Problem 39 - 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 \(0.520 \, \text{J/g}^{\circ}\text{C}\).

Step 1: Understand the Problem

We need to find the specific heat capacity \(c\) of titanium using the principle of conservation of energy. The heat lost by the titanium must equal the heat gained by the water.

Step 2: Use the Heat Transfer Formula

The formula for heat transfer is \(q = mc\Delta T\), where \(q\) is the heat exchanged, \(m\) is the mass, \(c\) is the specific heat capacity, and \(\Delta T\) is the temperature change.

Step 3: Calculate Heat Gained by Water

For water: \(m = 75.0 \, \text{g}\), \(c = 4.18 \, \text{J/g}^{\circ}\text{C}\), and \(\Delta T = 24.3 - 21.7 = 2.6^{\circ}\text{C}\). \[ q_{\text{water}} = 75.0 \, \text{g} \times 4.18 \, \text{J/g}^{\circ}\text{C} \times 2.6^{\circ}\text{C} = 815.1 \, \text{J} \]

Step 4: Calculate Heat Lost by Titanium

Let \(c_t\) be the specific heat capacity of titanium, \(m = 20.8 \, \text{g}\), and \(\Delta T = 99.5 - 24.3 = 75.2^{\circ}\text{C}\). Since the heat lost by titanium equals the heat gained by water: \[ 20.8 \, \text{g} \times c_t \times 75.2^{\circ}\text{C} = 815.1 \, \text{J} \]

Step 5: Solve for Specific Heat Capacity of Titanium

Rearrange to find \(c_t\): \[ c_t = \frac{815.1 \, \text{J}}{20.8 \, \text{g} \times 75.2^{\circ}\text{C}} = \frac{815.1}{1565.76} = 0.520 \, \text{J/g}^{\circ}\text{C} \]

Step 6: Conclusion

The specific heat capacity of titanium is approximately \(0.520 \, \text{J/g}^{\circ}\text{C}\).

Key Concepts

Thermal Equilibrium

When a system reaches thermal equilibrium, it means all its parts have arrived at the same temperature. In the exercise, when the titanium metal and the water are mixed, they initially have different temperatures. However, over time, they exchange heat until both reach the same final temperature of \(24.3^{\circ}\mathrm{C}\).
This state is crucial because it signals that no more heat is being transferred between the substances. At thermal equilibrium, the principle of conservation of energy ensures the heat lost by one body is precisely the heat gained by the other. This allows us to calculate unknown variables, such as the specific heat capacity of titanium, using known values from both substances involved.

Heat Transfer

Heat transfer is the flow of thermal energy from a hotter object to a colder one. It continues until thermal equilibrium is reached. This concept is based on the relationship between heat and temperature difference.
In our problem, the formula used is \( q = mc\Delta T \), where \( q \) represents the heat exchanged. This formula helps us quantify how much heat is transferred:

• \( m \) is the mass of the substance.
• \( c \) is the specific heat capacity.
• \( \Delta T \) is the change in temperature.

Heat transfer is fundamental in understanding how substances at different temperatures interact and influence each other.

Conservation of Energy

Conservation of energy is a principle stating that energy in an isolated system cannot be created or destroyed, only transferred or converted from one form to another.
In the context of our calorimetry problem, the conservation of energy ensures that the total heat lost by the titanium is equal to the total heat gained by the water. This is expressed by the equation:

• Heat lost by titanium = Heat gained by water
• \( 20.8 \, \text{g} \times c_t \times 75.2^{\circ}\text{C} = 815.1 \, \text{J} \)

This equation allows us to solve for the unknown variable, which in this case is the specific heat capacity of titanium, \( c_t \). Understanding energy conservation is key for a broad range of physics and chemistry applications.

Coffee-Cup Calorimeter

A coffee-cup calorimeter is a simple device used to measure the heat exchange of a system. It typically involves two nested Styrofoam cups, which provide insulation to minimize heat loss to the environment. This setup allows for the accurate measurement of the heat exchanged between substances inside the calorimeter.
In our exercise, the coffee-cup calorimeter holds the water and the heated titanium. As the two substances interact, the calorimeter system isolates the thermal exchange, making it easier to calculate the specific heat capacities and other thermal properties without external interference.
While basic, coffee-cup calorimeters are excellent educational tools for demonstrating principles like heat transfer and thermal equilibrium. They highlight the practical application of theoretical concepts by allowing precise energy measurements in seemingly simple systems.