Since there are 6366 hours in a year, 1930 kWh is about 0.3 kW per hour. (1930 / 6366)
Most 120 Volt appliances have their Watts listed somewhere on the appliance. Divide this number by 1,000 to get kilowatts. Determine the rate you are charged for electric power. Your electric bill will tell you how many kWh(kilowatt hours) you use, and what the total charge is. Divide your total charge ($?) by the kWh used. In northern Illinois, we are charged about $0.11 per kWh. Multiply the number of hours the appliance is used by its power consumption in kilowatts. This gives you the number of kWh the appliance used. Then multiply that kWh value by the rate for your area that you determined from your electric bill. That will be the cost for the number of hours the appliance was on. Example: If I toast 2 slices of bread in my toaster, it takes about 3 minutes, or 0.05 hours. My electric bill is $61.38 and I used 558 kwh for that month. Dividing $61.38 by 558, I get about $0.11 per kWh. So, I divide my toaster Wattage (750) by 1,000 to get 0.75 kW. So now I can multiply the hours that the appliance was used (0.05) by its power consumption rate in kW (0.75). This result is 0.0375 kWh of energy used. Then I can multiply the rate the electric company charges ($0.11) by the energy used (0.0375 kWh). The final result is $0.004125.
Your question should read, "How many kW.h are there in 0.07 GW.h?" It's important to use the correct symbols. There are 1000 000 000 kilowatt hours in one gigawatt hour -so you can now work it out for yourself.
There are two measures of electricity to know about: kWh, and kW. kWh, or kilowatt-hours, is a measure of total energy use over time. It's like how far you've driven your car. kW, or kilowatts, is a measure of instantaneous consumption, like how fast you are driving at any given second. The average home in Phoenix might max out at around 3 kW or a little more when everything - your computers, refrigerator, lights, air conditioning, etc. - are running. (An individual hair dryer might use 1.5 kW, so don't run those too long.) kW, however, is not how you get billed. Almost all residential properties get billed on kWh, the total amount of electricity they use each month. kWh can be calculated by multiplying kW by the number of hours you used that kW. This is the same as figuring out how far you've driven by multiplying your speed by how much time you spent driving at that speed. In Phoenix, an average house your size might use around 30-40 kWh every day. It could be twice this in the summer when you're air conditioning, and much less in the fall or spring when you're not.
It depends on where you live. You want to know how much Sun you get (solar hours). If you live in Southern California or Arizona, you might get 6 hours per day. If you live in Northern CA or Oregon, it might be closer to 4. First, you take your 1,000 kWh/mo and divide that by 30 to get your kWh/day. 1,000 / 30 = 33.3 kWh/day Then you divide this by the number of solar hours per day your area gets. Let's say 5. 33.3 / 5 = 6.67 kW Last you need to adjust for real world inefficiencies. A good rule of thumb is to suppose approximately 75% - 80% inefficiencies. So increase your system a little. 6.67 kW / 0.75 = 8.9 kW. This is a fairly large residential system, but you can reasonably expect it to cover 100% of your electricity needs if you use 1,000 kWh per month. Want to know how much an 8.9 kW solar system might cost? Go to and play the solar calculator in the related link below. (Yes, it's free.)
You need to understand the difference between power - as in watts - and energy - as in Watt hours. Watts/kiloWatts only tells you what it's doing at the moment, at that very instant. While Wh/kWh tell you the sum of what it's been doing over a certain time. 1 kW = 1000 w So 90 W / 1000 W = 0.09 kW As soon at the light is turned on, the lamp starts to use energy at the rate of 90 W = 0.09 kW If you leave it on for 12 hours it will have used up 0.09 x 12 = 1.08 kWh
40-60 depending on what time of year it is.
100 kWh
The amount of KWh used by an oil boiler in a year can vary based on factors such as the efficiency of the boiler, the size of the home, and how often the boiler is used. On average, a residential oil boiler may use around 2,000 to 3,000 kWh per year for heating.
The amount of kWh an industry uses in a year can vary widely depending on the size of the industry, its operations, and energy efficiency measures in place. On average, an industrial facility can use anywhere from 1,000,000 to 10,000,000 kWh per year.
New energy star models are about 470 kWh per year.
The annual kWh usage of an air source heat pump (ASHP) can vary based on factors such as size, efficiency, climate, and usage patterns. A typical ASHP for a residential home might use around 2,000-3,000 kWh per year for heating.
In ten hours, a 200W bulb will use: 10 * 200 = 2000 Watt-hours = 2 kwh
none, if it is unplugged
It depends on the btu of the unit
Probably near the Russian average of 2400 kWh per year or 6.6 kWh per day. That is an average load of 276 watts so the supply would need to be rated at about 4 kW.
The average freezer uses around 100-400 watts of electricity, depending on its size and efficiency. This translates to about 2.4-9.6 kWh per day, or around 876-3504 kWh per year.
On average, a fan oven typically uses between 2-3 kWh per hour when in use. This can vary depending on the temperature setting and cooking time.