1 / 58

Properties of light, EM spectrum; Heat, radiation and sunlight

Properties of light, EM spectrum; Heat, radiation and sunlight. Wavelength. The different colors of light can be expressed in terms of wavelength. 10 000 km. Long Radio Waves. 1 000 km. 100 km. 10 km. 1 km. AM Radio Waves. 100 m. 10 m. Short Radio Waves. TV. FM. 1 m. Weather.

kiri
Download Presentation

Properties of light, EM spectrum; Heat, radiation and sunlight

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Properties of light, EM spectrum; Heat, radiation and sunlight M. Tedesco – Fall 2008

  2. Wavelength • The different colors of light can be expressed in terms of wavelength. 10 000 km Long Radio Waves 1 000 km 100 km 10 km 1 km AM Radio Waves 100 m 10 m Short Radio Waves TV FM 1 m Weather 100 mm 10 mm Microwaves Radar 1 mm Infrared Visible Light Ultraviolet M. Tedesco – Fall 2008 X Rays etc

  3. The sun emits ~ 9 % of its energy as ultraviolet, 46 % as a visible light and 45 % as infrared radiation • Within the visible spectrum the wavelength ranges for the different colors are: Violet  0.40 – 0.45 Blue  0.45 – 0.5 Green  0.5 – 0.57 Yellow  0.57 – 0.59 Orange  0.59 – 0.62 Red  0.62 – 0.7 M. Tedesco – Fall 2008

  4. Wavelength Low Frequency Long Wave Amplitude High Frequency Short Wave M. Tedesco – Fall 2008

  5. Important Wavelengths for Climate • Short wave radiation from the Sun0.48 microns • Long Wave Radiation from the Earth/Atmosphere10 micronsmeasurements in microns:1 micron = 10-6 m = 1/1000mm M. Tedesco – Fall 2008

  6. The particle nature of light was first demonstrated by Einstein showing that light beamed on a metal plate could free electrons by acting like impacting particles called ‘photons’, whose energy depends only on the wavelength according to the law E(L) = 1.975*10^-25/L Joule Where L is given in meters. M. Tedesco – Fall 2008

  7. Thus a shortwave has more energy than a longwave. • Example: The energy of a photon with L = 0.1*10^-6 m E(L) = 1.975*10^-18 J M. Tedesco – Fall 2008

  8. Question: What would be the speed of an electron if all the energy were converted to kinetic energy ? (Note that the mass of the electron is 9.11*10^-31 Kg M. Tedesco – Fall 2008

  9. Kinetic Energy KE = 0.5*m*v^2 Solving for v : v = [2*KE/m]^0.5= [2*1.975*10^-18/(9.11*10^-31]]^0.5 = 2.08*10^6 m/s Which is fast enough to free an electron from a hydrogen atom or ionize the hydrogen M. Tedesco – Fall 2008

  10. RadiationTotal energy of phenomena relative to total solar energy per day: • Solar energy received per day 1 • Strong earthquake 10-2 • Average Hurricane 10-4 • Krakatoa explosion 1883 10-5 • nuclear bomb 10-5 • summer thunderstorm 10-8 • tornado 10-11 M. Tedesco – Fall 2008

  11. Terrestrial Radiation lost to space • 90% of the terrestrial radiation emitted to space comes from the atmosphere • 10% comes through the window from the underlying land and sea M. Tedesco – Fall 2008

  12. Radiation Budget • Shortwave Radiation In • Shortwave and Longwave Radiation Out • Shortwave Radiation Out (reflected) • Longwave Radiation Out (emitted) • Radiation Budget: • (Sin + Lin) - (Sout + Lout) M. Tedesco – Fall 2008

  13. M. Tedesco – Fall 2008

  14. M. Tedesco – Fall 2008

  15. M. Tedesco – Fall 2008

  16. Latitude M. Tedesco – Fall 2008

  17. Sketch showing the variations in the 3 orbital components: obliquity, orbital eccentricity and precession of the perihelion. M. Tedesco – Fall 2008

  18. Variations in the 3 components over the past 500 000 years and D record from Vostok Ice Core McGuffie and Henderson-Sellers, 2005 M. Tedesco – Fall 2008

  19. Blackbody Radiation • All objects whose temperature are above absolute zero Kelvin (-273.15oC) emit radiation at all wavelengths • A “blackbody” is one that is a perfect absorber and perfect emitter (hypothetical, though Earth and Sun are close) M. Tedesco – Fall 2008

