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22-2 Using Electric Energy

22-2 Using Electric Energy. Energy Transfer. A lot of devices are used to convert electrical energy into some other useable form Light bulb, hair dryer, stove A lot of energy lost in heat Light bulb & engines get warm. Heating a Resistor.

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22-2 Using Electric Energy

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  1. 22-2 Using Electric Energy

  2. Energy Transfer • A lot of devices are used to convert electrical energy into some other useable form • Light bulb, hair dryer, stove • A lot of energy lost in heat • Light bulb & engines get warm

  3. Heating a Resistor • What are some appliances used to convert almost all energy into thermal energy • Hair dryer, space heater, hot plate • To determine how much total energy is converted to thermal energy…. • P = E/t  E = Pt • P = IV  V = IR • So… • E = I2Rt

  4. A heater has a resistance of 10.0 Ω. It operates on 120.0 V • What is the current through the resistance • Known Unknown • R = 10.0 Ω I = ??? • V = 120.0 V • I = V / R • = 120.0 V / 10.0 Ω • = 12.0 A

  5. A heater has a resistance of 10.0 Ω. It operates on 120.0 V • What thermal Energy is supplied by the heater in 10.0 s • Known Unknown • I = 12.0 A E = ??? • R = 10.0 Ω • t = 10.0 s • E = I2Rt • (12.0 A)2(10.0 Ω)(10.0 s) • =14.4 x 103 J • =14.4 kJ

  6. Transmission of Electric Energy • Niagara Falls & Hoover dam produce a lot of electricity w/o much pollution • Has to transmitted long distances • Thermal energy lost is P=I2R • Engineers want to reduce I or R • **1km of wire = 0.20 Ω

  7. Transmission of Electric Energy • Supposed a farm house is 3.5 km away and has a 41 A electric stove. • The power dissipated (loss) in the wires = • 2(3.5 km)(0.20 Ω/km) • = 1.4 Ω • P = I2R • = (41 A)2 x 1.4Ω = 2400 W

  8. Kilowatt-Hour • Electric Companies charge you for energy • Energy used by any device is its rate of energy consumption • Joules/second (J/s)(sec) • Too small for commercial use so companies use the Kilowatt-Hour or kWh • = 1000 watts delivered continuously for 3600 sec • = 3.6 x 106 J

  9. Appliances • Other than heating appliances, most don’t need more than 1000-W of power • Evident in your lab yesterday

  10. Cost of operating • A television set draws 2.0 A when operated on 120 V. • A) How much power does the TV use? • B) If the set operated for an average of 7.0 h/day, what energy in kWh does it consume/month(30 days) • C) At 11 cents per kWh, what is the cost of operating the set per month? • Known Unkown • I = 2.0 A E = ??? • V = 120.0 V Total Cost = ??? • t = (7.0 h/day)(30 days) • Cost = 11 cents / kWh

  11. A) How much power does the TV use? • P = IV • = (2.0 A)(120.0 V) • = 240 W

  12. B) If the set operated for an average of 7.0 h/day, what energy in kWh does it consume/month(30 days) • E = Pt • (240 W)(7.0 h/day)(30 d) • = 5.0 x 104Wh • = 5.0 x 101 kWh

  13. C) At 11 cents per kWh, what is the cost of operating the set per month? • Cost = (5.0 x 101 kWh)($0.11/kWh) • = $5.50

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