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GEU 0047: Meteorology Lecture 02 Heat Energy

GEU 0047: Meteorology Lecture 02 Heat Energy. Heat Energy. Temperature is our way of quantifying matter’s internal kinetic energy .

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GEU 0047: Meteorology Lecture 02 Heat Energy

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  1. GEU 0047: MeteorologyLecture 02Heat Energy

  2. Heat Energy Temperature is our way of quantifying matter’s internal kinetic energy. It is a macroscopic measurement of the average kinetic energyfound in the random and microscopic motions (vibration, rotation, and collision) of countless atoms and molecules. Temperature is related to heat energy.

  3. it is a measure of the average translational kinetic energy associated with the disordered microscopic motion of atoms and molecules. Here, Temp. is measured in Kelvin in the SI units. using  Kinetic temperature definition What is temperature? K: Boltzman const. = 13.805 x 10-24 J deg-1 K = nR; R is the gas law constant in units of 1.98 cal/deg mole

  4. Temperature Conversions CELCIUS C = 5/9 (F-32) Fahrenheit F = (9/5 C) + 32 Absolute or Kelvin K = C + 273

  5. Boiling Point of Water Celcius C = 5/9 (F-32) C = 100 Fahrenheit F = (9/5 C) + 32 F = (9/5 100) + 32 = 180 + 32 Absolute = 212 or Kelvin K = C + 273 K = 100 + 273 = 373

  6. Freezing Point of Water Celcius C = 5/9 (F-32) C = 5/9 (32-32) = 0.0 Fahrenheit F = (9/5 C) + 32 F = 32 Absolute or Kelvin K = C + 273 K = 0 + 273 = 273

  7. Analog Temperature Conversion Plot

  8. Celsius to Fahrenheit Conversion

  9. Distribution of Speeds

  10. Maxwell-Boltzman Velocity Distribution E E E Notice that the above plot is not symmetric => probability kinetic energy ≠ average kinetic energy.

  11. Temperature and Kinetic Energy T = a mwv2 a = 4.0 x 10-5 Ks2/m2 v2 = average molecular speed (KE) mw = molecular weight Atmospheric Molecule 28.01 N2 32.00 O2 18.02 H2O 44.01 CO2 28.96 Weighted Mean Composition

  12. Molecular Velocity For a cold day (-15 oC) what is the average speed of a nitrogen molecule as compared to a hot day (32 oC) v2 = T / a mw = 258 / (4E-5 x 28.01) vcold = 480 m/s v2 = T / a mw = 305 / (4E-5 x 28.01) vhot = 522 m/s

  13. Day Ground Temperature Official Temperature is read at a height of 1.5 meter above the ground, in the shade, and out of the wind.

  14. Night Ground Temperature • The ground radiates away the daytime heat faster than the air above it. Air is a very poor conductor.

  15. Air Turbulence • Air motion causes mixing, removing stagnant boundary air Hence, larger temperature gradients are possible without the wind

  16. Radiation Shield • Thick forest, or Clouds can provide a radiation shield

  17. Thermal Insulation • Thick Forest, Low Clouds can provide a thermal blanket

  18. Temperature Data Mean Daily Temperature: average over 24 hours Mean Annual Temperature: average of 12 months Average mean daily Temp.: average of mean daily temperatures over 30 years Annual Temperature Range: Difference between largest monthly mean and smallest monthly mean temperature.

  19. Growing Days • Number of days when the mean daily temperature is 1 degree above the base temperature for the particular crop.

  20. Cooling Degree Days • Used to estimate energy and power consumption needs for cooling indoor air during summer. Base Temp = 65oF

  21. Heating Degree Days • Used to estimate energy and power consumption needs for heating indoor air during winter. Base Temp = 65oF

  22. Controls of Temperature (important) • Solar Insolation • Date & Time • Latitude • Exposure (wind, humidity) • Geographic • Land • Water • Oceanic • Currents • Topography • Elevation

  23. Heat Index (apparent temp. due to RH) ● ● ● ● ● ● ● ● ● ● ● ● ● c.f. Table D.2 in Appendix D

  24. Relative humidity (RH)(important) • Amount of water vapor Amount required for saturation • Water vapor pressure Saturation water vapor pressure • Hence, RH depends on the ambient Temp (and Pressure)

  25. Wind Chill (apparent temp. due to wind) • Wind Chill Equivalent Temperature • If the air temperature is 10。F and the wind is 25 mph, the wind chill equivalent temperature is -29 。F . Wind chill = 35.74 + 0.6215 T - 35.75 V0.16 + 0.4275 T V0.16 T: air temp. (F) V: wind speed (mph)

