Contemporary Challenges: Fall-2026
Homework 2 (SOLUTION): Due Week 2 Friday (10/9)
- Gasoline Sun
S1 5611S
Electromagnetic radiation energy from the Sun arrives at the upper atmosphere of our planet at a rate of about 1350 J/(s \(\cdot\) m\(^2\)). Use this information, together with the average radius of the Earth's orbit, to show that the Sun radiates energy at a rate of about \(4 \times 10^{26}\) J/s.
We want to know the rate that energy radiates from the Sun. At a distance of \(1.5 \times 10^{11}\) m, the energy arrives at a rate of \(1350\) J/(s \(\cdot\) m\(^2\)). By conservation of energy, the rate that at which the energy leaves the sun, must also equal the rate at which it passes through the sphere at radius \(r\) from the sun. Using that the Area of a sphere is \(4\pi r_{\text{orbit}}^2\), we can find the energy per time.
\begin{align} \frac{\text{Energy}}{\text{time}} &= (1350 \text{ J/s.m\(^2\)} )(4 \pi (1.5 \times 10^{11} \text{ m})^2) \\ &= 4 \times 10^{26} \text{ J/s} \end{align}
We know from radiometric dating of rocks on Earth (and the Moon and Mars) that our solar system is about 4 billion years old. Let's make a naïve hypothesis (like scientists did in the early 1900s) that the Sun is powered by burning hydrocarbons. What mass of gasoline would be needed to power the Sun at a rate of \(4 \times 10^{26}\) J/s for 4 billion years? Compare to the actual mass of the Sun.
Note: The energy density of hydrocarbon fuels, including gasoline, natural gas, dry logs of wood, chocolate, croissants, gummy bears, etc. etc. is \(\approx\) 40 MJ/kg.
Total energy emitted as light during the 4 billion year lifetime: \begin{align} E_{\text{light}} &= (4 \times 10^{26} \text{ J/s})(3\times 10^7 \text{ s/y})(4\times 10^9 \text{ y}) \\ &= 50\times10^{42} \text{ J} = 50\times10^{36} \text{ MJ} \end{align} How much gasoline would it take to produce this energy? \begin{align} m_{\text{gasoline}} = \frac{50\times10^{36}\text{ MJ}}{40 \text{ MJ/kg}} = 1.25\times10^{36} \text{ kg} \end{align} We can compare this to the mass of the sun, \(M_{\text{sun}} = 2\times 10^{30}\) kg. The sun would have to be almost 1 million times more massive if it was running on hydrocarbon fuels.
- Hot showers and standard deviation
S1 5611S
(a) Estimate the energy used during a typical 10-minute shower in the United States. The dominant variables in this problem are the temperature rise of the water and the flow rate of the showerhead. Develop a coarse-grained model that incorporates these two variables. For simplicity, you may assume: (i) The water heater is 100% efficient at converting electrical energy into heat, and (ii) the water heater is located close to the shower, so heat losses in the pipes can be neglected.
(b) Develop a reasonable estimate for the standard deviation of your answer to part a. That is, imagine measuring the energy used for a 10-minute shower in a thousand randomly selected U.S. households. How much would the energy use vary? To develop a reasonable estimate, spend a little time doing internet research (or in-person research) on the variability of shower flow rates, the variability of ground temperature, and the variability of shower water temperature (personal preference). Once you can justify reasonable numbers for the variability of these inputs, propogate the uncertainty through your coarse-grained model using the methods we discussed in class.
(c) Sense making: Compare the typical energy used for a 10-minute shower to the typical energy used to drive an elecric car at 70 mph for 10 minutes.
Shower
The flow rate of a shower can vary significantly. The average flow rate is about 2 gal/min, but you can easily find showers using 1.5 gal/min to 2.5 gal/min. The standard deviation is about 0.5 gal/min (this is estimate).
