01

Use kWh = watts ÷ 1,000 × hours

Multiply energy by the all-in price per kWh.

A 1,500-watt heater used two hours a day for 30 days uses 90 kWh. At $0.16 per kWh, the energy portion is about $14.40. Fixed charges and tiered rates can make the bill differ.

02

Choose the best available power value

Measured average power is better than a maximum nameplate rating.

Nameplates may show maximum input. Cycling compressors, heaters, and motors vary over time. A plug-in energy meter can capture representative energy for suitable plug loads; follow its ratings and instructions.

03

Measure run time across a representative period

A single hour can miss cycles and standby behavior.

Use several ordinary days, then separate weekdays, weekends, and seasons when behavior changes. For central or hardwired equipment, use built-in monitoring, utility interval data, or qualified measurement.

04

Treat the result as a scenario, not a bill promise

Rates, taxes, demand charges, and cycles vary.

Run low and high scenarios. Compare the estimate with actual daily kWh from the bill or utility portal, and investigate only differences large enough to matter.

05

A powerful short task can cost less than a modest all-day load

Compare energy over the same time window, then compare dollars.

These are invented inputs for checking the arithmetic, not typical appliance ratings. A 1,500 W device used for ten minutes daily consumes 0.25 kWh a day. A 100 W device operating continuously consumes 2.4 kWh a day. At the same illustrative $0.20/kWh price over 30 days, the costs are $1.50 and $14.40 respectively. The lower-wattage device costs more in this scenario because it runs far longer.

Do not infer either rating from a product category. Inspect the actual equipment specification or use a compatible, correctly rated energy meter. Convert minutes to hours before calculating: ten minutes is 10 ÷ 60 hours, not 0.10 hours. Keep full precision until the displayed result.

06

Account for cycling without counting it twice

Scheduled hours and active-power time are different inputs.

Suppose a fictional load draws 200 W while active and is active for one quarter of a 24-hour observation window. Ignoring standby for this example, that is 200 ÷ 1,000 × 24 × 0.25 = 1.2 kWh/day. Over 30 days at $0.20/kWh, the estimate is $7.20.

In the worksheet, enter 24 scheduled hours and 25% duty cycle. Alternatively, enter six active hours and 100%. Do not enter six hours and 25%; that reduces the time twice. If a measurement already gives a 50 W average over the whole day, use that average with 24 hours and 100% duty. A fixed duty estimate misses variable-speed operation and changing conditions.

07

Give a proposed change a break-even test

Energy cost is one part of an equipment decision.

Suppose a verified comparison suggests $5 less electricity use each month, and the proposed change costs $180. At the same use pattern, $60 annual savings would give a three-year simple payback. That is an arithmetic scenario, not a return guarantee. If the expected remaining service life is shorter, or the replacement adds maintenance cost, the apparent savings may not justify the purchase.

Before buying, compare the same useful service, check whether the change merely shifts energy to gas or another device, and measure the actual baseline. In the multi-appliance worksheet, the one-time cost is compared with a 365-day projection. For a seasonal device, calculate its actual season separately; do not treat a January daily pattern as a full-year forecast.

08

Use a measured kWh total when a nameplate is misleading

A measurement over a known interval can replace guessed active hours.

Illustrative case, not a product test. These inputs are invented to make the reasoning reproducible. Replace them with your own observations; the result is not a typical household or product benchmark.

Suppose a compatible energy meter records 1.8 kWh over 36 hours for one plug-in appliance during ordinary use. Divide by 36 and multiply by 24 to get 1.2 kWh/day for that observation. Repeating that pattern for 30 days gives 36 kWh; at a hypothetical $0.20/kWh, the cost is $7.20.

The equivalent average power is 1.8 ÷ 36 × 1,000 = 50 W. In the simple calculator, enter 50 W and 24 hours/day. Do not enter the appliance’s maximum label power as well, and do not apply an additional duty-cycle reduction to this already averaged value.

Choose one input method
Evidence availableWhat to enterDo not do this
Measured average powerAverage watts × full observation scheduleApply duty cycle again
Active watts and active hoursActive watts × actual active hoursUse plugged-in hours as active hours
Only maximum nameplate powerLabel the result as a rough scenarioPresent it as a measured monthly bill

This extrapolation assumes the observed period represents the future period. Door openings, ambient conditions, settings and seasonal operation can change the result. Use a longer representative observation when practical. Only use a meter rated for the appliance and follow its instructions; this exercise does not authorize measuring hardwired equipment, opening a panel or connecting an unsuitable high-current load.

HS TOOL / 03

Turn watts into cost.

Build a transparent energy-use scenario from power, time, days, and price.

Formula and assumptions: kWh = watts ÷ 1,000 × hours/day × days; cost = kWh × price. Cycling loads and fixed bill charges need separate treatment.

RESULTEnter the measurements to calculate.
Source: Virginia Cooperative Extension energy-use method ↗

QUESTIONS THIS ANSWERS

Questions this answers

  • How much does a 1,500-watt heater cost to run?
  • How do I convert appliance watts into monthly electricity cost?

Found something wrong? Report an error or read the corrections policy.

2 SOURCESEvidence ledger

Sources

  1. 01
    Estimating Appliance and Home Electronic Energy Use ↗

    Virginia Cooperative Extension · accessed 5 Sept 2026

  2. 02
    Use of Electricity ↗

    U.S. Energy Information Administration · accessed 5 Sept 2026