Question

A typical American household uses approximately \(30 \mathrm{kWh}\) per day of electricity. Convert this to Joules and then imagine building a water tank \(10.8 \mathrm{~m}\) above the house \(^{37}\) to supply one day's worth of electricity. \(^{38}\) How much mass of water is this, in kg? At a density of \(1,000 \mathrm{~kg} / \mathrm{m}^{3}\), what is the volume in cubic meters, and what is the side length of a cube \(^{39}\) having this volume? Take a moment to visualize (or sketch) this arrangement.

Short answer

To find the side length of the cubic water tank required, first perform the following calculations: 1. Convert daily electricity usage to Joules: 30 kWh * 3.6 * 10^6 J/kWh = 108,000,000 J 2. Calculate the mass of water needed: m = (108,000,000 J) / (9.81 m/s² * 10.8 m) ≈ 1,081,552 kg 3. Calculate the volume of water: V = 1,081,552 kg / 1,000 kg/m³ = 1,081.552 m³ 4. Determine the side length: a = ∛(1,081.552 m³) ≈ 10.58 m So, the side length of a cubic water tank needed to store one day's worth of electricity for a typical American household is approximately 10.58 meters.

Step-by-step solution

Convert daily electricity usage to Joules

To convert daily electricity usage from 30 kWh to Joules, use the following formula: $$\text{Energy (J)} = \text{Energy (kWh)} \cdot 3.6 \times 10^{6}$$ We'll multiply 30 kWh by the conversion factor to obtain Joules: $$\text{Energy (J)} = 30\,\text{kWh} \cdot 3.6 \times 10^{6}\,\text{J}\,\text{kWh}^{-1}$$

Calculate potential energy

Next, we'll calculate the potential energy of the water at a height of 10.8 m as: $$\text{Potential Energy (J)} = m\,g\,h$$ where \(m\) is the mass of water in kg, \(g\) is the gravitational acceleration (\(9.81 \mathrm{~m/s^2}\)), and \(h\) is the height in meters.

Determine the mass of water needed

Now, we equate the daily electricity requirement (converted to Joules) with the potential energy, to find the mass of water needed: $$30\,\text{kWh} \cdot 3.6 \times 10^{6}\,\text{J}\,\text{kWh}^{-1} = m \cdot 9.81\,\text{m/s²} \cdot 10.8\,\text{m}$$ Solve for the mass of water (m): $$m = \frac{30\,\text{kWh} \cdot 3.6 \times 10^{6}\,\text{J}\,\text{kWh}^{-1}}{9.81\,\text{m/s²} \cdot 10.8\,\text{m}}$$

Calculate the volume of water

After finding the mass of the water (m), we now calculate the volume (V) using the given water density (\(\rho = 1,000\,\text{kg/m³}\)) and the following formula: $$V = \frac{m}{\rho}$$ Use the result from Step 3 to find the volume of water: $$\text{Volume (m³)} = \frac{\text{Mass of water (kg)}}{1,000\,\text{kg/m³}}$$

Determine the side length

Finally, we will determine the side length (a) of a cubic water tank with the volume obtained in Step 4: $$a = \sqrt[3]{\text{Volume}}$$ By plugging in the volume from Step 4, we obtain the side length of the cube: $$\text{Side Length (m)} = \sqrt[3]{\text{Volume (m³)}}$$ After completing these steps, the student should have calculated the mass of water required, the volume of the water, and the side length of a cubic water tank needed to store one day's worth of electricity for a typical American household.

Key concepts

Energy Conversion

Understanding energy conversion is fundamental in physics, especially when we talk about different forms of energy and how one form can be transformed into another. This principle is at the heart of many real-world applications, from power plants to the exercise you've encountered.

In your exercise, you've converted electrical energy, measured in kilowatt-hours (kWh), to mechanical energy in the form of potential energy. This is a direct example of energy conversion. The formula you used for conversion, \( \text{Energy (J)} = \text{Energy (kWh)} \cdot 3.6 \times 10^{6} \), transforms electrical energy (useful for powering appliances) to Joules, which is the standard unit of energy in physics representing the ability to perform work.

Whenever we're talking about hydropower, solar power, or wind turbines, energy conversion is at play. These systems convert natural forces into electrical energy, which we then convert for various uses—just like converting the kWh to Joules, which could then be used to calculate the potential energy stored in water elevated above a house.

Potential Energy Calculation

Potential energy is the stored energy an object possesses because of its position or state. For your exercise, the potential energy is due to the water's position above the ground.

The potential energy (PE) of an object can be calculated using the formula \( \text{PE} = m \cdot g \cdot h \), where \(m\) is mass, \(g\) is the acceleration due to gravity (on Earth, \(9.81\,\text{m/s}^2\)), and \(h\) is height above the reference point. Think of it as the amount of work needed to lift the water to that height or the work that gravity could do if the water were allowed to fall.

In your problem, you've equated the potential energy of water at a given height to the energy needs of a household, represented in Joules. This concept not only demonstrates the energy potential of positional states but also helps highlight the relative energy efficiency or inefficiency of different energy storage methods.

Kinetic Energy

While kinetic energy wasn't directly calculated in your exercise, it's a critical concept to grasp in the realm of energy discussions. Kinetic energy is the energy of motion. Any object in motion has kinetic energy, which can be calculated by the formula \( KE = \frac{1}{2}mv^2 \), where \(m\) is the mass of the object and \(v\) is its velocity.

If the water in our problem were to flow down from the tank, as it would in a hydropower plant, it would convert its potential energy into kinetic energy—the energy of flowing water—which could then be converted into electrical energy. This transformation of energy from potential to kinetic and then back to electrical is what powers hydroelectric dams around the world. Understanding this concept is key in grasping how energy is harnessed and utilized, and it can also help in visualizing different scenarios where energy conservation matters.