Surface 3 T= 1200K E = 0.8 %3D Surface 1 Surface 2 q= 100W/m? E = 0.6 Perfectly insulated E = 0.8
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- Earth absorbs solar energy and radiates infrared energy. The intensity of the solar radiation incident on earth is J = 1350 Wm-2, also known as the solar constant. Assume earth’s surface (ground) temperature to be uniform at Ts, and that the ground and atmosphere are black (emissivity = 1) for infrared radiation. The radius of the earth is 6.378 x 106 m. The diagram shows the ground at the surface temperature Ts and the atmosphere, represented as a thin black layer, at temperature Ta . Suppose the atmosphere absorbs 100% of the infrared radiation emitted by the ground. Assume that the ground absorbs 47.5% of the incident solar energy, and that the atmosphere absorbs 17.5% of the incident solar energy (for a total of 65% absorbed by the planet). Calculate the "steady state” numerical values of the earth’s ground temperature Ts and the atmospheric temperature Ta taking into account the “greenhouse effect” of atmospheric infrared absorption and emission described above.Charge-coupled device (CCD) image sensors, which are common in modern digital cameras, respond differently to light sources with different spectral distributions. Daylight and incandescent light may be approximated as a blackbody at the effective surface temperatures of 5800 K and 2800 K, respectively. Determine the fraction of radiation emitted within the visible spectrum wavelengths, from 0.40 mm (violet) to 0.76 mm (red), for each of the lighting sources.Light from an ideal spherical blackbody 15.0 cm in diameteris analyzed by using a diffraction grating that has 3850 lines/cm.When you shine this light through the grating, you observe that thepeak-intensity wavelength forms a first-order bright fringe at +-14.4from the central bright fringe. (a) What is the temperature of the blackbody?(b) How long will it take this sphere to radiate 12.0 MJ of energyat constant temperature?