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 gal of water beginning with water at ${20}^{\circ }\mathrm{C}$ . (b) If the energy is provided by electricity costing $\mathrm{\$}0.085/\mathrm{kWh}$ , calculate its cost. (c) If distilled water sells in a grocery store for $\mathrm{\$}1.26$ per gal, what percentage of the sales price is represented by the cost of the energy?

Short answer

The total heat energy required to vaporize 1.00 gal of water at 20°C is $8.92480\cdot {10}^6\text{J}$, which is equivalent to 2.47911 kWh. The cost of this energy is $0.21073. If distilled water sells for $1.26 per gallon, the cost of the energy represents approximately 16.72% of the sales price.

Step-by-step solution

Convert gallons to liters

To work with the given data, we need to convert the volume of water from gallons to liters (1 gal = 3.78541 L): $V=1.00\text{gal}\times \frac{3.78541\text{L}}{1\text{gal}}=3.78541\text{L}$

Determine the mass of water in kilograms

Using the density of water (ρ = 1.00 g/mL or 1.00 kg/L), we can find the mass of water: $m=\rho \times V=1.00\frac{\text{kg}}{\text{L}}\times 3.78541\text{L}=3.78541\text{kg}$

Calculate the heat energy to raise the temperature to 100°C

We use the specific heat of water (c = 4.18 J/g°C or 4186 J/kg°C) and the initial temperature (Ti = 20°C) to calculate the heat energy (Q1) required to raise the temperature to the boiling point (Tf = 100°C): $Q_1=m\times c\times (T_f-T_i)=3.78541\text{kg}\times 4186\frac{\text{J}}{\text{kg}{\cdot }^{\circ }\text{C}}\times ({100}^{\circ }\text{C}-{20}^{\circ }\text{C})=3.78541\cdot {10}^5\text{J}$

Calculate the heat energy to vaporize the water

Now we use the heat of vaporization of water (Lv = 2.257 kJ/g or 2.257 x 10^6 J/kg) to calculate the heat energy (Q2) required to vaporize the water: $Q_2=m\times L_v=3.78541\text{kg}\times 2.257\times {10}^6\frac{\text{J}}{\text{kg}}=8.54616\cdot {10}^6\text{J}$

Calculate the total heat energy required

Add the energies calculated in Steps 3 and 4 to find the total heat energy (Q_total) required to heat and vaporize the water: $Q_{\text{total}}=Q_1+Q_2=3.78541\cdot {10}^5\text{J}+8.54616\cdot {10}^6\text{J}=8.92480\cdot {10}^6\text{J}$

Convert the energy to kilowatt-hours

We need to convert the energy from joules to kilowatt-hours (kWh) to determine its cost: $Q_{\text{kWh}}=\frac{Q_{\text{total}}}{3.6\cdot {10}^6\frac{\text{J}}{\text{kWh}}}=\frac{8.92480\cdot {10}^6\text{J}}{3.6\cdot {10}^6\frac{\text{J}}{\text{kWh}}}=2.47911\text{kWh}$

Calculate the cost of the energy

Use the cost of electricity ($0.085 / kWh) to calculate the cost of the energy required to vaporize 1.00 gal of water: $\text{Cost}=Q_{\text{kWh}}\times \frac{\text{\$0.085}}{\text{kWh}}=2.47911\text{kWh}\times \frac{\text{\$0.085}}{\text{kWh}}=\mathrm{\$}0.21073$

Calculate the percentage of the sales price represented by the energy cost

Divide the energy cost by the sales price of the distilled water ($1.26 per gal) and multiply by 100 to get the percentage: $\text{Percentage}=\frac{\text{Cost}}{\text{Sales Price}}\times 100=\frac{\mathrm{\$}0.21073}{\mathrm{\$}1.26}\times 100=16.72\mathrm{\%}$ So, the cost of the energy represents approximately 16.72% of the sales price of the distilled water.

Key concepts

Specific Heat

When it comes to heating up water during distillation, a crucial factor is its specific heat. Specific heat is defined as the amount of heat energy required to raise the temperature of a unit mass of a substance by one degree Celsius. Water has a high specific heat, meaning it needs a lot of energy to increase its temperature.

For instance, to calculate the energy required to heat 1.00 gal of water from 20°C to the boiling point, the specific heat of water, which is 4186 J/kg°C, is used. This means for every kilogram of water, 4186 joules are needed for each degree Celsius increase in temperature. This substantial energy input for temperature change is why heating water forms a significant portion of distillation's energy consumption.

Heat of Vaporization

Proceeding to the phase change, the heat of vaporization accounts for the energy needed to convert water from liquid to gas without changing its temperature. This energy is substantial because it has to break the intermolecular forces holding the water molecules together.

In the distillation process, after the water is heated to 100°C, the heat of vaporization becomes the next critical energy hurdle. For water, this value is exceptionally high at 2.257 x 10^6 J/kg. During vaporization, this quantity of energy is required for every kilogram of water to turn it into steam. Therefore, the energy needed for the phase change is one of the primary contributors to the high energy cost of distillation.

Energy Cost Analysis

Conducting an energy cost analysis involves determining the total energy required for a process and translating this into monetary terms to understand its economic impact. In our example, the distillation of 1.00 gal of water necessitates a considerable amount of energy, summed up from heating and vaporization.

By converting the total energy required to kilowatt-hours (kWh) and multiplying by the cost per kWh of electricity, the cost to distill that gallon of water can be determined. This step links the physical energy consumption directly with its financial implications, showing us that the energy cost represents a significant fraction of the retail price of distilled water, which is important for both producers and consumers to realize.

Unit Conversion

Unit conversion is the process of converting different units of measure to a common system to facilitate calculations and comparisons. It's essential for accurately calculating energy values and costs in scientific and engineering contexts.

In distillation calculations, you often start with measurements like gallons for volume and need to convert to liters or kilograms to match with constants like specific heat or heat of vaporization, which are expressed using the metric system. Similarly, energy in joules needs to be converted to kilowatt-hours to calculate the electricity cost, since utility companies charge for electricity based on kWh. Proper unit conversion ensures precision and avoids costly mistakes in both the experiment and the associated energy cost analysis.