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?

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: \(m = 1.00 \cdot 10^{-3} \text{m}^3 \times 1000 \frac{\text{kg}}{\text{m}^3} = 1.00 \, \text{kg}\)

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: \(Q_{h} = 1.00 \, \text{kg} \times 4.186 \, \frac{\text{kJ}}{\text{kg} \cdot \text{K}} \times (100^{\circ} \mathrm{C} - 25^{\circ} \mathrm{C}) = 313.95 \mathrm{~kJ}\)

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: \(Q_v = 1.00 \, \text{kg} \times 2260 \frac{\text{kJ}}{\text{kg}} = 2260 \mathrm{~kJ}\)

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

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: \(\text{Cost} = \text{Energy} \times \text{Price per kWh} = 0.715 \mathrm{~kWh} \times \$0.085 /\mathrm{kWh} \approx \$0.061\)

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.