Question

The main reason that distillation is a costly method for purifying water is the high energy required to heat and vaporize water. (a) Using the density, specific heat, and heat of vaporization of water from Appendix B, calculate the amount of energy required to vaporize \(1.00 \mathrm{~L}\) of water beginning with water at \(25^{\circ} \mathrm{C}\). (b) If the energy is provided by electricity costing \(\$ 0.085 / \mathrm{kWh},\) calculate its cost. \((\mathbf{c})\) If distilled water sells in a grocery store for \(\$ 0.49\) per \(L,\) what percentage of the sales price is represented by the cost of the energy?

Short answer

The total energy required to vaporize 1.00 L of water at 25°C is approximately 2573.95 kJ, which costs around \$0.061 using electricity at a rate of \$0.085/kWh. This energy cost represents approximately 12.4% of the sales price of distilled water (\$0.49 per L).

Step-by-step solution

Finding the mass of water

First, we need to find the mass of 1.00 L of water. We can do this by using the density of water, which is given as \(1000 \frac{\text{kg}}{\text{m}^3}\). Since there are \(1000 \text{ L}\) in \(1 \text{ m}^3\), the mass of 1.00 L water is: \(m = 1.00 \cdot 10^{-3} \text{m}^3 \times 1000 \frac{\text{kg}}{\text{m}^3} = 1.00 \, \text{kg}\)

Calculating the energy required to raise the temperature

Next, we need to calculate the amount of energy needed to raise the temperature of 1.00 kg of water from \(25^{\circ} \mathrm{C}\) to \(100^{\circ} \mathrm{C}\) (the boiling point of water). We can use the specific heat of water, given as \(c_p = 4.186 \frac{\text{kJ}}{\text{kg} \cdot \text{K}}\). The energy required to heat the water, given by \(Q_{h}\), can be calculated as follows: \(Q_{h} = mc_p \Delta T\) Where m is the mass of the water, \(c_p\) is the specific heat of the water, and \(\Delta T\) is the change in temperature. Plugging in the values, we get: \(Q_{h} = 1.00 \, \text{kg} \times 4.186 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}} \times (100^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C}) = 313.95 \mathrm{~kJ}\)

Calculating the energy required to vaporize water

Now, we need to find the energy required to vaporize the water. We can do this by using the heat of vaporization, given as \(L_v = 2260 \frac{\text{kJ}}{\text{kg}}\) The energy required to vaporize the water, given by \(Q_v\), can be calculated as follows: \(Q_v = mL_v\) Where m is the mass of the water, and \(L_v\) is the heat of vaporization. Plugging in the values, we get: \(Q_v = 1.00 \, \text{kg} \times 2260 \frac{\text{kJ}}{\text{kg}} = 2260 \mathrm{~kJ}\)

Total energy required

We now combine the energy to heat and vaporize the water to find the Total energy required (\(Q_t\)): \(Q_t = Q_h + Q_v = 313.95 \mathrm{~kJ} + 2260 \mathrm{~kJ}= 2573.95 \mathrm{~kJ}\)

Calculating the cost of energy

We are given the cost of electricity as \(\$0.085/\mathrm{kWh}\). We first need to convert our energy from kJ to kWh by dividing by \(3.6 \times 10^3 \mathrm{~kJ/kWh}\): \(Q_t = \frac{2573.95 \mathrm{~kJ}}{3.6 \times 10^3 \mathrm{~kJ/kWh}} = 0.715 \mathrm{~kWh}\) We can now calculate the cost of the energy: \(\text{Cost} = \text{Energy} \times \text{Price per kWh} = 0.715 \mathrm{~kWh} \times \$0.085 /\mathrm{kWh} \approx \$0.061\)

Calculating the percentage of the sales price

We can now find the percentage of the sales price that is represented by the cost of the energy: \(\text{Percentage} = \frac{\text{Cost of Energy}}{\text{Sales Price}} \times 100\% = \frac{\$0.061}{\$0.49} \times 100\% \approx 12.4\% \) So the cost of the energy represents approximately 12.4% of the sales price of distilled water.

Key concepts

Specific Heat Capacity

Specific heat capacity is a measure of how much energy is needed to raise the temperature of a unit mass of a substance by one degree Celsius. For water, this value is notably high, which means it can absorb a lot of heat without a significant rise in temperature. This property makes water an excellent coolant and a mainstay of many industrial processes. In our specific problem, we calculated the energy required to heat 1 kilogram of water from 25°C to 100°C using the specific heat capacity of water, which is 4.186 kJ/kg·K. This required 313.95 kJ of energy.

The formula used for this calculation is:

• {{ \(Q = mc_p \Delta T \) }}

where:

• Q is the heat energy in joules (J)
• m is the mass in kilograms (kg)
• c_p is the specific heat capacity in kJ/kg·K
• ΔT is the change in temperature

Even though this seems straightforward, the high specific heat capacity of water signals the considerable energy needed in heating, which can be costly depending on the volume of water being processed.

Heat of Vaporization

The heat of vaporization is the amount of energy required to convert a given quantity of liquid into gas at a constant temperature. This is an important factor in processes like boiling water, where the molecules need enough energy to break the intermolecular forces binding them in the liquid state. For water, the heat of vaporization is 2260 kJ/kg.

In the exercise, to vaporize 1 kg of water, 2260 kJ is needed, which is significantly higher than simply heating it to the boiling point. Using the formula:

• {{ \(Q_v = mL_v \) }}

where:

• Q_v is the energy for vaporization
• m is the mass of the water (kg)
• L_v is the heat of vaporization (kJ/kg)

The calculation shows that vaporization requires a large energy input, emphasizing why processes like distillation can be energy-intensive and expensive.

Cost of Electricity

The cost of electricity is an important consideration when discussing energy-intensive processes like water distillation. This cost is calculated based on how much energy is used, and the unit cost of electricity. With energy measured in kilowatt-hours (kWh), understanding the conversion from joules or kilojoules is key.

From the exercise, after calculating the total energy required to vaporize 1 liter of water (2573.95 kJ), we convert this to kWh by dividing by 3600 kJ/kWh. The resulting energy use was found to be 0.715 kWh. With electricity priced at $0.085 per kWh, the total cost for this energy use comes out to $0.061.

This highlights the significance of efficiency in industrial processes, as minimizing energy use can lead to considerable cost savings, especially with large-scale operations.

Distillation Process

Distillation is a method of separating components based on differences in their boiling points, often used for purifying or concentrating liquids. It involves heating the liquid to form vapor and then cooling it back to liquid form.

In water distillation, the process efficiently removes dissolved salts and impurities, but it requires both heating the water and providing enough energy for vaporization. As highlighted by the example, this energy requirement can drive up operating costs, especially when dealing with large volumes of water.

The total energy calculated in our exercise was 2573.95 kJ, demonstrating the energy-intensive nature of water distillation. Coupling this with the cost analysis, it shows why this method could be less preferable when cheaper alternatives for water purification are available. Understanding and potentially optimizing this process can lead to energy savings and reduced costs, which is particularly beneficial in industrial water treatment.