A 95.0 g sample of copper (c = 0.20 J/°C·g) is heated to 82.4°C and then placed in a container of water (c = 4.18 J/°C·g) at 22.0°C. The final temperature of the water and the copper is 25.1°C. What was the mass of the water in the original container? Assume that all heat lost by the copper is gained by the water.

Use the formulas below to help in your problem-solving:

[tex]\[
\begin{aligned}
-q_{\text{copper}} &= q_{\text{water}} \\
- c_{\text{copper}} m_{\text{copper}} \Delta T_{\text{copper}} &= c_{\text{water}} m_{\text{water}} \Delta T_{\text{water}}
\end{aligned}
\][/tex]

A. 0.246 g H₂O
B. 4.73 g H₂O
C. 84.0 g H₂O
D. 36,700 g H₂O



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].