Answer :
To solve the problem of finding the mass of water in the original container, we will use the specific heat capacity formula for both copper and water and the principle of conservation of energy, which states that the heat lost by the copper will equal the heat gained by the water.
Here’s the detailed step-by-step solution:
1. Given Data
- Mass of copper ([tex]\(m_{\text{copper}}\)[/tex]): [tex]\(95.0 \, \text{g}\)[/tex]
- Specific heat capacity of copper ([tex]\(c_{\text{copper}}\)[/tex]): [tex]\(0.20 \, \text{J/g} \cdot \degree\text{C}\)[/tex]
- Initial temperature of copper ([tex]\(T_{\text{initial, copper}}\)[/tex]): [tex]\(82.4 \, \degree\text{C}\)[/tex]
- Final temperature of the system ([tex]\(T_{\text{final}}\)[/tex]): [tex]\(25.1 \, \degree\text{C}\)[/tex]
- Initial temperature of water ([tex]\(T_{\text{initial, water}}\)[/tex]): [tex]\(22.0 \, \degree\text{C}\)[/tex]
- Specific heat capacity of water ([tex]\(c_{\text{water}}\)[/tex]): [tex]\(4.18 \, \text{J/g} \cdot \degree\text{C}\)[/tex]
2. Calculate the change in temperature for copper:
[tex]\[ \Delta T_{\text{copper}} = T_{\text{final}} - T_{\text{initial, copper}} = 25.1 \, \degree\text{C} - 82.4 \, \degree\text{C} = -57.3 \, \degree\text{C} \][/tex]
3. Calculate the heat lost by the copper ([tex]\(q_{\text{copper}}\)[/tex]):
[tex]\[ q_{\text{copper}} = -(c_{\text{copper}} \times m_{\text{copper}} \times \Delta T_{\text{copper}}) = -(0.20 \, \text{J/g} \cdot \degree\text{C} \times 95.0 \, \text{g} \times -57.3 \, \degree\text{C}) = 1088.7 \, \text{J} \][/tex]
4. Change in temperature for water:
[tex]\[ \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 25.1 \, \degree\text{C} - 22.0 \, \degree\text{C} = 3.1 \, \degree\text{C} \][/tex]
5. Heat gained by water ([tex]\(q_{\text{water}}\)[/tex]):
The heat gained by the water is equal to the heat lost by the copper:
[tex]\[ q_{\text{water}} = q_{\text{copper}} = 1088.7 \, \text{J} \][/tex]
6. Calculate the mass of water ([tex]\(m_{\text{water}}\)[/tex]):
We use the relationship:
[tex]\[ q_{\text{water}} = c_{\text{water}} \times m_{\text{water}} \times \Delta T_{\text{water}} \][/tex]
[tex]\[ m_{\text{water}} = \frac{q_{\text{water}}}{c_{\text{water}} \times \Delta T_{\text{water}}} = \frac{1088.7 \, \text{J}}{4.18 \, \text{J/g} \cdot \degree\text{C} \times 3.1 \, \degree\text{C}} \approx 84.0 \, \text{g} \][/tex]
Therefore, the mass of the water in the original container is approximately [tex]\(84.0 \, \text{g}\)[/tex].
The correct answer is:
[tex]\[ 84.0 \, \text{g H}_2 \text{O} \][/tex]
Here’s the detailed step-by-step solution:
1. Given Data
- Mass of copper ([tex]\(m_{\text{copper}}\)[/tex]): [tex]\(95.0 \, \text{g}\)[/tex]
- Specific heat capacity of copper ([tex]\(c_{\text{copper}}\)[/tex]): [tex]\(0.20 \, \text{J/g} \cdot \degree\text{C}\)[/tex]
- Initial temperature of copper ([tex]\(T_{\text{initial, copper}}\)[/tex]): [tex]\(82.4 \, \degree\text{C}\)[/tex]
- Final temperature of the system ([tex]\(T_{\text{final}}\)[/tex]): [tex]\(25.1 \, \degree\text{C}\)[/tex]
- Initial temperature of water ([tex]\(T_{\text{initial, water}}\)[/tex]): [tex]\(22.0 \, \degree\text{C}\)[/tex]
- Specific heat capacity of water ([tex]\(c_{\text{water}}\)[/tex]): [tex]\(4.18 \, \text{J/g} \cdot \degree\text{C}\)[/tex]
2. Calculate the change in temperature for copper:
[tex]\[ \Delta T_{\text{copper}} = T_{\text{final}} - T_{\text{initial, copper}} = 25.1 \, \degree\text{C} - 82.4 \, \degree\text{C} = -57.3 \, \degree\text{C} \][/tex]
3. Calculate the heat lost by the copper ([tex]\(q_{\text{copper}}\)[/tex]):
[tex]\[ q_{\text{copper}} = -(c_{\text{copper}} \times m_{\text{copper}} \times \Delta T_{\text{copper}}) = -(0.20 \, \text{J/g} \cdot \degree\text{C} \times 95.0 \, \text{g} \times -57.3 \, \degree\text{C}) = 1088.7 \, \text{J} \][/tex]
4. Change in temperature for water:
[tex]\[ \Delta T_{\text{water}} = T_{\text{final}} - T_{\text{initial, water}} = 25.1 \, \degree\text{C} - 22.0 \, \degree\text{C} = 3.1 \, \degree\text{C} \][/tex]
5. Heat gained by water ([tex]\(q_{\text{water}}\)[/tex]):
The heat gained by the water is equal to the heat lost by the copper:
[tex]\[ q_{\text{water}} = q_{\text{copper}} = 1088.7 \, \text{J} \][/tex]
6. Calculate the mass of water ([tex]\(m_{\text{water}}\)[/tex]):
We use the relationship:
[tex]\[ q_{\text{water}} = c_{\text{water}} \times m_{\text{water}} \times \Delta T_{\text{water}} \][/tex]
[tex]\[ m_{\text{water}} = \frac{q_{\text{water}}}{c_{\text{water}} \times \Delta T_{\text{water}}} = \frac{1088.7 \, \text{J}}{4.18 \, \text{J/g} \cdot \degree\text{C} \times 3.1 \, \degree\text{C}} \approx 84.0 \, \text{g} \][/tex]
Therefore, the mass of the water in the original container is approximately [tex]\(84.0 \, \text{g}\)[/tex].
The correct answer is:
[tex]\[ 84.0 \, \text{g H}_2 \text{O} \][/tex]