The average 10-minute shower uses 20 gallon of water, which has a mass \(m\) = 80 kg, the standard deviation in water mass is \(\sigma_m=\) 20 kg. Therefore \(\sigma_m/m=\) 25%)
The water must be heated from ground temperature (pipes are buried in the ground where the temperature is about 12 C) to comfortable shower temperature (body temperature is 38 C, water burns skin at 50 C). Most people like the temperature in the range 38 - 42 C. Let's say the average value of \(\Delta T\) = 28 C, and the standard deviation is about 2 C. Therefore, \(\sigma_{\Delta T}/\Delta T=\) 10%)
The heat required is then \(Q = m c \Delta T\), where \(c = 4200\) J/(K.kg) and \(m\) is the mass of the water. \begin{align} Q_\text{avg} &= (80 \text{ kg})(4.2\times10^3 \text{ J/(K.kg)})(28 \text{ K}) = 10 \text{ MJ} \end{align}
Using pythagorium addition of the percentage uncertainties I get a final uncertainty of 27%. My final answer for a 10-minute shower in then \begin{align} Q &= 10 \pm 3\text{ MJ} \end{align}
If I refined my coarse-grained model, I could include the energy required to pump this water up a hill. For a hill of 100 m, let's compare the change in gravitational potential energy to the heat energy: \begin{align} \frac{PE}{Q} = \frac{m g h}{m c \Delta T} = \frac{(10\text{ m/s})(100\text{ m})}{(4200\text{ J/(K kg)})(30\text{ K})} = \frac{10^3}{126\times10^3} = \frac{1}{126} \end{align} For a hill of 100 m, the gravitational potential energy cost is less than \(1\%\) of the heat energy.
Car
In the class notes we estimated 20,000 J/s. Driving the car for 10 minutes would use \begin{align} \text{Energy used } = (2 \times 10^4 \text{ J/s})(6 \times 10^2 \text{ s}) \approx 10 \text{ MJ} \end{align}
The rate that we use electrical energy to heat shower water is very similiar to the rate that we use electrical energy to drive a car at 70 mph.
- Wave energy
S1 5611S
The ocean swell lifts water against the force of gravity and makes water move in circular patterns. Energy can be harvested from this ocean swell: gravitational potential energy and kinetic energy are both available (see Chapter F of MacKay). Imagine I draw a line in the ocean that magically absorbs all of the energy associated with the swell, effectively flattening the waves.
If the line is parallel to the crest of the waves and has length \(L\), the time-averaged rate that energy must be absorbed to flatten the swell (energy per time) is \begin{align} \text{energy per time} &= \frac14 \rho g h^2vL \end{align} where \(\rho\) is the density of water, \(g\) is acceleration due to gravity, \(h\) is the height of the wave crest compared to flat water, \(v\) is the velocity of the wave crest. This equation is derived on page 307 of Sustainable Energy by David McKay.
A real device for harvesting wave energy can achieve an efficiency \begin{align} \text{efficiency} &= \frac{0.5\text{ J (electrical energy)}}{1\text{ J (wave energy)}} \end{align} Assume that typical Oregon deep-water waves have \(v = 15\) m/s and \(h = 1\) m. Estimate the electrical power that Oregon could capture if we built wave farms along the entire coast. Give your answer in J/s. There is 10-20% uncertainty in the estimates of \(v\) and \(h\), so the precision of your calculation and final answer should be consistent with respect to this uncertainty.
We want to find the electrical energy that ocean waves could generate for Oregon. We're using the relationship \begin{align} P_{\text{wave}} &= \frac14 \rho g h^2vL \end{align} which was given. Assuming \(v=\)15 m/s and \(h=\)1 m, the energy per time per length of the wave crest is \(P_{\text{wave}} = 37\) kJ/(s\(\cdot\)m).
We need to know the length of the Oregon coast, measured as a straight line. It takes about 5 hours to drive from the southern border of Oregon to the northern border, driving at 100 km/h. So I'll guess 500 km. The total electrical power generated by the coastline is \begin{align} P_{\text{coast}} &= \left(3.7\times 10^4\frac{\text{J}}{\text{s}\cdot\text{m}}\right) \cdot \left(\frac{0.5 \text{ J (electrical)}}{1 \text{ J (wave)}}\right) \cdot 5\times 10^5\text{ m}\\ &= 9\times 10^9\frac{\text{J}}{\text{s}} \end{align} which is 9 gigawatts.
