Issue 006 - Energy - Power and thermal physics
How many power plants of extra AC demand can a heat dome create?
AP reported on July 10, 2026 that much of the Lower 48 was about to face an unusually large, long-lasting heat dome, with temperatures 15 to 25 degrees F above normal in many areas and the event potentially affecting as much as two-thirds of the continental United States.
The problem
Estimate the extra peak electrical power demand caused by air conditioners during this heat wave.
If a large power plant is roughly 1 gigawatt, is the added AC demand closer to the output of 10, 50, or 100 large power plants?
Because Fermi problems target an order of magnitude, I normally use no more than two significant digits and write most calculations in scientific notation; the Fermi reference explains both conventions.
Before checking sources
Matt's first pass
I have not done an instantaneous power-demand problem in a while, so my equation choice may not be ideal, and my pegs might not be either.
The news item said as much as two-thirds of the United States might be affected. But as the weather pattern moves, I assumed maybe 10% of U.S. households would be warmed enough that they would use AC when otherwise they might not have.
new AC households ~= (1.3 x 10^8 households) x 0.10
~= 1.3 x 10^7 householdsTo keep the math easy, I did not separate central AC from window or wall units. I assumed the average household in the affected areas might operate one window-mounted AC unit.
For per-unit demand, I remembered AC units as power-hungry. I guessed the average unit might pull about 50 amps, and I used 100 volts as easy household power.
power per AC ~= current x voltage
~= 50 amps x 100 volts
~= 5,000 watts
~= 5 kWThen I multiplied that per-household power by the number of households I thought would newly run AC:
extra peak demand ~= (1.3 x 10^7 AC units) x (5 kW/unit)
~= 6.5 x 10^7 kW
~= 6.5 x 10^4 MW
~= 65 GWAt 1 GW per large power plant, my first-pass answer was about 65 additional power plants. That felt very high to me, so I suspected I had overestimated.
Calibration Score
Matt's Calibration Score: 70 / 100
Higher is better: earn points for accurate pegs, sound models, correct math, and a result close to the sourced answer. The image shows percent full of it: 100 minus the Calibration Score.
Pegs: 10/30. Affected-household and AC-draw pegs were the weak spots.
Model: 30/30. Households times extra AC power is the right peak-demand model.
Math: 10/10. The arithmetic was clean.
Result: 20/30. Canceling errors brought the result near the corrected scale.
Grounding facts
PJM's July 2, 2026 hot-weather update said it expected to serve a peak load of about 166,000 MW, near or above its historic summer peak. That is only one large regional grid, not the entire country. So a national heat-wave increment of tens of gigawatts is not outlandish.
Also remember the unit distinction: power is instantaneous draw in kW, MW, or GW. Energy is power accumulated over time in kWh, MWh, or GWh. A 50 GW extra peak lasting 5 hours is 250 GWh of extra energy.
After checking sources
Check and recalibrate
Matt's final number is plausible, but the route needs adjustment. A 5 kW draw is high for a window unit; it is closer to a large central AC system. EnergySage's examples put a 3-ton central system around 2 to 2.6 kW depending on efficiency, with larger systems around 3 to 4 kW. Window units are commonly much lower.
The offsetting issue is that 10% of households is probably low for a heat dome that can affect as much as two-thirds of the continental U.S. EIA reports about 88% of U.S. households use AC.
affected households ~= (1.3 x 10^8 households) x (2 / 3)
~= 8.7 x 10^7 households
affected AC households ~= (8.7 x 10^7 households) x 0.88
~= 7.7 x 10^7 AC householdsThe key thermal-physics idea is that cooling load roughly rises with the indoor-outdoor temperature difference, plus sun, humidity, air leaks, and internal heat. A 15 to 25 degree F hotter-than-normal day can make many systems run longer at the peak hour. Instead of assuming each affected home adds a full AC unit from zero, use an extra peak average of about 0.5 to 1 kW per AC household.
extra residential peak demand
~= (7.7 x 10^7 AC households) x (0.5 to 1 kW/household)
~= 3.9 x 10^7 to 7.7 x 10^7 kW
~= 39 to 77 GWCommercial buildings matter too. EIA reports 254 billion kWh per year for residential AC and 170 billion kWh per year for commercial cooling. A rough commercial add-on of 10 to 30 GW during a widespread heat-wave peak is reasonable for a national-scale estimate.
total extra peak demand ~= residential extra + commercial extra
~= (39 to 77 GW) + (10 to 30 GW)
~= 50 to 110 GWA good calibrated answer is therefore on the order of 50 to 100 GW, or roughly 50 to 100 large 1-GW power plants. Matt's 65 GW answer was not too high; the answer is genuinely grid-scale.
Post-check reflection
Matt's reflection
The assumptions that mattered most were affected households and draw per unit. I expected my appliance-draw estimate to be shaky. I do not have a lot of experience calculating instantaneous power draw, and my intuitions are not well-honed for the exact draw of specific appliances, even if I can usually get them roughly ordered. As for homes that use AC, the almost-90% U.S. adoption rate is a useful number to remember moving forward.
The corrected answer captures my actual answer pretty well. That means two things at once: I was close because of canceling errors, which seems common in these problems, and the extra demand on the existing grid is genuinely very high.
Energy equivalent to the production of 50 to 100 extra power plants is significant. Alarmingly so. With that context, I feel like this should be a much bigger issue in public discourse.
Recommended memory peg
Remember that a large power plant is roughly 1 GW, a central home AC system often draws a few kW while running, and about 9 out of 10 U.S. households use air conditioning. Also remember P = I x V for electrical power.
Reader results
Bars show how submitted estimates sort into the answer choices from the gut-check prompt.