OK. We can do this.
3 months = 90 days
The unit cost is ($0.085) per kilowatt-hour
Here we go:
($ 425/90da) · (1da/24hr) · (1 da/86,400sec) · (1 kW-hr/$ 0.085) =
(425 · 1 · 1 · 1) / (90 · 24 · 86,400 · 0.085) · ($-da-da-kW-hr / da-hr-sec-$) =
2.679 x 10⁻⁵ day-kW/sec
This is not too useful yet. It needs some more unit conversion.
I think all it needs is another (1 day = 86,400 sec) .
(2.679 x 10⁻⁵ day-kW/sec) · (86,400 sec / da) =
(2.679 x 10⁻⁵ · 86,400) · (da-kW-sec / sec-da) =
<em>2.315 kW</em>
If I have not gang aglay somewhere in all of that, then this is the answer to the question: <em>2.315 kW .</em>
I know that you probably didn't follow everything that I did, and if you did, you don't have a lot of confidence in this answer. I have to admit that I have serious doubts too. So I have to check the answer somehow.
What would it cost if the household was using power steadily at the rate of 2.315 kilowatts, and the family went away for three months and just left everything cooking like that ?
2.315 kilowatts ===> <u>2.315 kilowatt-hours</u> every hour
(2.315 kW hours per hr) x (24 hrs per day) = <u>55.56 kWh per day</u>
(55.56 kWh per day) x (90 days) = <u>5,000.4 kWh</u>
(5,000.4 kWh) x (8.5¢ for each kWh) = <u>42,503.4¢ on the bill</u>
42,503.4¢ = $425.034 That's good enough for me ! ! I rest my case.
