Answer :
Let's solve this in a step-by-step manner:
1. Identify the given data:
- Mass of copper ([tex]\(m_c\)[/tex]): [tex]\(95.0 \, \text{g}\)[/tex]
- Specific heat capacity of copper ([tex]\(c_c\)[/tex]): [tex]\(0.20 \, \text{J/g} \cdot ^\circ\text{C}\)[/tex]
- Initial temperature of copper ([tex]\(T_{i,c}\)[/tex]): [tex]\(82.4^\circ \text{C}\)[/tex]
- Final temperature ([tex]\(T_f\)[/tex]): [tex]\(25.1^\circ \text{C}\)[/tex]
- Specific heat capacity of water ([tex]\(c_w\)[/tex]): [tex]\(4.18 \, \text{J/g} \cdot ^\circ\text{C}\)[/tex]
- Initial temperature of water ([tex]\(T_{i,w}\)[/tex]): [tex]\(22.0^\circ \text{C}\)[/tex]
2. Calculate the change in temperature for copper ([tex]\(\Delta T_c\)[/tex]):
[tex]\[ \Delta T_c = T_f - T_{i,c} = 25.1^\circ \text{C} - 82.4^\circ \text{C} = -57.3^\circ \text{C} \][/tex]
3. Calculate the change in temperature for water ([tex]\(\Delta T_w\)[/tex]):
[tex]\[ \Delta T_w = T_f - T_{i,w} = 25.1^\circ \text{C} - 22.0^\circ \text{C} = 3.1^\circ \text{C} \][/tex]
4. Calculate the heat lost by the copper ([tex]\(q_c\)[/tex]):
[tex]\[ q_c = c_c \times m_c \times \Delta T_c \][/tex]
Plug in the values:
[tex]\[ q_c = 0.20 \, \text{J/g} \cdot ^\circ\text{C} \times 95.0 \, \text{g} \times (-57.3^\circ \text{C}) = 1088.7 \, \text{J} \][/tex]
Note: The heat lost by the copper is typically considered negative, but when calculating the absolute value of energy transferred, we use the positive value [tex]\(1088.7 \, \text{J}\)[/tex].
5. Relate the heat gained by the water to the heat lost by the copper:
[tex]\[ q_w = q_c \][/tex]
6. Set up the equation for the heat absorbed by the water:
[tex]\[ q_w = c_w \times m_w \times \Delta T_w \][/tex]
Since [tex]\( q_w = q_c \)[/tex]:
[tex]\[ 1088.7 \, \text{J} = 4.18 \, \text{J/g} \cdot ^\circ\text{C} \times m_w \times 3.1^\circ \text{C} \][/tex]
7. Solve for the mass of the water ([tex]\(m_w\)[/tex]):
[tex]\[ m_w = \frac{1088.7 \, \text{J}}{4.18 \, \text{J/g} \cdot ^\circ\text{C} \times 3.1^\circ \text{C}} \][/tex]
[tex]\[ m_w = \frac{1088.7 \, \text{J}}{12.958 \, \text{J/g}} = 84.01759530791786 \, \text{g} \][/tex]
8. Conclusion:
The mass of the water in the container was approximately [tex]\(84.0 \, \text{g}\)[/tex].
Therefore, the correct answer is [tex]\(84.0 \, \text{g } \text{H}_2 \text{O}\)[/tex].
1. Identify the given data:
- Mass of copper ([tex]\(m_c\)[/tex]): [tex]\(95.0 \, \text{g}\)[/tex]
- Specific heat capacity of copper ([tex]\(c_c\)[/tex]): [tex]\(0.20 \, \text{J/g} \cdot ^\circ\text{C}\)[/tex]
- Initial temperature of copper ([tex]\(T_{i,c}\)[/tex]): [tex]\(82.4^\circ \text{C}\)[/tex]
- Final temperature ([tex]\(T_f\)[/tex]): [tex]\(25.1^\circ \text{C}\)[/tex]
- Specific heat capacity of water ([tex]\(c_w\)[/tex]): [tex]\(4.18 \, \text{J/g} \cdot ^\circ\text{C}\)[/tex]
- Initial temperature of water ([tex]\(T_{i,w}\)[/tex]): [tex]\(22.0^\circ \text{C}\)[/tex]
2. Calculate the change in temperature for copper ([tex]\(\Delta T_c\)[/tex]):
[tex]\[ \Delta T_c = T_f - T_{i,c} = 25.1^\circ \text{C} - 82.4^\circ \text{C} = -57.3^\circ \text{C} \][/tex]
3. Calculate the change in temperature for water ([tex]\(\Delta T_w\)[/tex]):
[tex]\[ \Delta T_w = T_f - T_{i,w} = 25.1^\circ \text{C} - 22.0^\circ \text{C} = 3.1^\circ \text{C} \][/tex]
4. Calculate the heat lost by the copper ([tex]\(q_c\)[/tex]):
[tex]\[ q_c = c_c \times m_c \times \Delta T_c \][/tex]
Plug in the values:
[tex]\[ q_c = 0.20 \, \text{J/g} \cdot ^\circ\text{C} \times 95.0 \, \text{g} \times (-57.3^\circ \text{C}) = 1088.7 \, \text{J} \][/tex]
Note: The heat lost by the copper is typically considered negative, but when calculating the absolute value of energy transferred, we use the positive value [tex]\(1088.7 \, \text{J}\)[/tex].
5. Relate the heat gained by the water to the heat lost by the copper:
[tex]\[ q_w = q_c \][/tex]
6. Set up the equation for the heat absorbed by the water:
[tex]\[ q_w = c_w \times m_w \times \Delta T_w \][/tex]
Since [tex]\( q_w = q_c \)[/tex]:
[tex]\[ 1088.7 \, \text{J} = 4.18 \, \text{J/g} \cdot ^\circ\text{C} \times m_w \times 3.1^\circ \text{C} \][/tex]
7. Solve for the mass of the water ([tex]\(m_w\)[/tex]):
[tex]\[ m_w = \frac{1088.7 \, \text{J}}{4.18 \, \text{J/g} \cdot ^\circ\text{C} \times 3.1^\circ \text{C}} \][/tex]
[tex]\[ m_w = \frac{1088.7 \, \text{J}}{12.958 \, \text{J/g}} = 84.01759530791786 \, \text{g} \][/tex]
8. Conclusion:
The mass of the water in the container was approximately [tex]\(84.0 \, \text{g}\)[/tex].
Therefore, the correct answer is [tex]\(84.0 \, \text{g } \text{H}_2 \text{O}\)[/tex].
