Question

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.

Answer 1

• 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.

Step-by-step 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