Problem 75 - Chemistry: The Central Science

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 \(\mathrm{B}\), calculate the amount of energy required to vaporize \(1.00 \mathrm{gal}\) of water beginning with water at \(20^{\circ} \mathrm{C}\). (b) If the energy is provided by electricity costing $$\$ 0.085 / \mathrm{kWh},$$ calculate its cost. (c) If distilled water sells in a grocery store for \(\$ 1.26\) per gal, what percentage of the sales price is represented by the cost of the energy?

Step-by-Step Solution

Answer

The energy required to vaporize 1.00 gal of water at 20°C is approximately 9.84 × 10^6 J, which translates to 2.73 kWh of electricity. The cost of this energy is $0.23. Thus, the energy cost represents about 18.3% of the sales price ($1.26) for a gallon of distilled water in a grocery store.

Step 1: Convert volume

We need to convert the volume of water from 1.00 gal to cubic meters: 1.00 gal × \(\frac{3.78541 \times 10^{-3}\,m^3}{1\,gal}\) = 3.78541 × 10^{-3} \(m^3\) #Step 2: Calculate the mass of water using its density#

Step 2: Find the mass of water

Next, we'll find the mass of water using its density (1000 kg/m³): mass = density × volume mass = 1000 kg/m³ × 3.78541 × 10^{-3} \(m^3\) mass ≈ 3.79 kg #Step 3: Calculate the energy required to raise the temperature of water to its boiling point using its specific heat#

Step 3: Calculate the energy for heating

To raise the temperature of water to its boiling point (100°C), we'll use its specific heat (4.186 J/g•°C): \(\Delta\)T = 100°C - 20°C = 80°C Energy = mass × specific heat × \(\Delta\)T Energy = 3.79 kg × 4.186 × 10^3 J/kg°C × 80°C Energy ≈ 1.27 × 10^6 J #Step 4: Calculate the energy required to vaporize water using its heat of vaporization#

Step 4: Calculate the energy for vaporization

We'll now find the energy required to vaporize the water using its heat of vaporization (2.26 × 10^6 J/kg): Energy = mass × heat of vaporization Energy = 3.79 kg × 2.26 × 10^6 J/kg Energy ≈ 8.57 × 10^6 J #Step 5: Calculate the total energy required#

Step 5: Calculate the total energy

Finally, calculate the total energy required: Total Energy = Energy for heating + Energy for vaporization Total Energy ≈ 1.27 × 10^6 J + 8.57 × 10^6 J Total Energy ≈ 9.84 × 10^6 J #b. Calculate the cost of the energy# #Step 1: Convert the total energy from Joules to kWh#

Step 6: Convert energy to kWh

Next, we need to convert the total energy from Joules to kilowatt-hours (kWh): 1.00 kWh = 3.6 × 10^6 J Total Energy = 9.84 × 10^6 J. Total Energy = 9.84 × 10^6 J × \(\frac{1\,kWh}{3.6\times10^6\,J}\) ≈ 2.73 kWh #Step 2: Calculate the cost of energy#

Step 7: Calculate the energy cost

Now, find the cost of the energy using the given cost per kWh ($0.085/kWh): Energy Cost = Total Energy × Cost per kWh Energy Cost = 2.73 kWh × \(0.085/kWh ≈ \)0.23 #c. Calculate the percentage of the sales price represented by the energy cost#

Step 8: Calculate the percentage

Finally, find the percentage of the sales price represented by the energy cost: Percentage = \(\frac{Energy\,Cost}{Sales\,Price}\) × 100 Percentage = \(\frac{\$0.23}{\$1.26}\) × 100 ≈ 18.3%

Key Concepts

Energy Calculation

Distillation requires significant energy input to convert liquid water into steam. This process involves two primary steps: heating the water to its boiling point and then vaporizing it. Each step requires a distinct energy calculation.
First, when heating water from 20°C to its boiling point (100°C), we must calculate the energy required, which is determined by the specific heat of water and the temperature change.
Second, once the water reaches 100°C, additional energy is needed to convert it from liquid to steam, which is quantified by the heat of vaporization. The total energy is the sum of these two separate calculations, revealing the high energy costs associated with distilling water efficiently.
Understanding these calculations allows us to anticipate the energy requirements and associated costs of water purification through distillation.

Specific Heat of Water

The specific heat of water is an essential concept in energy calculations involving temperature changes. Specific heat refers to the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius. For water, this value is 4.186 J/g°C.
This means that for every gram of water, 4.186 joules of energy is needed to increase the temperature by just one degree Celsius. Knowing this value helps calculate the total energy needed to heat water during processes like distillation.
For example, when heating 3.79 kg of water from 20°C to 100°C, the energy required can be found using the formula: \[Energy = \text{mass} \times \text{specific heat} \times \Delta T\]where \(\Delta T\) is the change in temperature. Plugging in the numbers gives the energy needed for the heating component of distillation.

Heat of Vaporization

Converting water from a liquid to vapor requires a specific amount of energy known as the heat of vaporization. This intrinsic property of water is the amount of energy necessary to transform one kilogram of liquid water into vapor at its boiling point, without changing the temperature.
For water, this value is significant: 2.26 × 10^6 J/kg, indicating a large energy requirement.
To calculate the vaporization energy for a given mass of water, use the formula: \[Energy = \text{mass} \times \text{heat of vaporization}\]In our example of 3.79 kg of water, the energy calculation results in the substantial amount of 8.57 × 10^6 joules, highlighting distillation's energy intensity.
Understanding the heat of vaporization is crucial for effectively estimating the total energy costs in processes where water is boiled and converted into steam.

Joules to kWh Conversion

Joules and kilowatt-hours (kWh) are common units of energy, but they measure quantities differently. Joules are a metric unit, while kWh are typically used in energy billing and consumption contexts.
To convert energy from joules to kWh, use the relation:\[ \text{1 kWh} = 3.6 \times 10^6 \text{ J} \]This conversion is crucial when translating physical energy calculations into practical, everyday contexts such as electricity costs.
For instance, if heating and vaporizing 1 gallon of water requires 9.84 million joules, convert this to kWh by dividing by 3.6 million: \[\frac{9.84 \times 10^6 \text{ J}}{3.6 \times 10^6 \text{ J/kWh}} \approx 2.73 \text{ kWh}\]This figure helps compute the cost of electricity needed for the distillation process, using the local cost per kWh. Such conversions allow energy use to be meaningfully analyzed and tracked.