Problem 84 - Chemistry and Chemical Reactivity

Question

A 192 -g piece of copper is heated to \(100.0^{\circ} \mathrm{C}\) in a boiling water bath and then dropped into a beaker containing \(751 \mathrm{g}\) of water (density \(=1.00 \mathrm{g} / \mathrm{cm}^{3}\) ) at \(4.0^{\circ} \mathrm{C} .\) What was the final temperature of the copper and water after thermal equilibrium was reached? \(\left(C_{\mathrm{Cu}}=0.385 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\right)\)

Step-by-Step Solution

Answer

The final temperature of the system is approximately 6.2°C.

Step 1: Identify Known Values

We have the following information:- Mass of copper (\(m_{\text{Cu}}\)) = \(192\, \text{g}\)- Initial temperature of copper (\(T_{\text{Cu,i}}\)) = \(100.0\, ^{\circ}\text{C}\)- Specific heat capacity of copper (\(C_{\text{Cu}}\)) = \(0.385\, \text{J/g\cdot K}\)- Mass of water (\(m_{\text{w}}\)) = \(751\, \text{g}\)- Initial temperature of water (\(T_{\text{w,i}}\)) = \(4.0\, ^{\circ}\text{C}\)- Specific heat capacity of water (\(C_{\text{w}}\)) = \(4.18\, \text{J/g\cdot K}\)

Step 2: Set Up Heat Balance Equation

At thermal equilibrium, the heat lost by copper equals the heat gained by water:\[-m_{\text{Cu}} \cdot C_{\text{Cu}} \cdot (T_{\text{f}} - T_{\text{Cu,i}}) = m_{\text{w}} \cdot C_{\text{w}} \cdot (T_{\text{f}} - T_{\text{w,i}})\]where \(T_{\text{f}}\) is the final equilibrium temperature.

Step 3: Solve for the Final Temperature

1. Plug in the known values: \[-192 \cdot 0.385 \cdot (T_{\text{f}} - 100) = 751 \cdot 4.18 \cdot (T_{\text{f}} - 4)\]2. Simplify and solve for \(T_{\text{f}}\): - Left Side: \[-73.92(T_{\text{f}} - 100) = -73.92T_{\text{f}} + 7392\] - Right Side: \[3135.18(T_{\text{f}} - 4) = 3135.18T_{\text{f}} - 12540.72\]3. Rearrange and solve:\[-73.92T_{\text{f}} + 7392 = 3135.18T_{\text{f}} - 12540.72\]4. Combine like terms: \[7392 + 12540.72 = 3135.18T_{\text{f}} + 73.92T_{\text{f}}\]5. Solve for \(T_{\text{f}}\):\[T_{\text{f}} = \frac{19932.72}{3209.1} \approx 6.2^{\circ}\text{C}\]

Key Concepts

Specific Heat Capacity

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius (or Kelvin). It is a property that tells us how much heat energy a material can store. Different substances have different specific heat capacities.
For instance, water has a high specific heat capacity of 4.18 J/g·K. This means it can absorb a lot of heat before it exhibits a noticeable change in temperature. Copper, on the other hand, has a specific heat capacity of 0.385 J/g·K, indicating it heats up and cools down quickly with smaller amounts of heat transfer.
Understanding specific heat capacity allows us to predict how different materials will react when heated or cooled, which is crucial for processes like calorimetry and understanding thermal equilibrium.

Heat Transfer

Heat transfer is the process by which thermal energy moves from a hotter entity to a cooler one. There are three primary methods of heat transfer: conduction, convection, and radiation. In our scenario involving copper and water, the focus is on conduction – the direct transfer of heat through a material.
When the hot copper is immersed in cooler water, the copper loses heat, and the water gains it until both reach thermal equilibrium. This means the heat given off by the copper equals the heat absorbed by the water, and they subsequently attain the same temperature. This transfer is governed by the law of energy conservation, which states that energy cannot be created or destroyed, only converted from one form to another. Thus, precise calculations involving heat balances are crucial to solving such problems.

Calorimetry

Calorimetry is the science of measuring the amount of heat involved in a chemical reaction or other process. The main tool for this measurement is a calorimeter, which consists of a well-insulated container. The goal is to measure changes in temperature caused by a particular reaction or physical transformation.
By employing the principles of calorimetry, we can precisely calculate the heat exchange in our copper-water situation. The formula used, relates the mass, specific heat capacity, and temperature changes of the substances involved. Here, the equation reflects heat lost by copper as equal to the heat gained by water, facilitating determination of their final shared temperature.
Through calorimetry, we effectively harness these measurements to understand and predict the thermodynamic behavior during interactions, making it a foundational tool in physical science.

• Provides insights into energy exchange
• Utilizes equations to relate materials' properties
• Ensures measurement accuracy with well-constructed calorimeters