Electricity

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#14516

Electricity

1 KW = 1ooo watts

KWH = KW used in an hour

1 joule/second = 1 watt

1 J/s = 1 W

1 kWh = 3.6 MJ

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#18722

A 2400-watt heater is run for 6 hours each day. If electricity is charged at 26.3 c/kWh, what is the cost of running the heater for 8 days, to the nearest cent?

Find the watts per day: 2400 × 6 = 14 400watts

Find the watts for 8 days 14 400 × 8 = 115 200

Change to kW by dividing by 1000 (remember kilo = 1000)

115 200 ÷ 1000 = 115.2 kWh

Find the cost 115.2 × 26.3 = 3029.76 cents

divide by 100 to change from cents to dollars

3029.76 ÷ 100
= $30.30 (rounded to the nearest cent)

[Admin]

#18723

Energy is used at the rate of 25 joules per second. How many watts of power are used?
Watts = 25 × 1 = 25 W	since 1 joule per second equals 1 Watt, then 25 joules per second will equal 25 watts this just means we multiply by 1 to get the answer

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#18726

If 5 75W light bulbs are running on average for 6 hours each day, and power costs 23.6 cents per kWh, what would be the cost of keeping these lights on for a week?
5 × 75 × 6 × 7 = 15,750 Wh	there are 5 lights, using 75 watts for 6 hours a day for 7 days in the week multiply to get the total amount of watt hours used
15,750 ÷ 1000 = 15.75kWh	convert to kilowatt hours by dividing by 1000
cost = 15.75 × 23.6 = 371.7c	now we can find the cost by multiplying the number of kWh and the cost per kWh
371.7c ÷ 100 = $3.18	convert to dollars by dividing by 100 ($1 = 100c) round to the nearest cent

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