. Heat gained by water: Since the water's temperature increases, it absorbs heat from the metal. The heat QwaterQ_{\text{water}} absorbed by the water can be calculated by: Qwater=mwatercwater(Tfinal waterTinitial water)Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{\text{final water}} - T_{\text{initial water}})

    Substituting values: Qwater=1.3kg4200J/(kg\cdotp°C)(7117)°CQ_{\text{water}} = 1.3 \, \text{kg} \cdot 4200 \, \text{J/(kg·°C)} \cdot (71 - 17) \, \text{°C} Qwater=1.3420054Q_{\text{water}} = 1.3 \cdot 4200 \cdot 54 Qwater=294840JQ_{\text{water}} = 294840 \, \text{J}

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

    Qmetal=mmetalcmetal(Tinitial metalTfinal water)Q_{\text{metal}} = m_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{initial metal}} - T_{\text{final water}})

    Setting Qmetal=QwaterQ_{\text{metal}} = Q_{\text{water}}, we get: mmetalcmetal(Tinitial metalTfinal water)=Qwaterm_{\text{metal}} \cdot c_{\text{metal}} \cdot (T_{\text{initial metal}} - T_{\text{final water}}) = Q_{\text{water}}

    Substitute values: 0.2500(Tinitial metal71)=2948400.2 \cdot 500 \cdot (T_{\text{initial metal}} - 71) = 294840

  3. Solve for Tinitial metalT_{\text{initial metal}}: 100(Tinitial metal71)=294840100 \cdot (T_{\text{initial metal}} - 71) = 294840 Tinitial metal71=294840100T_{\text{initial metal}} - 71 = \frac{294840}{100} Tinitial metal71=2948.4T_{\text{initial metal}} - 71 = 2948.4 Tinitial metal=2948.4+71T_{\text{initial metal}} = 2948.4 + 71 Tinitial metal3019°CT_{\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?


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.

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Math Problem Analysis

Mathematical Concepts

Heat transfer
Specific heat capacity
Thermodynamics

Formulas

Q = mcΔT (heat energy formula)
Q_water = Q_metal (heat gained by water equals heat lost by metal)

Theorems

Conservation of energy in isolated systems

Suitable Grade Level

High School