Problem 128 - General Chemistry

Question

A piece of iron was heated to \(95.4^{\circ} \mathrm{C}\) and dropped into a constant-pressure calorimeter containing \(284 \mathrm{~g}\) of water at \(32.2^{\circ} \mathrm{C}\). The final temperature of the water and iron was \(51.9^{\circ} \mathrm{C}\). Assuming that the calorimeter itself absorbs a negligible amount of heat, what was the mass (in grams) of the piece of iron? The specific heat of iron is \(0.449 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), and the specific heat of water is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).

Step-by-Step Solution

Answer

The mass of the piece of iron is 1184 g.

Step 1: Identify Known Quantities

We have the initial temperature of the iron, \(T_{i,\text{iron}} = 95.4^{\circ} \mathrm{C}\), and the initial temperature of the water, \(T_{i,\text{water}} = 32.2^{\circ} \mathrm{C}\). The final temperature of both is \(T_f = 51.9^{\circ} \mathrm{C}\). The mass of the water is 284 g, and the specific heats are \(c_{\text{iron}} = 0.449 \mathrm{~J/g^{\circ}C}\) and \(c_{\text{water}} = 4.18 \mathrm{~J/g^{\circ}C}\).

Step 2: Write Heat Transfer Equation

Since no heat is lost to the calorimeter, the heat lost by the iron equals the heat gained by the water. Thus, we write: \[ m_{\text{iron}} \cdot c_{\text{iron}} \cdot (T_{f} - T_{i,\text{iron}}) = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{f} - T_{i,\text{water}}) \]where \(m_{\text{iron}}\) is the mass of the iron.

Step 3: Calculate Heat Gained by Water

Calculate the heat gained by the water using:\[ Q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot (T_{f} - T_{i,\text{water}}) \]Substitute known values:\[ Q_{\text{water}} = 284 \times 4.18 \times (51.9 - 32.2) \]\[ Q_{\text{water}} = 284 \times 4.18 \times 19.7 \]\[ Q_{\text{water}} = 23355.764 \text{ J}\]

Step 4: Solve for Mass of Iron

Using the heat transfer equation,\[ m_{\text{iron}} \cdot 0.449 \cdot (51.9 - 95.4) = -23355.764 \]Solve for \(m_{\text{iron}}\):\[ m_{\text{iron}} \cdot 0.449 \cdot (-43.5) = -23355.764 \]\[ m_{\text{iron}} = \frac{-23355.764}{0.449 \times -43.5} \]\[ m_{\text{iron}} = 1184.0 \text{ g}\].

Step 5: Confirm Negative Heat Flow

Ensure the concept is sound: heat lost by iron is negative of what is gained by water, confirming the calculations work logically.

Key Concepts

Specific Heat Capacity

Specific heat capacity is a fundamental property that describes how much heat energy is needed to raise the temperature of a substance by one degree Celsius. It tells us how a material absorbs heat and is measured in units like J/g°C. Different substances have different specific heat capacities due to their unique molecular structures.

For instance, water has a high specific heat capacity of 4.18 J/g°C, meaning it can absorb or release a lot of heat without much change in temperature. This is why water is great at stabilizing temperatures. Conversely, iron has a specific heat capacity of 0.449 J/g°C, making it more susceptible to temperature changes with the same amount of heat.

• High specific heat: Takes more energy to change temperature.
• Low specific heat: Temperature changes with less energy.

Understanding these values is crucial in heat transfer calculations, indicating how different materials react to gaining or losing heat.

Calorimetry

Calorimetry is the science of measuring the heat of chemical reactions or physical changes. It's a powerful method for studying energy transfer. A common tool in calorimetry is the calorimeter, an insulated container used to measure heat changes between substances in thermal contact.

In our exercise, we use a constant-pressure calorimeter. It assumes all the heat lost by the iron is absorbed by the water, without any heat leaving the system. This assumption simplifies the calculations and is essential for accurate calorimetry.

• The system includes all materials exchanging heat inside the calorimeter.
• No heat is lost to the surroundings, simplifying calculations.

Calorimetry allows us to connect the heat absorbed or released by materials and quantify energy changes during interactions.

Thermal Equilibrium

Thermal equilibrium occurs when two substances reach the same temperature and no net heat flows between them. In our heat transfer study, when iron and water reach a final temperature of 51.9°C, this demonstrates they are in thermal equilibrium.

The concept of thermal equilibrium is important for understanding heat transfer calculations. It helps to establish the ending condition of the system, simplifying the analysis since the energy changes can be equated.

• Energy is conserved: Heat lost by one material equals heat gained by another.
• This condition is used to set up the heat transfer equation.

Knowing when a system is in thermal equilibrium helps define and confirm the accuracy of heat transfer models.

Iron and Water Interactions

The interaction between iron and water in heat transfer provides a practical application of the concepts discussed. Iron initially at a higher temperature transfers heat to cooler water, until both reach the same temperature, indicating thermal equilibrium.

This interaction showcases:

• The role of specific heat capacities in determining the rate of temperature change.
• How different materials transfer heat among each other.
• The resulting equilibrium temperature as a balance point.

When calculating such a scenario, we use the heat transfer equation to quantify the energy changes involved. Iron's lower specific heat capacity leads to a larger temperature change compared to water for the same energy exchange.

Understanding these interactions is critical in fields ranging from materials science to engineering, helping predict how substances will behave in thermal environments.