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Sagot :
Sure, let's break down this problem step-by-step:
### Step 1: Convert Masses to Kilograms
- Mass of ice: [tex]\(3.20 \, \text{g} = 0.0032 \, \text{kg}\)[/tex]
- Mass of water: [tex]\(85 \, \text{g} = 0.085 \, \text{kg}\)[/tex]
- Mass of copper calorimeter: [tex]\(50 \, \text{g} = 0.050 \, \text{kg}\)[/tex]
### Step 2: Given Data
- Initial temperature of ice: [tex]\(-15^{\circ} \text{C}\)[/tex]
- Initial temperature of water: [tex]\(40^{\circ} \text{C}\)[/tex]
- Specific heat capacity of ice: [tex]\(2,100 \, \text{J/kg} \cdot \text{K}\)[/tex]
- Specific heat capacity of water: [tex]\(4,186 \, \text{J/kg} \cdot \text{K}\)[/tex]
- Specific heat capacity of copper: [tex]\(400 \, \text{J/kg} \cdot \text{K}\)[/tex]
- Latent heat of fusion of ice: [tex]\(3.34 \times 10^5 \, \text{J/kg}\)[/tex]
### Step 3: Heat Required to Warm Ice from [tex]\(-15^{\circ} \text{C}\)[/tex] to [tex]\(0^{\circ} \text{C}\)[/tex]
[tex]\[ Q_1 = m_{\text{ice}} \times c_{\text{ice}} \times \Delta T_{\text{ice}} \][/tex]
[tex]\[ Q_1 = 0.0032 \, \text{kg} \times 2,100 \, \text{J/kg} \cdot \text{K} \times (0^{\circ} \text{C} - (-15^{\circ} \text{C})) \][/tex]
[tex]\[ Q_1 = 0.0032 \times 2,100 \times 15 \][/tex]
[tex]\[ Q_1 = 100.8 \, \text{J} \][/tex]
### Step 4: Heat Required to Melt Ice at [tex]\(0^{\circ} \text{C}\)[/tex]
[tex]\[ Q_2 = m_{\text{ice}} \times L_f \][/tex]
[tex]\[ Q_2 = 0.0032 \, \text{kg} \times 3.34 \times 10^5 \, \text{J/kg} \][/tex]
[tex]\[ Q_2 = 1,068.8 \, \text{J} \][/tex]
Now, the total heat required for the ice to become liquid water at [tex]\(0^{\circ} \text{C}\)[/tex]:
[tex]\[ Q_{\text{total ice}} = Q_1 + Q_2 \][/tex]
[tex]\[ Q_{\text{total ice}} = 100.8 + 1,068.8 \][/tex]
[tex]\[ Q_{\text{total ice}} = 1,169.6 \, \text{J} \][/tex]
### Step 5: Heat Lost by Water and Calorimeter
The water and copper calorimeter will lose heat as they cool down to the final temperature [tex]\( T_f \)[/tex].
### Final Balance Equation
The heat gained by ice (to warm up and melt) will be equal to the heat lost by the water and the copper calorimeter combined:
[tex]\[ Q_{\text{ice to } T_f} = m_{\text{ice}} \times c_{\text{water}} \times (T_f - 0) \][/tex]
[tex]\[ Q_{\text{water to } T_f} = m_{\text{water}} \times c_{\text{water}} \times (40 - T_f) \][/tex]
[tex]\[ Q_{\text{copper to } T_f} = m_{\text{copper}} \times c_{\text{copper}} \times (40 - T_f) \][/tex]
Using the heat balance equation:
[tex]\[ Q_{\text{total ice}} + Q_{\text{ice to } T_f} = Q_{\text{water to } T_f} + Q_{\text{copper to } T_f} \][/tex]
And solving numerically, we find that:
[tex]\[ T_f \approx 35.618 \, ^\circ \text{C} \][/tex]
### Final Results:
1. Heating ice to [tex]\(0^{\circ} \text{C}\)[/tex]:
- Energy required: [tex]\(100.8 \, \text{J}\)[/tex]
2. Melting ice at [tex]\(0^{\circ} \text{C}\)[/tex]:
- Energy required: [tex]\(1,068.8 \, \text{J}\)[/tex]
3. Final temperature of the resulting mixture:
- [tex]\(35.618 \, ^\circ \text{C}\)[/tex]
This concludes our detailed step-by-step solution for the problem.
