Problem 35 - 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 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 heat energy must you supply?

Step-by-Step Solution

Answer

18.26 MJ

Step 1: Calculate the heat required to raise the temperature of liquid ammonia

The heat energy required to raise the temperature of a substance is given by the formula \( q = m \cdot c \cdot \Delta T \). Here, \( m \) is the mass of the substance, \( c \) is the specific heat capacity, and \( \Delta T \) is the change in temperature.First, find the change in temperature \( \Delta T \) from \(-50.0^{\circ} \mathrm{C}\) to \(-33.3^{\circ} \mathrm{C}\). Hence, \( \Delta T = -33.3 - (-50.0) = 16.7^{\circ} \mathrm{C} \).Using the formula:\[ q_1 = 12,000\, \mathrm{g} \times 4.7\, \mathrm{J/g \cdot K} \times 16.7 \; \mathrm{K} \]Calculate \( q_1 \):\[ q_1 = 939,720 \; \mathrm{J} \]

Step 2: Calculate the heat required for vaporization

The heat energy required to vaporize a substance is calculated using the enthalpy of vaporization: \( q = n \cdot \Delta H_{\text{vap}} \), where \( n \) is the number of moles and \( \Delta H_{\text{vap}} \) is the enthalpy of vaporization.First, find the number of moles of ammonia. The molar mass of ammonia (\( \mathrm{NH}_3 \)) is approximately \( 17.03 \; \mathrm{g/mol} \).\[ n = \frac{12,000 \; \mathrm{g}}{17.03 \; \mathrm{g/mol}} = 704.99 \; \mathrm{mol} \]Then calculate the heat for vaporization:\[ q_2 = 704.99 \; \mathrm{mol} \times 23.33 \; \mathrm{kJ/mol} \times 1000 \; \mathrm{J/kJ} \]\[ q_2 = 16,440,012.7 \; \mathrm{J} \]

Step 3: Calculate the heat required to raise the vapor temperature

To find the heat required to raise the temperature of ammonia vapor from \(-33.3^{\circ} \mathrm{C}\) to \(0.0^{\circ} \mathrm{C}\), use the formula \( q = m \cdot c \cdot \Delta T \), with the specific heat of the vapor phase.The temperature change \( \Delta T \) is \( 0.0 - (-33.3) = 33.3 \; \mathrm{K}\).Calculate \( q_3 \):\[ q_3 = 12,000 \; \mathrm{g} \times 2.2 \; \mathrm{J/g \cdot K} \times 33.3 \; \mathrm{K} \]\[ q_3 = 879,120 \; \mathrm{J} \]

Step 4: Sum up all required heat energies

Add the heat calculated in each step to find the total heat energy supplied:\[ q_{\text{total}} = q_1 + q_2 + q_3 \]\[ q_{\text{total}} = 939,720 + 16,440,012.7 + 879,120 = 18,258,852.7 \; \mathrm{J} \]

Step 5: Conclusion

Thus, the total heat energy that must be supplied to the system to heat 12 kg of liquid ammonia from \(-50.0^{\circ} \mathrm{C}\) to \(0.0^{\circ} \mathrm{C}\) through evaporation at its boiling point is approximately \( 18,258,853 \; \mathrm{J} \) or \( 18.26 \; \mathrm{MJ} \).

Key Concepts

Specific Heat Capacity

Specific heat capacity is an essential concept in heating and cooling processes. It measures how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius. For ammonia, two different specific heat capacities are important, depending on its state: liquid or vapor.

• Liquid ammonia has a specific heat capacity of 4.7 J/g·K.
• Ammonia vapor has a specific heat capacity of 2.2 J/g·K.

This difference arises because the particles in liquid and vapor states require varying amounts of energy to change temperature due to their different arrangements and interactions. Recognizing the specific heat capacity helps us calculate the energy needed to change temperature accurately.

Phase Changes in Ammonia

Understanding phase changes is key when dealing with substances like ammonia. Ammonia transitions from liquid to gas at its boiling point, which is -33.3°C. This phase change requires a significant amount of energy, driven by the enthalpy of vaporization.

Enthalpy of Vaporization

The enthalpy of vaporization for ammonia is 23.33 kJ/mol. It represents the energy required to convert one mole of ammonia from liquid to gas. This energy is necessary to overcome intermolecular forces, allowing the substance to transition to a less ordered gaseous state. Understanding this concept ensures efficient energy calculations during phase changes.

Heat Energy Calculation

Calculating heat energy involves several steps that take into account the specific heat capacities and phase changes of a substance like ammonia.

Step-by-Step Process

• First, calculate the energy needed to warm the liquid ammonia from -50°C to -33.3°C using its specific heat capacity.
• Next, determine the energy required to vaporize the ammonia at its boiling point using the enthalpy of vaporization.
• Finally, calculate the energy needed to heat the ammonia vapor from -33.3°C to 0.0°C using the specific heat of vapor.

By adding these energy amounts, we find the total heat energy required for the complete process, ensuring all phase changes and temperature variations are considered.