  26. Matter Phases • In order of increasing Temperature (Energy): • CRYSTAL Occurring at the coldest temperatures • SOLID • LIQUID • GAS • PLASMA Occurring at the highest temperatures

  27. Matter Phases • In order of decreasing Organization (Symmetry): • CRYSTAL Highly Ordered • SOLID • LIQUID • GAS • PLASMA Highly Disorganized

  28. Phase Transitions

  29. State Changes Energy increased and absorbed by substance: • SOLID to LIQUID Melting • LIQUID to GAS Boiling • SOLID to GAS Sublimation Energy decreased and released by substance: • GAS to SOLID Deposition • GAS to LIQUID Condensation • LIQUID to SOLID Freezing

  30. Water Crystals Atomic and Molecular Structures Lead to Macroscopic Order Heat Energy must be absorbed by the solid to break the highly ordered ice crystals. Heat Energy is released by a liquid in order to crystallize.

  31. Phase Diagram

  32. Latent Heat T = const T = const

  33. Latent Heat of Fusion (Lf) Heat Energy required to convert solid to liquid

  34. Latent Heat of Evaporation (Lv) Heat Energy required to convert liquid to gas.

  35. Water Latent Heat Exchange Condensation yield 6.75 times more heat energy than Fusion (Evaporation require 6.75 times more heat energy than Melting). For water, Lf = 80 Cal /gram; Lv = 540 Cal/gram

  36. Heat/energy Units • Calorie: the amount of heat required to raise the temperature of 1 gram of water by 1 degree Celsius. • 1 cal. = 4.186 Joules • (1 Food calorie = 1,000 calories = 4186 J) (see Appendix A for more)

  37. Specific Heat (比熱) • Q = m cDT Q = Heat Energy (human perception) m = mass DT = Temperature difference c = specific heatresponsible for the thermal properties of the substance (J/kg/oC) • DT = Q/mc

  38. Specific Heat DT = Q/mc For a given amount of heat energy, say 10,000 Joules, what is the temperature change for 1 kg of water and 1 kg of sand? Csand = 838 J/kgoC Cwater = 4180 J/kgoC DTsand = 10,000/1(838) = 11.9 oC DTwater = 10,000/1(4180) = 2.4 oC

  39. Northern Hemisphere Southern Hemisphere.

  40. Land Versus Sea Land masses in the North cause more temperature variations than in the South where oceans keep the temperature more even and moderate.

  41. Melting DT = Q/mc Amount of heat energy needed to bring a 0.25 kg ice block to a temperature of 50oC? (Starting Temp = 0oC Ending Temp = 50oC) Q = heat needed to make transition from ice to water + heat needed to heat water from 0 to 50 oC Q= mLf + mcDT

  42. Melting DT = Q/mc Q = heat needed to make transition from ice to water + heat needed to heat water from 0 to 50 oC Q = (250 g) (80 Cal/g) (4.186 J/Cal) + (0.25 kg) (4180 J/kgoC) (50-0 oC) = 83720 + 52250 = 135970 Joules Can you calculate it ? (important)

  43. The amazing water molecule Liquid/gas state only Plus high heat capacity (thermal inertial) and solvent power (Covalent bond) 共價鍵結 (hydrogen bonding) (in electricity)

  44. Freezing This latent heat energy is released when water droplets freeze. Water vapor that condenses also gives off latent heat. Both processes help heat the atmosphere. The opposite processes (melting or evaporation) cause heat energy to be removed from the atmosphere.

  45. CONDENSATION • Gas to Liquid (or Freezing, Liquid to Solid) • ENERGY IS RELEASED, Gas has a higher internal energy than the liquid state. • A WARMING PROCESS

  46. EVAPORATION • LIQUID to GAS • ENERGY IS REMOVED, Liquid has a lower internal energy than the gaseous state. • A COOLING PROCESS

  47. Radiation Energy transport via electromagnetic waves

  48. Convection Energy transport by mass motion

  49. Conduction Energy transport by vibrational translation The jostling of atoms and molecules in close proximity in a solid, especially one with high conductivity.

  50. 電磁波譜 電磁波依波長可分為： • γ射線、x射線、紫外線、可見光、紅外線與無線電波 --無線電波又可進一步細分成微波、超短波、短波和長波 --紅外線有時也細分為近紅外線、遠紅外線與次毫米波

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