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Warm-up

Warm-up. Agree. Disagree. The temperature of boiling water does not increase even though energy is supplied to it continually. . Warm-up. Separating magnets that stick to each other. Allowing magnets to stick to each other. Some water molecules taken from liquid water are put in air.

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Warm-up

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  1. Warm-up Agree. Disagree. The temperature of boiling water doesnot increase even though energy issupplied to it continually. 

  2. Warm-up Separating magnets that stick toeach other. Allowing magnets to stick to eachother. Some water molecules taken from liquid water are put in air. Water molecules in air come together and form dew. Do these requireenergy supply?  

  3. Warm-up Ice molecules moves ________ (more/ less) freely than water molecules; less so they have ________ (more / less) KE than water molecules. less

  4. Introduction Change of state Matter exists in 3 states: solid, liquid, gas e.g. water vaporization (takes place at boiling point) Fusion (at melting point) ice water steam solidification condensation at freezing point at boiling point

  5. temperature gas liquid solid time Cooling curve When a hot liquid is cooled down, its temperature drops. Graph of temperature vs time? ??

  6. Experiment 3a Cooling curve of octadecan-1-ol Record the temperature of the melted octadecan-1-ol as it cools down. Video Video

  7. Cooling curve • When solid is heated, it melts to a liquid • On cooling, the variation of temperature with time from A to B • The temperature remains constant from B to C, the energy evolved is called the latent heat • The latent heat is used to change from liquid to solid at constant temperature, which is called the melting point

  8. Why temperature remains constant? • From liquid to solid, energy is released to the surroundings due to the decrease in the kinetic energy of the particles • From solid to liquid, energy is absorbed from the surroundings due to the increase in the kinetic energy of the particles

  9. Specific latent heat of fusion and vaporization • Definition • Latent heat is required is proportional to the mass of the substance that undergoes state change i.e. Latent heat (H)  mass (m) H = L m L is a constant called the specific latent heat • Unit: J kg-1

  10. temperature / C A B C D Here is a cooling curve of octadecan-1-ol . Cooling curve AB drops steadily — liquid cooling (temperature falling )

  11. temperature / C A B C D Cooling curve BC is flat — liquid solidifying (temperature unchanged)

  12. temperature / C A B C D Cooling curve CD drops steadily — solid cooling to room temperature (temperature falling)

  13. temperature / C A B C D Cooling curve melting point: melting point read from the flat part BC

  14. Cooling curve of water temperature liquid-solid mixture Simulation

  15. Latent heat The cooling curve shows: When a substance is solidifying, • it loses energy continuously but... • its temperature remains unchanged

  16. Latent heat During change of state: The energy given out/absorbed is called latent heat means‘hidden’

  17. Latent heat Ice-water mixture stays at 0 oC until all the ice is melted. This energy is called latent heat of fusion of ice. • temperature unchanged • energy is absorbed from air to change the ice to water

  18. Latent heat Energy supplied continuously to keep water boils… This energy is called latent heat of vaporizationof water. • temperature unchanged • energy is absorbed to change the water to stream

  19. steam condensation vaporization releases latent heat of vaporization water solidification fusion ice

  20. steam vaporization condensation absorbs latent heat of vaporization water solidification fusion ice

  21. steam condensation vaporization releaseslatent heat of fusion water solidification fusion ice

  22. steam vaporization condensation water absorbs latent heat of fusion solidification fusion ice

  23. State Change solid liquid gas liquid

  24. strongattraction weakattraction 2 Latent heat and particle motion molecule Regular arrangement breaks up

  25. PE  2 Latent heat and particle motion Energy has to be suppliedto oppose the attractive forceof the particles. PE related to the forces of attraction between the particles solid  liquidor liquid  gas average potential energy 

  26. 2 Latent heat and particle motion The transfer of energy does not change the KE. Temperature does not change. latent heat = change in PE during change of state Simulation

