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gregori [183]
3 years ago
8

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.
Chemistry
1 answer:
Vanyuwa [196]3 years ago
7 0

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

1 W=\frac{1}{1000}kW

Thus,

100 W =\frac{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\times 365=2190 h

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

E=P\times t=0.1 kW\times 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\times 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=\frac{1}{1000}kW

Thus,

100 W =\frac{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\times 365=2190 h

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

E=P\times t=0.025 kW\times 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\times 0.15)=\$ 8.21

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

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