Problem 39 - Chemistry and Chemical Reactivity

Question

Liquid ammonia, \(\mathrm{NH}_{3}(\ell),\) was once used in home refrigerators as the heat transfer fluid. The specific heat capacity of the liquid is \(4.7 \mathrm{J} / \mathrm{g} \cdot \mathrm{K}\) and that of the vapor is \(2.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .\) The enthalpy of vaporization is \(23.33 \mathrm{kJ} / \mathrm{mol}\) at the boiling point. If you heat \(12 \mathrm{kg}\) of liquid ammonia from \(-50.0^{\circ} \mathrm{C}\) to its boiling point of \(-33.3^{\circ} \mathrm{C},\) allow it to evaporate, and then continue warming to \(0.0^{\circ} \mathrm{C},\) how much energy must you supply?

Step-by-Step Solution

Answer

The energy required is approximately 18,510.5 kJ.

Step 1: Calculate the Energy to Heat Liquid Ammonia

First, we need to calculate the energy required to heat the liquid ammonia from \(-50.0^{\circ} \mathrm{C}\) to the boiling point of \(-33.3^{\circ} \mathrm{C}\). This is done using the specific heat capacity formula:\[ q_1 = m \cdot C_\ell \cdot \Delta T \]Where:- \( m \) is the mass of ammonia, \( m = 12000 \, \mathrm{g} \) (since \( 12 \, \mathrm{kg} = 12000 \, \mathrm{g} \))- \( C_\ell \) is the specific heat capacity of liquid ammonia, \( 4.7 \, \mathrm{J} / \mathrm{g} \cdot \mathrm{K} \)- \( \Delta T \) is the temperature change, \( -33.3^{\circ} \mathrm{C} - (-50.0^{\circ} \mathrm{C}) = 16.7^{\circ} \mathrm{C} \)Substitute the values:\[ q_1 = 12000 \, \mathrm{g} \cdot 4.7 \, \mathrm{J/g \cdot K} \cdot 16.7 \]\[ q_1 = 1191960 \, \mathrm{J} \]

Step 2: Calculate Energy for Vaporization

Next, calculate the energy required for the vaporization of ammonia. This is given by:\[ q_2 = n \cdot \Delta H_v \]Where:- \( n \) is the number of moles, \( n = \frac{12000 \, \mathrm{g}}{17.03 \, \mathrm{g/mol}} \approx 704.76 \, \mathrm{mol}\) (Molar mass of ammonia is 17.03 g/mol)- \( \Delta H_v \) is the enthalpy of vaporization, \( 23.33 \, \mathrm{kJ/mol} = 23330 \, \mathrm{J/mol} \)Substitute the values:\[ q_2 = 704.76 \, \mathrm{mol} \times 23330 \, \mathrm{J/mol} \]\[ q_2 = 16440310.8 \, \mathrm{J} \]

Step 3: Calculate the Energy to Heat Vaporized Ammonia

Finally, calculate the energy required to heat the vaporized ammonia from the boiling point to \(0.0^{\circ} \mathrm{C}\). This is done using:\[ q_3 = m \cdot C_v \cdot \Delta T \]Where:- \( C_v \) is the specific heat capacity of vapor ammonia, \( 2.2 \, \mathrm{J} / \mathrm{g} \cdot \mathrm{K} \)- \( \Delta T \) is \(0.0^{\circ} \mathrm{C} - (-33.3^{\circ} \mathrm{C}) = 33.3^{\circ} \mathrm{C} \)Substitute the values:\[ q_3 = 12000 \, \mathrm{g} \cdot 2.2 \, \mathrm{J/g \cdot K} \cdot 33.3 \]\[ q_3 = 879120 \, \mathrm{J} \]

Step 4: Total Energy Required

Now, add all the energies together to find the total energy required:\[ q_{total} = q_1 + q_2 + q_3 \]Substitute the values from previous steps:\[ q_{total} = 1191960 \, \mathrm{J} + 16440310.8 \, \mathrm{J} + 879120 \, \mathrm{J} \]\[ q_{total} = 18510490.8 \, \mathrm{J} \]Therefore, the total energy required is approximately \( 18510.5 \, \mathrm{kJ} \).

Key Concepts

Specific Heat Capacity

Specific heat capacity is a fundamental property of materials that indicates how much energy is needed to raise the temperature of a unit mass by one degree Celsius.
In the context of this exercise, we are looking at the specific heat capacity of liquid ammonia and its vapor. For liquid ammonia, the specific heat capacity is given as 4.7 J/g·K, which means each gram of liquid ammonia requires 4.7 Joules of energy to increase its temperature by one Kelvin (or degree Celsius, as the scale is the same for \( \Delta T \) when in increments).
Specific heat capacity is crucial when calculating the energy needed for heating processes. In heating the ammonia from \( -50.0^{\circ} \text{C} \) to \( -33.3^{\circ} \text{C} \), we can apply the equation \( q = m \cdot C \cdot \Delta T \). Here, \( q \) is the energy required, \( m \) is the mass, \( C \) represents the specific heat capacity, and \( \Delta T \) is the change in temperature.
Using these variables effectively helps us calculate how much total energy is needed to change the temperature of a given substance within different states or phases.

Enthalpy of Vaporization

Enthalpy of vaporization refers to the energy required to transform a given amount of a liquid into a gas at its boiling point. This energy change is a crucial part in thermodynamics when considering phase changes.
For ammonia, the enthalpy of vaporization at its boiling point is given as 23.33 kJ/mol. This indicates that for each mole of ammonia, 23.33 kJ of energy is needed to change it from a liquid to a vapor.
The formula \( q = n \cdot \Delta H_v \) helps us calculate the energy for vaporization, where \( q \) is the energy involved, \( n \) represents the number of moles, and \( \Delta H_v \) is the enthalpy of vaporization.
In our exercise scenario, the total energy to vaporize liquid ammonia is found by first determining the number of moles from the given mass and then using this formula. This demonstrates how thermal energy plays a key role in phase transitions, enabling us to predict or analyze the energy costs of physical processes in substances.

Energy Calculation

Energy calculation in thermodynamics often involves determining how much energy is transferred or transformed in various processes.
For this exercise, the energy calculation consists of three main parts: heating the ammonia to its boiling point, vaporizing it, and then heating the vapor.
The strategies used include applying specific formulas for each part:

• For heating the liquid, the energy \( q = m \cdot C_l \cdot \Delta T \).
• To vaporize, the energy \( q = n \cdot \Delta H_v \).
• And for heating the vapor, \( q = m \cdot C_v \cdot \Delta T \).

Each of these calculations involves specific heat capacity, changes in temperature, and enthalpy values appropriate for each step, creating a more holistic understanding of the energy required.
Adding up all these calculated energies gives the total amount of energy necessary for the entire process, in this case, approximately 18510.5 kJ. This cumulative approach ensures a precise and comprehensive energy accounting throughout the heating and phase change process.