  27. Specific latent heat Specific = for 1 kg of a substance e.g. energy E without temperature change 1 kg solid X 1 kg liquid X E= latent heat for 1 kg of X = specific latent heat of X

  28. Specific latent heat Energy transferred to change the state of 1 kg of the substance without a changein temperature. symbol: l or E = ml unit: J kg-1

  29. Specific latent heat of fusion of ice (lf ) lf = energy needed to change 1 kg ofice to water (without temperature change) Find (1) mass melted m and (2) energy transferred E  lf = E/m

  30. Experiment 3b Measuring the specific latent heat of fusion of ice Video Simulation

  31. Experiment 3b control apparatus experimental apparatus Ice also melts at roomtemperature, so a control is needed.

  32. Experiment 3b Measuring the specific latent heat of fusion of ice For ice, lf = 3.34  105 J kg-1 ice (0 C) water (0 C)

  33. b Specific latent heat of vaporization of water (lv) lv = energy needed to change 1 kg of water to steam (without change of temperature) Find (1) mass boiled away m and (2) energy transferred E  lv = E/m

  34. Experiment 3c Video Simulation

  35. Experiment 3c For water, lv = 2.26  106 J kg-1 steam (0 C) water (0 C)

  36. condensation energy out 2260 kJ energy out 334 kJ solidification Summary: change of state vaporization energy in 2260 kJ steam (1 kg) water (1 kg) energy in 334 kJ ice (1 kg) fusion

  37. ice (ice and water) melting boiling (water and stream) (334) (2260) water Summary: from ice to steam stream 100 0 energy / kJ (420) Energy involved in heating 1 kg of water

  38. Consider a cup of water... Consider a cup of water (mass m) being heated from0 °C until it starts to boil at 100°C. Since E = mcT and E = mlv mcT = mlv  lv= c T = 4200×100 = 420 kJ kg-1 Is the student correct? (Yes/No)

  39. When vapour condenses... When vapour condenses, is the surrounding air warmedor cooled? (warmed/cooled)

  40. Jimmy melts... Temperature / oC Z X Y time / s 0 0.5 1.2 2.3 3.7 Jimmy melts three materials X, Yand Zof equal mass at the same time and place.

  41. ... Temperature / °C Z X Y time / s 0 0.5 1.2 2.3 3.7 Which material(s) has/have the greatest melting point?( X / Y / Z )

  42. Temperature / °C Z X Y time / s 0 0.5 1.2 2.3 3.7 Which material(s) has/have the largest value of specificlatent heat of fusion? ( X / Y / Z )

  43. Temperature / °C Z X Y time / s 0 0.5 1.2 2.3 3.7 Which material(s) release(s) largest amount of energy(per kg) when they freeze? ( X / Y / Z )

  44. Fusion and Boiling • Boiling of water and fusion of ice • when water boils, the temperature is found to remained at 100oC. The energy supplied is only used to change from water to steam without any change in temperature • When the steam condenses, the temperature remains at 100oC and energy is given out

  45. Fusion and Boiling • Fusion of ice • when ice melts, the temperature is found to remained at 0oC. The energy supplied is only used to change from ice to water without any change in temperature • When the water freezes, the temperature remains at 0oC and energy is given out

  46. Latent Heat of Fusion and Vaporization • Energies used in fusion and vaporization are called the latent heat of fusion and vaporization • The constant temperatures are called the melting point and boiling point

  47. Example 1 Finding specific latent heat of fusion of ice Result of melting experiment Mass of water in experimental cup: m1 = 0.050 kg in control cup: m2 = 0.014 kg Joulemeter reading initial: j1 = 15 000 J final: j2 = 29 200 J

  48. Example 1 Finding specific latent heat of fusion of ice (a) Find the specific latent heat of fusion of ice. Results: m1 = 0.050 kg m2 = 0.014 kg j1 = 15 000 J j2 = 29 200 J lf =E /m = (j2 – j1) / (m2 – m1) = (29 200 – 15 000) / (0.050 – 0.014) = 3.94  105 J kg–1

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