Question

1 (a) Assume that the lights in your kitchen use 300 watts. How much energy and how much does it cost to leave the lights on 24 hours a day for a week if electricity is 8 cents/kilowatt hour?

(b) For a month (assume 30 days/month)?

(c) For a year?

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2 (a) How much energy and how much money do you use to run your window air conditioner rated at 1500 watts continuously for the month of July (assume 8¢/kWh)?

(b) If you assume that coal was used to produce the electricity for your air conditioner, how much coal was burned to produce the electricity used?

(c) How much CO2 was produced by the electricity used to run your air conditioner?

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3 (a) An average incandescent light bulb has a life expectancy of 1,000 hours. How much energy would a typical 60 watt bulb use in a lifetime, assuming it lasts for 1,000 hours?

(b) At 8¢/kWh, how much would it cost over its lifetime?

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4 (a) A compact fluorescent bulb uses 15 watts and has a life expectancy of 10,000 hours. How much energy and how much would it cost to use a compact fluorescent for 10,000 hours?

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5 (a) If your car gets 20 miles per gallon (MPG), and you drive an average of 10,000 miles each year, how many gallons of gas do you use a year?

(b) At $3.00 per gallon, how much will you spend on gasoline for the year?

(c) If the combustion of each gallon of gasoline produces 22 lbs of CO2, how much CO2 does your car produce each year?

(d) If you traded your car in & bought one that got 25 MPG, how much gasoline would you save in one year?

(e) How much money would you save?

(f) How much less CO2 would be emitted into the atmosphere from your improved car?

Answer 1

PLEASE DO NOT POST THESE MANY QUESTIONS AT ONCE....

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1) P = 300 W

For a week

as we need energy to be in kW/h

P = 300 / 1000 = 0.3 kW

t = 24 hours a days

so,

E = 0.3 * 7 * 24

E = 50.4 kWh

so,

cost will be

cost = 403.2 cents or 4.032 dollars

(b) For a month

E = 0.3 * 30 * 24

E = 216 kWh

so,

cost = 1728 cents or 17.28 dollars

For a year

E = 0.3 * 24 * 365

E = 2628 kwh

so

cost = 210.24 dollars

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2)

(a) P = 1500 W = 1.5 kW

t = 24 * 31 ( continuously means 24 hours per day for 31 days in july)

so,

E = 1.500 * 24 * 31

E = 1116 kWh

cost = 8928 dollars

(b) 1 ton of coal produces 2500 kwh of energy

so,

mass of coal needed = 0.4464 tons

(c) need data for C02 emisison

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3) P = 60 W = 0.06 kW

t = 1000 hrs

E = 0.06 * 1000

E = 60 kWh

so,

cost = 480 cents