Question

In a certain city, electricity costs $0.15 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.00 hours per day with (a) a 100.-watt incandescent light bulb (b) an energy efficient 25-watt fluorescent bulb that produces the same amount of light? Assume 1 year = 365 days.

Answer 1

(a) Power of bulb is 100 W, converting this into kW.

`1 W=(1)/(1000)kW`

Thus,

`100 W =(100)/(1000)kW=0.1 kW`

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

`t=6* 365=2190 h`

Electricity used will depend on power and number of hours as follows:

`E=P* t=0.1 kW* 2190 h=219 kW.h`

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 219 kW.h will be:

`Cost=\$ (219* 0.15)=\$ 32.85`

Therefore, annual cost of incandescent light bulb is
`\$ 32.85`

(b) Power of bulb is 25 W, converting this into kW.

`1 W=(1)/(1000)kW`

Thus,

`100 W =(25)/(1000)kW=0.025 kW`

The bulb is used for 6 hours per day for a year, in 1 year there are 365 days thus, total hours will be:

`t=6* 365=2190 h`

Electricity used will depend on power and number of hours as follows:

`E=P* t=0.025 kW* 2190 h=54.75 kW.h`

The cost of electricity is $0.15 per kW.h thus, cost of electricity for 54.75 kW.h will be:

`Cost=\$ (54.75* 0.15)=\$ 8.21`

Therefore, annual cost of fluorescent bulb is
`\$ 8.21`.