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 = $100 W \times 6.5 h = 650 Wh$

Consumption of power in one year = $100 W \times 6.5 h\times 365 = 237250 Wh$

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

$237250 Wh = 237.25 kWh$

Given that $1 kWh$ = $0.18$

So, for $237.25 kWh$:

$237.25 kWh\times$ $ $0.18/ kWh$ = $$42.705$

Hence, the annual cost is $$42.705$.

b)

Consumption of power in one day = $25 W \times 6.5 h = 162.5 Wh$

Consumption of power in one year = $25 W \times 6.5 h\times 365 = 59312.5 Wh$

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

$59312.5 Wh = 59.3125 kWh$

Given that $1 kWh$ = $0.18$

So, for $59.3125 kWh$:

$59.3125 kWh\times$ $ $0.18/ kWh$ = $$10.676$

Hence, the annual cost is $$10.676$.