A 95.0 g sample of copper $\left( c _y=0.20 J /{ }^{\circ} C \cdot g \right)$ is heated to $82.4^{\circ} C$ and then placed in a container of water $\left(c_y=4.18 J /{ }^{\circ} C \cdot g \right)$ at $22.0^{\circ} C$. The final temperature of the water and the copper is $25.1^{\circ} C$. 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.
$\begin{aligned}
-q_{\text {mas }} & =q_{\text {wate }} \\
-c_m m_m \Delta T_m & =c_w m_v \Delta T_w
$\end{aligned}$
A. 0.246 g H _2 O
B. 4.73 g H _2 O
C. 84.0 g H _2 O
D. 36,700 g H _2 O
Answer (2)
Calculate the change in temperature for the copper: Δ T c = 25. 1 ∘ C − 82. 4 ∘ C = − 57. 3 ∘ C .
Calculate the change in temperature for the water: Δ T w = 25. 1 ∘ C − 22. 0 ∘ C = 3. 1 ∘ C .
Use the formula − c c m c Δ T c = c w m w Δ T w and solve for m w .
Substitute the values and calculate the mass of water: m w = 84.0 g .
Explanation
Problem Analysis We are given a calorimetry problem where a piece of copper is heated and then placed into water. We need to find the mass of the water, assuming all the heat lost by the copper is gained by the water. We are given the mass and specific heat of the copper, the initial and final temperatures of the copper and water, and the specific heat of the water.
Calculating Temperature Changes First, we need to calculate the change in temperature for both the copper and the water.
For the copper: Δ T c = T f − T c , i = 25. 1 ∘ C − 82. 4 ∘ C = − 57. 3 ∘ C
For the water: Δ T w = T f − T w , i = 25. 1 ∘ C − 22. 0 ∘ C = 3. 1 ∘ C
Setting up the Equation Now we use the equation that relates the heat lost by the copper to the heat gained by the water: − c c m c Δ T c = c w m w Δ T w We want to solve for m w , the mass of the water. Rearranging the equation, we get: m w = − c w Δ T w c c m c Δ T c
Calculating the Mass of Water Now, we plug in the given values: m w = − ( 4.18 J / ∘ C ⋅ g ) ( 3. 1 ∘ C ) ( 0.20 J / ∘ C ⋅ g ) ( 95.0 g ) ( − 57. 3 ∘ C ) m w = − 12.958 − 1088.7 g m w = 84.01759530791786 g Rounding to one decimal place, we get: m w ≈ 84.0 g
Final Answer Therefore, the mass of the water in the original container was approximately 84.0 g.
Examples
Calorimetry is used in many real-world applications, such as determining the caloric content of food. By burning a food sample in a calorimeter and measuring the heat released, scientists can determine the number of calories the food contains. This information is then used to create nutrition labels, helping people make informed decisions about their diet. Similarly, calorimetry is used in the chemical industry to measure the heat of reactions, which is crucial for designing safe and efficient chemical processes.
The mass of the water in the container is approximately 84.0 g, calculated using the principles of calorimetry where the heat lost by the copper equals the heat gained by the water. This was done by comparing temperature changes and using specific heat values for both substances.
;
