How to Calculate the Electricity Cost of an Appliance

To calculate what an appliance costs to run, multiply its power in watts by the hours it runs, divide by 1,000 to get kilowatt-hours (kWh), then multiply by your electricity price per kWh. A 1,500-watt space heater running 6 hours a day for 30 days uses 1.5 × 6 × 30 = 270 kWh; at $0.17 per kWh, about the 2025 US residential average reported by the Energy Information Administration (EIA), that is $45.90 a month.

The formula

Your utility bills you for kilowatt-hours: one kWh is 1,000 watts used for one hour. The electricity cost calculator handles days per week, partial-power use and multiple units.

Step 1: Find the wattage

• Nameplate label: most appliances list watts (W) or volts and amps on a label or in the manual.
• Volts × amps: if only amps are listed, multiply by voltage: a 120-volt device rated 10 amps is 1,200 watts. The watts to amps calculator converts either way.
• EnergyGuide label: refrigerators, freezers, water heaters and some other appliances carry the FTC’s yellow EnergyGuide label with estimated annual kWh, which already accounts for typical use.
• Plug-in meter: an inexpensive power meter measures actual use, which is often lower than the rated maximum.

Step 2: Estimate hours of use

Use realistic daily or weekly hours. For appliances that cycle on and off, such as refrigerators, air conditioners and space heaters with thermostats, multiply by a duty cycle, the share of time the motor or element is actually running.

Step 3: Find your rate

Divide your bill’s total charges by the kWh used. If your bill is $153 for 850 kWh, your effective rate is $153 ÷ 850 = $0.18 per kWh. This includes delivery and fixed charges, so it slightly overstates the cost of an extra kWh if your plan has a large fixed monthly fee. If you are on a time-of-use plan, use the rate for the hours the appliance runs; running a dryer or charging an EV off-peak can cut the cost substantially.

Worked examples

Space heater

1,500 W, 6 hours a day, 30 days, thermostat holding it on full power the whole time:

• 1,500 × 6 × 30 ÷ 1,000 = 270 kWh
• 270 × $0.17 = $45.90 a month

At the EIA’s 2026 forecast of about 18.2 cents, the same use costs $49.14.

Window air conditioner with a duty cycle

1,000 W, 8 hours a day, compressor running about 60% of that time:

• 1,000 × 8 × 0.60 × 30 ÷ 1,000 = 144 kWh
• 144 × $0.17 = $24.48 a month

Refrigerator

A refrigerator using 1.5 kWh a day:

• 1.5 × 30 × $0.17 = $7.65 a month, or about $93 a year

Electric vehicle

An EV that goes 3.5 miles per kWh and drives 1,000 miles a month:

• 1,000 ÷ 3.5 = 286 kWh
• 286 × $0.17 = $48.57 a month, before charging losses (often around 10%)

The EV charging cost calculator compares home and public charging with gasoline.

Time-of-use example

Suppose a time-of-use plan charges $0.35 per kWh during the late-afternoon peak and $0.12 overnight. An electric dryer drawing 3,000 W for one hour per load, used for 13 loads a month, consumes 3 × 13 = 39 kWh. Running it at peak costs 39 × $0.35 = $13.65 a month; running it overnight costs 39 × $0.12 = $4.68. The energy is identical, but the timing cuts the cost by about two-thirds. The same logic applies to dishwashers, pool pumps and EV charging, which is why many utilities now offer these plans. Check your utility’s rate schedule for the exact peak hours, which often change between summer and winter.

Typical cost per hour

At $0.17 per kWh, using typical wattages (your appliance may differ):

High-wattage heating appliances (heaters, dryers, ovens, water heaters) dominate most bills. Electronics are cheap per hour but can add up if left on all day.

Standby (“phantom”) power

Many devices draw a few watts even when “off”: game consoles, TVs, chargers, cable boxes and anything with a clock or remote. Because standby runs 8,760 hours a year:

At $0.17, each continuous watt costs about $1.49 a year. A 5-watt standby load costs about $7.45 a year; ten such devices, about $75. Smart power strips or simply unplugging rarely used devices can trim this.

LED vs. incandescent: a quick payback check

Replacing a 60 W incandescent bulb with a 10 W LED that gives similar light, used 5 hours a day:

• Incandescent: 60 × 5 × 365 ÷ 1,000 = 109.5 kWh, or $18.62 a year
• LED: 10 × 5 × 365 ÷ 1,000 = 18.25 kWh, or $3.10 a year
• Savings: about $15.51 a year per bulb, so a $2 to $3 LED pays for itself within a few months

Ways to cut appliance costs

1. Target heat: lower the thermostat on heaters and water heaters, and use cold-water laundry cycles.
2. Shift to off-peak hours if you have a time-of-use plan.
3. Maintain equipment: clean refrigerator coils and AC filters; clear dryer lint every load.
4. Right-size: a smaller, efficient appliance used appropriately beats a large one running inefficiently.
5. Check EnergyGuide labels when buying; the annual cost estimate makes comparisons simple.
6. Consider solar: the solar panel calculator estimates the system size needed to offset your usage.

To convert between kWh and other energy units, such as BTUs for heating comparisons, use the energy converter. For comparing prices of other things you buy, how to calculate unit price applies the same “cost per unit” thinking.