Sense-making: Make a Comparison---Divide your answer by the population of Oregon. How does the wave energy production rate compare to the rate that people are using energy (see the class notes from Conservation of Energy: First Law of Thermodynamics)?
The population of Oregon is 4.1 million, which means that waves could generate about 2250 J/s per person. This is a little more than a quarter of the average energy usage for a US citizen. I conclude that a significant amount of energy is available from waves, but I worry that coverage of the entire Oregon coast could be expensive, and could also have major ecological repercussions.
It's notable that coverage of 10% of the Oregon coast could produce as much energy as a nuclear power plant. I wonder which solution would cost more (if you did a lifetime analysis).
- AI data centers energy usage in 2026
S1 5611S
AI data centers are operating and under development widely in the year 2026. A global data center tracker shows a known power capacity of 67000 MW in total. This includes some data centers that are not yet operational, but excludes some operational data centers for which power capacity is not reported. For rough estimation purposes, let's assume these effects balance out and the reported total is a good estimate for the current power capacity of AI data centers globally.
To begin building a coarse grained model of the energy flow, let's assume that all energy used by AI data centers are powering a major use-case of AI today: user prompting of AI models. Here is a rough sketch of an energy flow diagram:
[Energy from data centers] ---(100%)---> [Energy used in prompting AI]
The energy used in a single prompt was estimated in 2025 to be about 0.3 Wh (Watt-hours).
(a) What are at least 2 assumptions we are making as we build this coarse-grained model for the energy usage of AI data centers?
(b) Based on the coarse-grained model, estimate the average number of AI prompts one person uses per day.
(c) Sense-making: Does this number make sense based on what you expected to find? If not, what is an idea you have for improving the coarse-grained model?
(a) We assumed the unknown information about data center power capacity would balance out to about 67000 MW in total. We assumed 100% of energy from data centers went straight to prompting. You can probably think of other assumptions too.
(b) We want to find \(n\), the number of prompts per person per day: \(\frac{\text{prompts}}{\text{person}\cdot\text{day}}\).
We currently know \(P_{DC}\), the AI data centers total power globally (in Watts), and \(E_\text{p}\), the energy of a prompt (in Watt-hours). We want to be able to express these in terms of people and prompts, because those units are important for findings \(n\). We can write
\[P_{\text{DC, entire world}} = 67000 \text{ MW/world,}\]
which translates to
\begin{align} P_{\text{DC, per person}} &= 67000 \frac{\text{MW}}{\text{world}} \cdot \frac{\text{world}}{8 \cdot 10^9 \text{ people}} \\ &= 6.7 \cdot 10^{10} \frac{\text{W}}{\text{world}} \cdot \frac{\text{world}}{8 \cdot 10^9 \text{ people}} \\ &= 8.5 \text{ W/person.} \end{align}
To express \(E_\text{p}\) in terms of prompts, we can simply write it as
\[E_\text{p} = 0.3 \text{ Wh/prompt.}\]
Finally, to get close to the units we want, we can write
\begin{align} n &= \frac{P_{\text{DC, per person}}}{E_\text{p}} \\ &= \frac{8.5 \text{ W}}{\text{person}} \cdot \frac{\text{prompts}}{0.3 \text{ W hour}} \\ &= 30 \frac{\text{prompts}}{\text{person}\cdot\text{hour}}. \end{align}
After one final conversion, we can write
\[n = 30 \frac{\text{prompts}}{\text{person}\cdot\text{hours}} \cdot \frac{24 \text{ hours}}{\text{day}} = 720 \frac{\text{prompts}}{\text{person}\cdot\text{day}}.\]
720 prompts per person per day.
(c) This is way more than I thought the average person would be prompting. Some preliminary searching shows that much of the energy used in AI models goes into training and making inferences. Perhaps these processes should added to improve the accuracy of the model. You can probably think of other ideas as well.