Problem 58 - Chemistry An Atoms First Approach

Question

A \(110 .-\mathrm{g}\) sample of copper (specific heat capacity \(=\) the \(0.20 \mathrm{J} /^{\circ} \mathrm{C} \cdot \mathrm{g}\)) is heated to \(82.4^{\circ} \mathrm{C}\) and then placed in a container of water at \(22.3^{\circ} \mathrm{C} .\) The final temperature of the water and copper is \(24.9^{\circ} \mathrm{C} .\) What is the mass of the water in the container, assuming that all the heat lost by the copper is gained by the water?

Step-by-Step Solution

Answer

The mass of the water in the container is approximately \(116.5 \,\text{g}\).

Step 1: Write down the formula for heat exchange

The formula for heat exchange is given by: \(q = mcΔT\) where: - \(q\) = heat exchanged - \(m\) = mass - \(c\) = specific heat capacity - \(ΔT\) = change in temperature Since the heat lost by the copper is gained by the water, we can write the equation as: \(m_{Cu}c_{Cu}ΔT_{Cu} = m_{H_2O}c_{H_2O}ΔT_{H_2O}\), where the subscripts refer to copper and water.

Step 2: Calculate the heat lost by the copper

We are given the following values for the copper: - \(m_{Cu} = 110 \,\text{g}\) - \(c_{Cu} = 0.20 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C})\) - \(T_{initial(Cu)} = 82.4 \,^{\circ}\mathrm{C}\) - \(T_{final(Cu)} = 24.9 \,^{\circ}\mathrm{C}\) The change in temperature for copper is: \(ΔT_{Cu} = T_{final(Cu)} - T_{initial(Cu)} = 24.9 - 82.4 = -57.5^{\circ}\mathrm{C}\) Now, we can calculate the heat lost by the copper: \(q_{Cu} = m_{Cu} \cdot c_{Cu} \cdot ΔT_{Cu} = 110 \,\text{g} \cdot 0.20 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C}) \cdot (-57.5^{\circ}\mathrm{C}) = -1265 \,\text{J}\)

Step 3: Calculate the heat gained by the water

Since the heat lost by the copper is gained by the water, we have: \(q_{H_2O} = -q_{Cu} = 1265 \,\text{J}\) We know the specific heat capacity of water: \(c_{H_2O} = 4.18 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C})\) and the initial and final temperature of the water: - \(T_{initial(H_2O)} = 22.3 \,^{\circ}\mathrm{C}\) - \(T_{final(H_2O)} = 24.9 \,^{\circ}\mathrm{C}\) The change in temperature for water is: \(ΔT_{H_2O} = T_{final(H_2O)} - T_{initial(H_2O)} = 24.9 - 22.3 = 2.6^{\circ}\mathrm{C}\) Now we can calculate the mass of water in the container:

Step 4: Calculate the mass of water

Using the equation \(q_{H_2O} =m_{H_2O} \cdot c_{H_2O} \cdot ΔT_{H_2O}\), we solve for the mass of water: \(m_{H_2O} = \dfrac{q_{H_2O}}{c_{H_2O} \cdot ΔT_{H_2O}} = \dfrac{1265\,\text{J}}{4.18 \,\mathrm{J}/(\mathrm{g} \cdot ^{\circ}\mathrm{C}) \cdot 2.6^{\circ}\mathrm{C}} \approx 116.5\, \mathrm{g}\) So, the mass of the water in the container is approximately \(116.5 \,\text{g}\).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a property that tells us how much heat energy is needed to raise the temperature of a given mass of a substance by one degree Celsius. It is denoted by the symbol 'c' and is measured in units of joules per gram per degree Celsius (\( \text{J/(g \textdegree C)} \) or \( \text{J/(kg \textdegree C)} \) if using kilograms).

The higher the specific heat capacity, the more energy is required to cause a change in temperature. For instance, water has a high specific heat capacity of 4.18 \text{J/(g \textdegree C)}, which means it can absorb a lot of heat before its temperature changes significantly. This property is crucial in heat exchange problems, as it establishes a relationship between heat added or lost and the corresponding temperature change. When we apply this concept to our problem, we use the known specific heat capacity of copper to determine the heat exchanged during its temperature drop.

Temperature Change in Heat Transfer

In the context of heat transfer, the temperature change (\( \text{ΔT} \) represents the difference between the final temperature and the initial temperature of a substance. It is a crucial factor when calculating the heat absorbed or released by a substance during a thermal exchange. For positive values of \( \text{ΔT} \) , the substance has absorbed heat, leading to an increase in temperature. Conversely, a negative \( \text{ΔT} \) indicates heat loss and a temperature decrease.

In our textbook exercise, we see this concept in action as the copper cools down from \( 82.4\textdegree C \) to \( 24.9\textdegree C \) . This temperature change is used along with the specific heat capacity and mass to determine the amount of heat lost by the copper. Similarly, the water's temperature change from \( 22.3\textdegree C \) to \( 24.9\textdegree C \) allows us to calculate the heat it had gained.

Mass Calculation in Heat Exchange

Calculating the mass in a heat exchange problem often involves knowing the amount of heat transferred, the specific heat capacity, and the temperature change. The relationship between these variables is given by the formula \( q = mc\text{ΔT} \) , where 'm' represents mass. By rearranging the formula, \( m = \frac{q}{c\text{ΔT}} \) , we can solve for the unknown mass when the other values are known.

In our exercise, we look for the mass of the water that absorbs the heat lost by the copper. By using the calculated heat transfer (q) and the known specific heat capacity and temperature change for water, we arrive at the mass of the water. This step is critical for understanding heat exchange systems and energy balance in a variety of real-world applications, from everyday cooking to complex industrial processes.