  20. Graybody Radiation • an object that is not a perfect absorber/emitter • it reflects part of incident radiation • emissivity, e, is the ratio of graybody exitance to blackbody exitance M. Tedesco – Fall 2008

  21. Emissivity • Describes the actual energy absorption and emission properties of real objects (“graybodies”) • Is wavelength dependent (so, it’s actually a “colored body”) • Emissivity establishes the radiant temperature Trad of an object M. Tedesco – Fall 2008

  22. Radiant Temperature vs. Kinetic Temperature • Two objects can have the same kinetic temperature but different radiant temperatures M. Tedesco – Fall 2008

  23. Emissivities of common substances in the infrared band Water  .92 - .96 Snow  .92 - .995 Ice  .96 Frozen soil  .93 - .94 Dry sand  .89 - .90 Polished Silver  .02 (aluminum foil keeps food warm because they radiate heat away very slowly !) Desert  .90 Grass  .90 M. Tedesco – Fall 2008

  24. Kirchoff’s law of emissivity • Kirchoff discovered that at any given wavelength, when a light shines on a body, the percentage of light the body absorbs is equal to the emissivity Absorptivity = Emissivity M. Tedesco – Fall 2008

  25. The solar constant Satellites measurements have shown that the sun’s output is almost constant. This output is expressed as an irradiance or flux density The Solar Irradiance I is energy (or heat) divided by time and area I = dE/(A*dt) or dE=dQ=I*A*dt M. Tedesco – Fall 2008

  26. Example: If I = 1000 W/m2, how much energy can a pool with an area of 200 m2 absorb in 1 hour ? M. Tedesco – Fall 2008

  27. Solution: • E = Q = 1000*200*3600 = 7.2*10^8 Joules M. Tedesco – Fall 2008

  28. Sunlight, therefore, can either raise temperature or evaporate water. • The questions we want to answer are, among others: how much sunshine does it take to raise the temperature of the atmosphere by 1 deg ? Or to evaporate 1m3 of water ? • The 1st law of thermodynamics provides the answer and it is basically the law of energy conservation • Generally, only thermal energy is considered as potential and kinetic energies are relatively small M. Tedesco – Fall 2008

  29. The first law of thermodynamics states that: Q = m*c* T+L* m Q is heat, m is mass and m is the relative mass change due to phase change (e.g., melting or boiling), c is the specific heat capacity (E/(m*deg)), L is the latent heat (E/m) M. Tedesco – Fall 2008

  30. THIS IS A FUNDAMENTAL LAW ! • THOSE OF YOU WHO WILL NOT LEARN AND UNDERSTAND THIS WILL BE STRONGLY PENALIZED ! M. Tedesco – Fall 2008

  31. Examples of latent heat: ICE  3.34*10^5 J/Kg (melting) WATER  2.5*10^6 (vaporizing 0 C) • Examples of specific heat capacity: ICE  2100 J/(Kg*deg) WATER  4186 VAPOR  1463 Air  1004 at constant pressure 760 at constant volume Iron  210 Quartz  840 M. Tedesco – Fall 2008

  32. Example: The sun provides 10^6 Joules to heat 10^4 Kg of air. How much the temperature increase ? T = 0.0996 Why ? M. Tedesco – Fall 2008

  33. Example: The sun provides 10^6 Joules to evaporate water. How much water will evaporate if none of the heat is used to raise water temperature ? m = 0.4 Kg Why ? M. Tedesco – Fall 2008

  34. Often it is useful to know the amount of evaporation in meters rather than kilograms. It also important to know the rate of evaporation or the rate of temperature change as a result of heating. • Let us work, therefore, on the first law and give it another expression M. Tedesco – Fall 2008

  35. Let us divide all terms by surface A Q/A=m*c* T/A+L* m/A If we want the rate of heating or evaporation, let us divide by the time interval dt dQ/(A* dt)=m*c* dT/(A * dt )+L* dm/(A * dt) (Note the change in the differential operator) M. Tedesco – Fall 2008

  36. At this point we remember that: m = r*V= r * A * dz and rewrite the law as follows: dQ/(A* dt)=r * dz*c* dT/dt +L* r*dz/ dt Using the hydrostatic equation r*dz=dr/g M. Tedesco – Fall 2008

  37. And the definition of mixing ratio: w = r(vapor)/[r(air)-r(vapor)] ~ r(vapor)/r(air) We have: dQ/(A* dt)=dp* c*dT/(g*dt) +L*dw* dp /(g*dt) Note that all quantities have the same unit as irradiance M. Tedesco – Fall 2008

  38. Example: Bright sunlight (I = 1000 W/m2) hits the lowest 100 mb of the atmosphere for 1 hour. How much does the temperature rise ? I = dQ/(A* dt) = dp* c*dT/(g*dt) = dp* c*dT/(g*dt) Solving for dT gives: dT = 1000*9.8*3600/(10^4*1004) = 3.51 C M. Tedesco – Fall 2008

  39. Example: The sun shines for 5 hours with irradiance 700 w/m2. If the water is 10 m deep, what is the increase of temperature ? I = dQ/(A* dt) = 700 = r * dz*c* dT/dt = 1000*10*4186*dT/(5*3600) = 0.301 deg M. Tedesco – Fall 2008

  40. Example: melting snow If the snowpack is 0.5 m and snow density is r = 300 Kg/m3, snow albedo is 0.8 and winter sun irradiance is 300 W/m2. How long will it take to melt all snow? dQ/(A* dt)=r * dz*c* dT/dt +L* r*dz/ dt= 0 + L* r*dz/ dt = I*(1-albedo) 300*0.2=60=3.34*10^5*300*0.5/dT = 231.94 hours ~ 9.66 days M. Tedesco – Fall 2008

  41. Solar radiation There are three basic laws regarding the amount of sunlight received by any place on Earth. Two of these laws are geometrical and the last expresses how radiation is attenuated as it passes through the atmosphere M. Tedesco – Fall 2008

  42. 1) The inverse square law of irradiance I (x) = I (y)*[y/x]^2 As an example. On Earth the solar constant is S = 1367 W/m2 and the average distance between Sun and the Earth is 149.5 million of Km. What is the solar constant on Mars if the average distance is 228 millions of Km ? I (Mars) = I(Earth)([149.5/228]^2 = 588 W/m2 Note that the law can also be solved to obtain distances, once the solar irradiance is known. M. Tedesco – Fall 2008

  43. 2) Earth-Sun distance formula The distance between Earth and Sun can be expressed as : d(e)=149.5+2.5*cos(number of days from July 1) M. Tedesco – Fall 2008

  44. Cosine law of irradianceIf Z is the zenith angle (describing how far is the sun from overhead) then I(z) = I(0)*cos(Z) 1 Cos(Z) Z M. Tedesco – Fall 2008

  45. Zenith Angle Formula The formula for the zenith angle (or rather for its cosine) is cos(Z) = sin(d)*sin(phi)+cos(d)*cos(phi)*cos(h) Where h is 2*p/24 radians per hour = 15 deg per hour from noon, phi is the latitude and d is the declination angle given by d = 23.5*cos[number of days from June 21] deg M. Tedesco – Fall 2008

  46. Sunrise and sunset hours The formula for computing the sunrise and sunset can be derived from the zenith angle as Z =90 at sunset or sunrise H (hours) = arccos(-tan(d)*tan(phi))/15 M. Tedesco – Fall 2008

  47. Azimuth angle formula The azimuth angle formula is the following: tan(az) = {cos(d)*sin(h)/[cos(d)*sin(phi)*cos(h)-sin(d)*cos(phi)]} Where h is 2*p/24 radians per hour = 15 deg per hour from noon, phi is the latitude and d is the declination angle given by M. Tedesco – Fall 2008

  48. The atmosphere scatters or absorbs some of the radiation …. M. Tedesco – Fall 2008

  49. When the sun is not at the zenith, the effective path length becomes longer and, therefore, less light will penetrate to sea level. • The lower the sun, the longer the effective path length and that’s why we can look at the sun during sunset or sunrise • The law describing the way radiation penetrates the atmosphere when the sun is not overhead is the Lambert’s law or Beer’s law M. Tedesco – Fall 2008

  50. Lambert’s law or Beer’s law of Light Transmission I(Z,a) = I(Z)*a^[sec(Z)] Which becomes I(Z,a) = I(0)*cos(Z)*a^[sec(Z)] Where a is the atmosperic transparency. A typical value for it is 0.7. M. Tedesco – Fall 2008

More Related