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Ch. 5&6 - Energy

Learn about different forms of energy, the concept of work, conservation of energy, thermal energy, heat transfer, and temperature. Discover how energy can change forms and how heat flows between materials. Understand the relationship between temperature and thermal energy.

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Ch. 5&6 - Energy

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  1. Ch. 5&6 - Energy I. Energy and Work Energy Work Conservation of Energy

  2. THERMAL The ability to cause change. MECHANICAL NUCLEAR ELECTRICAL CHEMICAL joules (J) A. Energy internal motion of particles ENERGY motion of objects changes in the nucleus motion of electric charges bonding of atoms

  3. Which has the most KE? • Which has the least KE? 80 km/h 50 km/h 80 km/h A. Energy • Kinetic Energy (KE) • energy in the form of motion • depends on mass and velocity 80 km/h truck 50 km/h motorcycle

  4. A. Energy • Potential Energy (PE) • stored energy • depends on position or configuration of an object • Which boulder has greater gravitational PE? • What other ways can an object store energy?

  5. B. Work • Work • transfer of energy through motion • force exerted through a distance W = Fd W: work (J) F: force (N) d: distance (m) 1 J = 1 N·m Distance must be in direction of force!

  6. W d F B. Work • Brett’s backpack weighs 30 N. How much work is done on the backpack when he lifts it 1.5 m from the floor to his back? GIVEN: F = 30 N d = 1.5 m W = ? WORK: W = F·d W = (30 N)(1.5 m) W = 45 J

  7. W d F B. Work • A dancer lifts a 40 kg ballerina 1.4 m in the air and walks forward 2.2 m. How much work is done on the ballerina during and after the lift? GIVEN: m = 40 kg d = 1.4 m - during d = 2.2 m - after W = ? WORK: W = F·d F = m·a F =(40kg)(9.8m/s2)=392 N W = (392 N)(1.4 m) W = 549 J during lift No work after lift. “d” is not in the direction of the force.

  8. C. Conservation of Energy • Law of Conservation of Energy • Energy may change forms, but it cannot be created or destroyed under ordinary conditions. • EX: • PE  KE • mechanical  thermal • chemical  thermal

  9. C. Conservation of Energy PE  KE View pendulum animation. View roller coaster animation.

  10. C. Conservation of Energy Mechanical  Thermal View rolling ball animations. View skier animation.

  11. Ch. 5 - Energy II. Thermal Energy Temperature Thermal Energy Heat Transfer

  12. A. Temperature • Temperature • measure of the average KE of the particles in a sample of matter

  13. B. Thermal Energy • Thermal Energy • the total energy of the particles in a material • KE - movement of particles • PE - forces within or between particles due to position • depends on temperature, mass, and type of substance

  14. 80ºC 80ºC 400 mL A B 200 mL B. Thermal Energy • Which beaker of water has more thermal energy? • B - same temperature, more mass

  15. C. Heat Transfer • Heat • thermal energy that flows from a warmer material to a cooler material • Like work, heat is... • measured in joules (J) • a transfer of energy

  16. 80ºC 10ºC A B C. Heat Transfer • Why does A feel hot and B feel cold? • Heat flows from A to your hand = hot. • Heat flows from your hand to B = cold.

  17. C. Heat Transfer • Specific Heat (Cp) • amount of energy required to raise the temp. of 1 kg of material by 1 degree Kelvin • units: J/(kg·K)or J/(kg·°C)

  18. 50 g Al 50 g Cu C. Heat Transfer • Which sample will take longer to heat to 100°C? • Al - It has a higher specific heat. • Al will also take longer to cool down.

  19. – Q = heat loss + Q = heat gain T = Tf - Ti C. Heat Transfer Q = m T  Cp Q: heat (J) m: mass (kg) T: change in temperature (K or °C) Cp: specific heat (J/kg·K)

  20. Coffee cup Calorimeter C. Heat Transfer • Calorimeter • device used to measure changes in thermal energy • in an insulated system, heat gained = heat lost

  21. C. Heat Transfer • A 32-g silver spoon cools from 60°C to 20°C. How much heat is lost by the spoon? GIVEN: m = 32 g Ti = 60°C Tf = 20°C Q = ? Cp = 235 J/kg·K WORK: Q = m·T·Cp m = 32 g = 0.032 kg T = 20°C - 60°C = – 40°C Q = (0.032kg)(-40°C)(235J/kg·K) Q = – 301 J

  22. C. Heat Transfer • How much heat is required to warm 230 g of water from 12°C to 90°C? GIVEN: m = 230 g Ti = 12°C Tf = 90°C Q = ? Cp= 4184 J/kg·K WORK: Q = m·T·Cp m = 230 g = 0.23 kg T = 90°C - 12°C = 78°C Q = (0.23kg)(78°C)(4184 J/kg·K) Q = 75,061 J

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