In a certain city, electricity costs $0.16 per kW⋅h. What is the annual cost for electricity to power a lamppost for 7.00 h per day with a 100. W incandescent light bulb versus an energy efficient 25 W fluorescent bulb that produces the same amount of light? Assume 1 year=365 days.

The annual cost for electricity to power a lampost for 7.00 h per day with a 100 W incandescent bulb is $ 40.89.

The annual cost for electricity to power the same lampost for the same time with an energy efficient 25 W fluorescent bulb is $ 10.22.

So, you can save $ 40.89 - $ 10.22 = $ 30.67 per each lampost if you replace the incadescent bulb with the energy efficient fluorescent bulb.

Explanation:

kW. h means kilowatt. hour and is a unit of energy derived from the definition of power.

The definition of power is energy per time:

Power = Energy / time

Thus, you get:

Energy = Power * time.

For the units, you get:

[Energy] = [Power] [time] = KW. h.

1. Calculate the energy used by the 100 W incandescent light bulb that is used 7.00 h per day, in a year (365 days).

E = Power * time = 100 W * 7.00 h / day * 365 day / year * 24 h = 255,500 W. h/year Divide by 1,000 to convert to kW: 255.5 kW. h/year

2. Calculate the energy used by the 25 W flourescent bulb:

E = 25 W * 7.00 h / day * 365 day / year * 24 h = 63,875 W. h/year Divide by 1,000 to convert to kW: 63.875 kW. h/year

3. Now, you can calculate both costs:

Cost per year = cost per kW. h times number of kW. h per year

100 W bulb:

$ 0.16 / kW. h * 255.55 kW. h / year = $ 40.89 / year

25 W bulb:

$ 0.16 / kW. h * 63.875 kW. h / year = $ 10.22 / year