1. Page 404 SciencePower 9

2. Think about it In every step between mining coal, generating electricity, transmitting it to your home and lighting a lamp some energy is used or escapes in the form of heat in the process. In this activity we will calculate what percentage of energy stored in the coal is actually converted into light in the light bulb.

3. Summary of steps 1. Mining the coal: the process of extracting coal from the ground is about 99% efficient. Some energy is used by the machines in the mining process. 2. Transportation of the Coal: transporting the coal from the mine to the power plant is also quite efficient. Some energy is used to power the train. Efficiency about 97%

4. 3. Generation of Electricity: Thermo – electric generating plants are not very efficient most of the energy from the coal escapes from the stacks or from coolant water as heat. Only about 33% of the energy from coal that is burned is converted into electricity.

5. 4. Transmission of Electricity: When electrical energy travels through the power lines, some of the energy goes into heating the conductors. Electrical transmission is about 85% efficient. 5. Conversion to light: Incandescent light bulbs are very inefficient. Most of the electrical energy is converted into heat. At best an incandescent light bulb is about 15% efficient.

6. What to Do...... Make a Table with the headings shown below. STEPEFFICIENCY OF STEP (%) STORED ENERGY REMAINING (out of original 1.0 X 10 6 J) OVERALL EFFICIENCY (%) You will need enough space for the 5 steps

7. STEPEFFICIENCY OF STEP (%) STORED ENERGY REMAINING (out of original 1.0 X 10 6 J) OVERALL EFFICIENCY (%) 1. Mining the coal 99% 1, 000 000J X 99% 100 = 990 000J 990,000J X 100 1,000,000 J 99% 2. Transportation of the coal. 97% 990 000J X 97% 100 = 960300J 960300 X100 1,000 000J 96.03% 3. Generation of electricity 33% 960300 X 33% 100 = 316899 J 316899 X 100 1,000 000J 31.69%

8. STEPEFFICIENCY OF STEP (%) STORED ENERGY REMAINING (out of original 1.0 X 10 6 J) OVERALL EFFICIENCY (%) 4. Transmission of Electricity 85% 316899 J X 85% 100% 269364.15 J 269364.15 X 100 1, 000, 000 J 26.94% 5. Conversion to Light 15% 269364.15J X 15% 100% 40,404 J 40404 X 100 1,000,000J 4.04% 6. Fluorescent Light 90% 269364J X 90% 100 = 242427.6 J 242427.6J X 100 1,000,000J 24. 24%

9. Electrical Power & Efficiency Electrical devices are labelled with a power rating. Power is measured in units called Watts (W) after James Watt who discovered that steam could move things and then developed the steam engine.

10. Electrical Power Power is the rate at which energy is transformed, or the rate at which work is done. Electrical Power is the rate at which electrical energy is produced or consumed in a given time. 1 Watt = 1 Joule per second 1W = 1J/s

11. Measuring Electrical Usage Generating stations have power ratings in Mega Watts (MW) = 1 million watts OR Giga Watts (GW) = 1 billion watts The Kilowatt hour is the SI unit used to measure energy usage An average Canadian family consumes over 16, 000 kW-h of electrical energy in one year

12. Efficiency Not all appliances use energy efficiently. Efficiency is a measure of how much useful energy an electrical device produces compared to how much energy was supplied to the device. Older devices tend to be less efficient than newer models.

13. This is an example of an Energuide label found on newer appliances. It helps consumers make informed choices when buying a new appliance

14. The energy star program was introduced in 1972 by the US Environmental Protection Agency and the US department of Energy.

15. Cost of Electricity In Ontario the cost of electricity is regulated at 5.6¢ for the first 1000kW-h during the winter and the first 600kW-h during summer. After 1000kW-h the cost increase to 6.5¢/kW-h

16. Cost to operate a Lap top Computer Cost to operate = power used X time X cost of electricity A laptop uses a 75W adapter when it is plugged in. Electricity costs 5.6 ¢/kW-h Calculate the cost to operate the computer for 1 year for 24 hours a day. Use the G.R.A.S.S. method for solving problems

17. Given: Power = 75W (convert to kW =75W X 1kW = 0.075kW 1000W Time = 24 hours per day for 365 days = 8760 hours Cost of Electricity = 5.6 ¢/kW-h Required : cost to operate Analysis: Cost to operate = power used X time X cost of electricity

18. Solution: Cost to operate = 0.075kW X 8760h X 5.6 ¢ kW-h = 3679 ¢ Statement: It would cost 3679¢ or $36.79 to operate a laptop computer for 24 hours a day for 1 year.

19. Practice Calculate the cost of operating a 1500W hairdryer for 6 minutes per day for 3 days. The cost of electricity is 5.6¢/kW-h Use the G.R.A.S.S. method.

20. Given: Power = 1500W = 1.5kW Time = 6mins X 3 days = 18 mins 18/60 = 0.3h Cost = 5.6 ¢/kW-h Required: Cost to operate Analysis: cost to operate = power X time X cost of electricity

21. Solution: cost = 1.5kW X 0.3h X 5.6 ¢ = 2.52¢ Statement: the cost to operate the hairdryer for 6 mins for 3 days is 2.5 ¢

22. Calculate the difference between the operating cost of a 60W incandescent light bulb and a 13W CFL, each operating for 100 hours. The cost of electricity is 11¢ per kW-h

23. Given: Power = 60W X 1kW = 0.06kW (incandescent) 1000W = 13W X 1kW = 0.013kW (CFL) 1000W Time = 100 hours Cost = 11 ¢/kW-h

24. Required: Difference in cost between an incandescent light bulb and a CFL Analysis: cost = power X time X cost of electricity

25. Solution: incandescent bulb cost = 0.06kW X 100h X 11¢ = 66 ¢ CFL cost = 0.013kW X 100h X 11 ¢ = 14.3 ¢ Statement: Obviously the CFL is much cheaper to run than the incandescent bulb by a saving of 51.7 ¢

26. Cost of operating Calculate the cost of operating a refrigerator power 750W for 1 year. Cost of electricity is 12 ¢/kW-h Given: Power = 75W, time = 365 X 24 = 8760 hours cost = 12 ¢/kW-h Reguired: Cost to operate for one year Analysis: Power X Time X cost

27. Substitute: 0.75KW X 8760 hours X 12 ¢/kW-h Solution: $788.40

29. Calculating the Efficiency of a light bulb. A light bulb uses 100J of electrical energy and produces 35J of light energy. Calculate the percentage efficiency of the light bulb. Given: Energy out = 35J Energy in = 100J Required: Percent efficient (% efficiency) Analysis: % efficiency = E out X 100 E in

30. Solution: % efficiency = 35J X 100 100J = 0.35 X 100 = 35% efficient Statement: the light bulb is 35% efficient.

31. A toaster oven uses 1200J of energy to produce 850J of thermal energy. Calculate the percent efficiency of the oven. Given: Required: Analysis: Solution Statement: