Problem 33 - Chemistry and Chemical Reactivity

Question

A piece of titanium metal with a mass of \(20.8 \mathrm{g}\) is heated in boiling water to \(99.5^{\circ} \mathrm{C}\) and then dropped into a coffee-cup calorimeter containing \(75.0 \mathrm{g}\) of water at \(21.7^{\circ} \mathrm{C} .\) When thermal equilibrium is reached, the final temperature is \(24.3^{\circ} \mathrm{C} .\) Calculate the specific heat capacity of titanium.

Step-by-Step Solution

Answer

The specific heat capacity of titanium is approximately 0.54 J/g°C.

Step 1: Understand the Problem

We need to find the specific heat capacity of the titanium metal. The metal and the water reach a thermal equilibrium, meaning the heat lost by the titanium equals the heat gained by the water.

Step 2: Write the Heat Transfer Equations

The heat lost or gained by a substance is given by:\[ q = m \cdot c \cdot \Delta T \]Where:- \( q \) is the heat (in Joules)- \( m \) is the mass- \( c \) is the specific heat capacity- \( \Delta T \) is the change in temperature.

Step 3: Calculate Change in Temperature for Both Substances

For Titanium:- Initial temperature, \( T_i = 99.5^{\circ}\mathrm{C} \)- Final temperature, \( T_f = 24.3^{\circ}\mathrm{C} \)- Change in temperature, \( \Delta T_{\mathrm{Ti}} = T_f - T_i = 24.3 - 99.5 = -75.2^{\circ}\mathrm{C} \)For Water:- Initial temperature, \( T_i = 21.7^{\circ}\mathrm{C} \)- Final temperature, \( T_f = 24.3^{\circ}\mathrm{C} \)- Change in temperature, \( \Delta T_{\mathrm{H2O}} = T_f - T_i = 24.3 - 21.7 = 2.6^{\circ}\mathrm{C} \)

Step 4: Set up the Equation for Heat Transfer

Since the heat lost by titanium is equal to the heat gained by water:\[ m_{\mathrm{Ti}} \cdot c_{\mathrm{Ti}} \cdot \Delta T_{\mathrm{Ti}} = m_{\mathrm{H2O}} \cdot c_{\mathrm{H2O}} \cdot \Delta T_{\mathrm{H2O}} \]Where:- \( m_{\mathrm{Ti}} = 20.8\, \mathrm{g} \)- \( m_{\mathrm{H2O}} = 75.0\, \mathrm{g} \)- \( c_{\mathrm{H2O}} = 4.18\, \mathrm{J/g^{\circ}C} \)

Step 5: Solve for the Specific Heat Capacity of Titanium

Rearrange the equation to solve for \( c_{\mathrm{Ti}} \):\[ c_{\mathrm{Ti}} = \frac{m_{\mathrm{H2O}} \cdot c_{\mathrm{H2O}} \cdot \Delta T_{\mathrm{H2O}}}{m_{\mathrm{Ti}} \cdot \Delta T_{\mathrm{Ti}}} \]Substitute the known values:\[ c_{\mathrm{Ti}} = \frac{75.0 \cdot 4.18 \cdot 2.6}{20.8 \cdot (-75.2)} \approx 0.54 \, \mathrm{J/g^{\circ}C} \]

Step 6: Interpret the Result

The specific heat capacity of titanium is approximately \( 0.54 \, \mathrm{J/g^{\circ}C} \), indicating how much heat energy is needed to raise the temperature of one gram of titanium by one degree Celsius.

Key Concepts

Thermal Equilibrium

When we talk about thermal equilibrium, we're discussing a state in which two objects at different temperatures come into contact and eventually reach a common temperature. This happens because heat energy moves from the hotter object to the cooler one until their temperatures equalize.

In our exercise, the piece of titanium and the water inside the coffee-cup calorimeter reach thermal equilibrium as they share heat with each other. Initially, the titanium is hotter, so it loses heat, while the cooler water gains that heat. The process continues until both the titanium metal and the water stabilize at the same temperature, which is recorded as 24.3°C. This final common temperature is a signature of reaching thermal equilibrium, demonstrating that no more heat is transferred between the two substances.

Understanding thermal equilibrium is crucial because it helps in calculating changes in temperature and heat energy exchanges, as seen in our task to determine the specific heat capacity of titanium.

Heat Transfer Equations

The principle that governs the movement of heat between two objects is quantified using heat transfer equations. One of the most common equations used is:

• \( q = m \cdot c \cdot \Delta T \)

This equation helps us calculate the amount of heat (\( q \)) transferred to or from a substance.

Here, \( m \) represents the mass of the substance, \( c \) is its specific heat capacity, and \( \Delta T \) is the change in temperature from initial to final state. Each of these variables plays a pivotal role. For instance, a larger mass would require more heat energy to achieve the same temperature change compared to a smaller mass.

Using this equation within our specific problem, we analyze both the titanium and water. Heat lost by titanium equals the heat gained by water. By setting their heat equations equal, we're able to derive one from the other and solve for unknown quantities like the specific heat capacity of an unknown substance, which in this case is titanium.

Calorimetry

Calorimetry is the science of measuring heat transfer. It revolves around the use of a calorimeter, which is an insulated device used to measure the amount of heat absorbed or released during a chemical or physical process.

In our exercise, a coffee-cup calorimeter is used to track the thermal exchange between titanium metal and water. Inside the calorimeter, the heat released by the cooling titanium is absorbed by the warmer water, allowing for precise measurements.

This method is crucial as it enables us to explore and quantify the specific heat capacities of various substances, helping us comprehend how different materials react to the introduction or loss of heat. Understanding calorimetry gives insight into changes at a molecular level and showcases how energy conservation principles are applied in real-world scenarios.

Temperature Change

When analyzing heat transfer, a key factor is understanding the concept of temperature change, denoted by \( \Delta T \). This represents the difference between the final temperature and the initial temperature of a substance.

In our specific exercise example, the titanium metal and the water undergo changes in temperature. For titanium, it cools down as it loses heat, while the water heats up. Their temperature changes can be calculated as:

• Titanium: \( \Delta T_{\text{Ti}} = 24.3^{\circ}\text{C} - 99.5^{\circ}\text{C} = -75.2^{\circ}\text{C} \)
• Water: \( \Delta T_{\text{H2O}} = 24.3^{\circ}\text{C} - 21.7^{\circ}\text{C} = 2.6^{\circ}\text{C} \)

These changes help us determine how much heat energy is transferred, which is critical in solving for the specific heat capacity. A deep understanding of how temperature change reflects energy movement provides a more complete picture of heat dynamics in any given system.