One deterrent to burglary is to leave your porch light on all the time. If your fixture contains a 60-W(watt) bulb at 120 V(volt), and your local power utility sells energy at 8 cents per kilowatt-hour, how much will it cost to leave the bulb on for the whole month?
2007-03-12 19:24:59 · 5 answers · asked by aznboi4et3nity 1 in Science & Mathematics ➔ Physics
The 60 watt bulb will dissipate 60 watts * 24 hours = 1440 watt-hours of power in one day. In one 30 day month, the bulb will dissipate 1.44 kW-hours/day * 30 days = 43.2 kW-hours. So the cost to operate the bulb is:
43.2 kW-hours * 0.08 dollars/kW-hour = $3.46
It will be 31/30 * 3.46 = $3.58 if the month has 31 days.
2007-03-12 19:35:54 · answer #1 · answered by heartsensei 4 · 1⤊ 0⤋
The bulb uses 60W. In one month there are 30*24 hours, so the total energy is 60*30*24 watt-hours. Divide by 1000 to get
(60*30*24)/1000 kW-hr
Multiply this by 0.08 to get the monthly cost in dollars.
2007-03-13 02:42:42 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋
You don't need to know the voltage for this problem. 60 W is the power of the bulb, and you just need to multiply it by the number of hours in a month (and divide by a thousand) to find out how many kilowatt-hours are used. That's (60 W)*(1 month)*(30 day / 1 month)*(24 hr / 1 day) = 43200 W-hr = 43.2 kW-hr. Then multiply by the cost per kilowatt-hour to get (43.2 kW-hr)*(8 c/kW-hr) = 345.6 cents, which is $3.46.
2007-03-13 02:39:18 · answer #3 · answered by DavidK93 7 · 0⤊ 0⤋
60W=.06Kw
1month=30day=30*24hr=720hr
0.06*720=43.2KW.hr
43.2(KW.hr)x0.08(dollar for 1KW.hr)=$3.456
2007-03-13 05:34:25 · answer #4 · answered by arman.post 3 · 0⤊ 0⤋
$3.46
2007-03-13 02:31:21 · answer #5 · answered by wax 1 · 0⤊ 0⤋