### Step 1: Convert Masses to Kilograms
- Mass of ice: [tex]\(3.20 \, \text{g} = 0.0032 \, \text{kg}\)[/tex]
- Mass of water: [tex]\(85 \, \text{g} = 0.085 \, \text{kg}\)[/tex]
- Mass of copper calorimeter: [tex]\(50 \, \text{g} = 0.050 \, \text{kg}\)[/tex]
### Step 2: Given Data
- Initial temperature of ice: [tex]\(-15^{\circ} \text{C}\)[/tex]
- Initial temperature of water: [tex]\(40^{\circ} \text{C}\)[/tex]
- Specific heat capacity of ice: [tex]\(2,100 \, \text{J/kg} \cdot \text{K}\)[/tex]
- Specific heat capacity of water: [tex]\(4,186 \, \text{J/kg} \cdot \text{K}\)[/tex]
- Specific heat capacity of copper: [tex]\(400 \, \text{J/kg} \cdot \text{K}\)[/tex]
- Latent heat of fusion of ice: [tex]\(3.34 \times 10^5 \, \text{J/kg}\)[/tex]
### Step 3: Heat Required to Warm Ice from [tex]\(-15^{\circ} \text{C}\)[/tex] to [tex]\(0^{\circ} \text{C}\)[/tex]
[tex]\[ Q_1 = m_{\text{ice}} \times c_{\text{ice}} \times \Delta T_{\text{ice}} \][/tex]
[tex]\[ Q_1 = 0.0032 \, \text{kg} \times 2,100 \, \text{J/kg} \cdot \text{K} \times (0^{\circ} \text{C} - (-15^{\circ} \text{C})) \][/tex]
[tex]\[ Q_1 = 0.0032 \times 2,100 \times 15 \][/tex]
[tex]\[ Q_1 = 100.8 \, \text{J} \][/tex]
### Step 4: Heat Required to Melt Ice at [tex]\(0^{\circ} \text{C}\)[/tex]
[tex]\[ Q_2 = m_{\text{ice}} \times L_f \][/tex]
[tex]\[ Q_2 = 0.0032 \, \text{kg} \times 3.34 \times 10^5 \, \text{J/kg} \][/tex]
[tex]\[ Q_2 = 1,068.8 \, \text{J} \][/tex]
Now, the total heat required for the ice to become liquid water at [tex]\(0^{\circ} \text{C}\)[/tex]:
[tex]\[ Q_{\text{total ice}} = Q_1 + Q_2 \][/tex]
[tex]\[ Q_{\text{total ice}} = 100.8 + 1,068.8 \][/tex]
[tex]\[ Q_{\text{total ice}} = 1,169.6 \, \text{J} \][/tex]
### Step 5: Heat Lost by Water and Calorimeter
The water and copper calorimeter will lose heat as they cool down to the final temperature [tex]\( T_f \)[/tex].
### Final Balance Equation
The heat gained by ice (to warm up and melt) will be equal to the heat lost by the water and the copper calorimeter combined:
[tex]\[ Q_{\text{ice to } T_f} = m_{\text{ice}} \times c_{\text{water}} \times (T_f - 0) \][/tex]
[tex]\[ Q_{\text{water to } T_f} = m_{\text{water}} \times c_{\text{water}} \times (40 - T_f) \][/tex]
[tex]\[ Q_{\text{copper to } T_f} = m_{\text{copper}} \times c_{\text{copper}} \times (40 - T_f) \][/tex]
Using the heat balance equation:
[tex]\[ Q_{\text{total ice}} + Q_{\text{ice to } T_f} = Q_{\text{water to } T_f} + Q_{\text{copper to } T_f} \][/tex]
And solving numerically, we find that:
[tex]\[ T_f \approx 35.618 \, ^\circ \text{C} \][/tex]
### Final Results:
1. Heating ice to [tex]\(0^{\circ} \text{C}\)[/tex]:
- Energy required: [tex]\(100.8 \, \text{J}\)[/tex]
2. Melting ice at [tex]\(0^{\circ} \text{C}\)[/tex]:
- Energy required: [tex]\(1,068.8 \, \text{J}\)[/tex]
3. Final temperature of the resulting mixture:
- [tex]\(35.618 \, ^\circ \text{C}\)[/tex]
This concludes our detailed step-by-step solution for the problem.
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