Question

n a certain city, electricity costs $0.18 per kW·h. What is the annual cost for electricity to power a lamp-post for 6.50 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) 100 watt = 0.1 kW

Annual cost can be calculated as:

Annual cost=0.1 kW ×6.50 h×365×0.13

=30.8 $

So, now the annual cost will be 30.8 $

B) 25 watt = 0.025 kW

Annual cost can be calculated as:

Annual cost=0.025 kW ×6.50 h×365×0.13

=7.71 $

So, now the annual cost will be 7.71 $

Answer 2

a)

Consumption of power in one day =
`100 W * 6.5 h = 650 Wh`

Consumption of power in one year =
`100 W * 6.5 h* 365 = 237250 Wh`

Since,
`1 Wh = 0.001 kWh`. So,

`237250 Wh = 237.25 kWh`

Given that
`1 kWh` =
`$0.18`

So, for
`237.25 kWh`:

`237.25 kWh*` $
`0.18/ kWh` = $
`42.705`

Hence, the annual cost is $
`42.705`.

b)

Consumption of power in one day =
`25 W * 6.5 h = 162.5 Wh`

Consumption of power in one year =
`25 W * 6.5 h* 365 = 59312.5 Wh`

Since,
`1 Wh = 0.001 kWh`. So,

`59312.5 Wh = 59.3125 kWh`

Given that
`1 kWh` =
`$0.18`

So, for
`59.3125 kWh`:

`59.3125 kWh*` $
`0.18/ kWh` = $
`10.676`

Hence, the annual cost is $
`10.676`.