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Chapter 12

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A 50.0 g ball is dropped from an altitude of 2.0 km. Calculate: U i , K max , & W done through the fall. Chapter 12. Thermal Energy. Thermodynamics. The movement of heat. Kinetic Theory. All matter is made up of tiny particles All particles are in constant motion

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Chapter 12

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  1. A 50.0 g ball is dropped from an altitude of 2.0 km. Calculate: Ui, Kmax, & W done through the fall

  2. Chapter 12 Thermal Energy

  3. Thermodynamics • The movement of heat

  4. Kinetic Theory • All matter is made up of tiny particles • All particles are in constant motion • All collisions are elastic

  5. Temperature • A measure of average kinetic energy

  6. Temperature • A measure of heat intensity

  7. Thermal Equilibrium • When the average kinetic energy of two or more substances become equal; thus their particles have the same exchange rate

  8. Because it is a measure of average kinetic energy, temperature is related to the motion of particles (atoms, molecules, ions, etc)

  9. Thermometer • A device, calibrated to some temp scale, that is allowed to come to thermal equilibrium with something else

  10. Temperature Scales • Celcius (oC) • Based on MP & BP of water • Kelvin (K) • Based of absolute temperature

  11. Temperature Scales • K = oC + 273

  12. Convert Temperatures 100 K = ___ oC 100 oC = ___ K

  13. Heat • A form of energy that flows due to temperature differences

  14. Heat (Q) • Because particle at higher temp. move faster than particles at a lower temp., the net flow of heat is H  C

  15. Heat (Q) • Heat will continue to have net flow from H  C as long as there is a temperature difference

  16. Heat (Q) • When there is no temperature differences, the system has reached thermal equilibrium

  17. Work • The movement of energy by means other than temperature difference

  18. 1st Law of Thermo. • The increase in thermal energy = sum of heat added & work done to a system

  19. 1st Law of Thermo. DE = Q + W

  20. In Most Engines • Heat is added by some high energy source (gas) • Work is done by the engine

  21. In Most Engines DE = Q + W But W < 0

  22. Entropy • A measure of the disorder in a system

  23. 2nd Law of Thermo. • In natural processes, entropy increases

  24. Entropy • When fuel is burned, entropy is increased

  25. Specific Heat (C) • The thermal energy required to raise 1 unit mass of matter 1 degree

  26. Specific Heat (C) • The thermal energy required to raise 1 kg of matter 1 degree K

  27. Heat (Q or DH) • Heat transfer = mass x specific heat x the temperature change • Q = mCDT

  28. Calculate the heat required to raise 50.0 g of water from 25.0oC to 65.0oC. Cwater = 4180 J/kgK

  29. Calculate the heat required to raise 250.0 g of lead from -25.0oC to 175.0oC. Clead = 130 J/kgK

  30. 28 kJ of heat was required to raise the temperature of 100.0 g of a substance from -125oC to 575oC. Calculate: C

  31. 3.6 kJ of heat was required to raise the temperature of 10.0 g of a substance from -22oC to 578oC. Calculate: C

  32. Conservation of Heat • The total energy of an isolated system is constant

  33. Conservation of Heat Because the total amount of heat is constant q or DHsystem = 0

  34. Conservation of Heat q or DHsystem = 0 DHsys = DH1 + DH2 + .. qsys = q1 + Dq2 + ..= 0

  35. Conservation of Heat qsys = q1 + Dq2 = 0 mCDT1 + mCDT2= 0 mCDT1 = - mCDT2

  36. Conservation of Heat qsys = qgained + qlost qgained = - qlost mCDTgain = - mCDTlost

  37. A 50.0 g slug of metal at 77.0 oC is added to 500. g water at 25.0oC. Teq= 27.0oC. Calculate: Cmetal Cwater = 4180 J/kgK

  38. A 200.0 g slug of metal at 77.5 oC is added to 400. g water at 25.0oC. Teq= 27.5oC. Calculate: Cmetal Cwater = 4180 J/kgK

  39. Solving Mixture Temperatures qsystem = 0 qsystem = qhot + qcold mCDThot = -mCDTcold DT = Tf – Ti mC(Tf – Ti)hot = -mC(Tf – Ti)cold

  40. Conservation of Heat mChTf - mChTh +mCcTf - mCcTc = 0

  41. Conservation of Heat mChTf - mChTh = -mCcTf + mCcTc

  42. 20.0 g of water at 25.0oC is added to 30.0 g water at 75.0oC. Calculate: TeqCwater = 4180 J/kgK

  43. 500. g of water at 75.0oC is added to 300. g water in a 200. g calorimeter all at 25.0oC. Calculate: TeqCwater = 4180 J/kgK Ccal = 1000 J/kgK

  44. A 500.0 g slug of metal at 87.5.oC is added to 4.0 kg water in a 1.0 kg can at 25.0oC. Teq= 27.5oC. Calculate: Cmetal Cwater = 4180 J/kgK Ccan = 1.0 J/gK

  45. States of Matter • Solid • Liquid • Gas

  46. Solid • Has definite size & definite shape • Particles vibrate at fixed positions

  47. Liquid • Has definite size but no definite shape • Particles vibrate at moving positions

  48. Gas • Has neither size nor shape • Particles move at random

  49. Change of State • When a substance changes from one state of matter to another

  50. Change of State • Change of state involves an energy change

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