2007-11-19 02:32:57 · 8 answers · asked by hudi 1 in Science & Mathematics ➔ Engineering
1.5 x 10 kwh = 15 kwh
2007-11-19 02:47:18 · answer #1 · answered by Como 7 · 2⤊ 1⤋
It means that the electric heater uses 1500 Joules of energy per second. So, in 10 hours, it will use, 1500 x 10 x 60 x 60 = 54 Mega Joules ( Million Joules)
2016-03-13 23:53:10 · answer #2 · answered by Anonymous · 0⤊ 0⤋
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1st of all power is equal to the work done per second; looking at the units of work, it is Force x distance, force is kg or in newton and distance in SI is meter and for time its second analyzing the units Power = (Newton - Meter)/second so by cancellation of units 1500 N-m/s x 10 hrs x 3600 s/hr = 54,000,000 N-m or Joules in term of heat energy
2016-04-10 06:33:58 · answer #3 · answered by ? 4 · 0⤊ 0⤋
A Watt is a unit of power,that is energy per unit of time,a watt is a Joule per second,1500 W is then 1500 Joules/s.In 10 hours it will use 1500W*10hours*3600s,(an hour has 3600 seconds),that is 54000000 Joules or 54MJ.I hope it will be useful for you.
2007-11-19 04:53:10 · answer #4 · answered by kamuss 2 · 0⤊ 3⤋
1500 w = 1.5kilowatts
1.5 kw X 10 hours = 15 kilowatt-hours
15kwh x $.10/kwh = $1.50
2007-11-19 03:08:01 · answer #5 · answered by mike 5 · 1⤊ 1⤋
The formula is simple. Multiply the watts x 24 hours, then divide by 1000 for kilowatts. Multiply that number by your price per kilowatt hour which will equal your daily cost. Then multiply the number of days in the month to get your monthly total.
Example; a 115 volt, 2.8 amp, 322 watts. To figure the monthly operation cost; 322 watts X 24 hours ÷ by 1000 kilowatts = 7.72 kilowatts per day it's consuming. Multiply 7.72 kilowatts by your kilowatt cost ( so if it's $0.08 per kilowatt hour) to get the daily cost of running it. 7.72 x $0.08 = $0.62 per day to run. Multiply the daily rate by the number of days in the month, 30 for example, and you get the monthly operation cost! $0.62 per day X 30 days = $18.60 per month to operate this. Once again, the formula is:
watts X 24 hours ÷ 1000 kilowatts X kilowatt cost X days in the month = monthly operational cost.
or
amps X volts X 24 hours ÷ 1000 kilowatts X kilowatt cost X days in the month = monthly operational cost.
2007-11-19 02:46:19 · answer #6 · answered by rob89434 4 · 2⤊ 2⤋
1500 x 10/1000 =15kWHr
2007-11-19 04:20:59 · answer #7 · answered by A.V.R. 7 · 0⤊ 1⤋
E=P*t
2007-11-19 02:42:52 · answer #8 · answered by morphriz 3 · 0⤊ 1⤋