Sample Solutions to Statue of Liberty HW

12,000 kilowatt-hours = 12,000,000 watt-hours

As noted in the directions, there are 3600 joules in one watt-hour thus we type

12,000,000 *3600

into Google and Google Calculator responds with:

12 000 000 * 3 600 = 43 200 000 000

Thus there are 43,200,000,000 joules used by the average household per year. Now we delete the blank spaces in "43 200 000 000" and type into Google:

43200000000 joules in foot-pounds

To which Google Calculator responds with:

43 200 000 000 joules = 3.18626848 × 1010 foot-pounds

We cannot copy/paste this number, since we would get "3.18626848 x 1010" but we can replace the "1010" with "10^10" which google understands (Google does evidently understand "x" in place of "*" for multiplication).

Together with its base, the Statue of Liberty weighs 54 million pounds, so we can type:

3.18626848 W 10^10 / 54,000,000

to which Google Calculator responds with:

(3.18626848 x (10^10)) / 54 000 000 = 590.049719

Thus the energy used by the average U.S. household could lift the Statue of Liberty plus its base about 590 feet in the air. What about just the statue itself?

We type 3.18626848 x 10^10 / 450,000 and Google Calculator responds with:

(3.18626848 x (10^10)) / 450 000 = 70 805.9662

That's a lot of feet! Let's delete the blank space after the "70" and try typing 70805.9662 feet in miles. We get:

70,805.9662 feet = 13.4102209 miles

It would lift the Statue of Liberty way up there! That's a lot of energy! You can create your own statistics and ask yourself other questions. For example, how many hours do you need to leave a 60 watt light bulb on (e.g. forgetting to turn it off when you leave the room) for the energy lost to be enough to lift your car 10 feet in the air? If spread out over a month, how much time "left on needlessly" per day would that be?