Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Discover a wealth of knowledge from professionals across various disciplines on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
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.
We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
