According to Planck's law of black- body radiation, the spectral energy density R as a function of wavelength A (m) and temperature T (K) is given by: 2nc h R = 1 hc/NkT e -1 where c=3x10° m/s is the sneed of 8 4. 3. 2. ctral Energy Density (W/m)
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- 1.26 Repeat Problem 1.25 but assume that the surface of the storage vessel has an absorbance (equal to the emittance) of 0.1. Then determine the rate of evaporation of the liquid oxygen in kilograms per second and pounds per hour, assuming that convection can be neglected. The heat of vaporization of oxygen at –183°C is .11.68 Two infinitely large, black, plane surfaces are 0.3 m apart, and the space between them is filled by an isothermal gas mixture at 811 K and atmospheric pressure. The gas mixture consists of by volume. If one of the surfaces is maintained at 278 K and the other at 1390 K, calculate (a) the effective emissivity of the gas at its temperature, (b) the effective absorptivity of the gas to radiation from the 1390 K surface, (c) the effective absorptivity of the gas to radiation from the 278 K surface, and (d) the net rate of heat transfer to the gas per square meter of surface area.A tungsten filament is heated to 2700 K. At what wavelength is the maximum amount of radiation emitted? What fraction of the total energy is in the visible range (0.4to0.75m)? Assume that the filament radiates as a graybody.
- The emissive power of a black body depends on the fourth power of the temperature and is given by W = A T4 Where, W = emissive power, BTUs per square feet per hour A = Stefan Boltzmann constant, 0.171 x 10-8 T = temperature,° Rankine what is the value of A units in J/(m2.7. The emissive power of a blackbody, at 0.8 µm wavelength is measured as 1E5 W/m², µm. Find the blackbody temperature. [Ans. 1739 K]The following figure was generated from experimental data relating to spectral black body emissive powen to the wavelength at three termperatures T1 T2 and T3 Ti > 12 > T3). E(W/m².um) 2(um The conclusion is that the measurements are: a Correct because the maxima in Ena show the correct trend Oa Correct because Planck's law is satisfied CC Wrong because the Stefan Boltzmann law is not satisfied O d. Wrong because Wien's displacement law is not satisfied
- 1) The spectral, hemispherical absorptivity of an opaque surface and the spectral irradiation at the surface are as shown. d' 1.0 G₂ (W/m².um) 500 0.2 0 0 0 2 4 6 8 10 12 14 16 0 2 4 6 8 10 12 14 16 λ(μm) λ(μm) a) How does the spectral, hemispherical reflectivity vary with wavelength? b) What is the total, hemispherical absorptivity of the surface? c) If the surface is initially at 500 K and has a total, hemispherical emissivity of 0.8, how will its temperature change upon exposure to the irradiation?A typical car's exterior consists of a thin layer of silica (SiO2) over an opaque painted metal panel. Silica is transparent in the visible wavelengths but offers high reflectance in the near- to mid- infrared wavelengths. The plot on the next page depicts the diffuse spectral reflectivity (pa) of the car's surface: Spectral reflectivity, P₂ 0.8 0.6 0.4 ལ 0.2 0 0.1 1 1 10 Wavelength, λ(μm) 100 If the car's exterior temperature is T₁ = 77°C, determine both the total absorptivity (a) and the total emissivity (a) of the silica-covered panel. Assume that the Sun's temperature is Tsun = 5800 K.1. Continuum wave equation. Show that for long wavelengths the equation of motion, d²u M s = dt² = C(us+1+ug-1-2u), reduces to the continuum elastic wave equation where v is the velocity of sound. d'u ôt² d²u əx²
- A DVST, or Direct View Storage Tube, is a kind of optical storage medium.The sun is the only star whose size we can easily measure directly; astronomers therefore estimate the sizes of other stars using Stefan's law. The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon energy of 0.8 eV, while Betelgeuse is approximately 10,000 times as luminous as the sun. How does the radius of Betelgeuse compare to the sun's radius? Why is Betelgeuse called a "red supergiant"?The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 um. If the peak wavelength of emitted radiation changes to 2.90 um, then the temperature (in K) of the black body is A. 500 B. 1000 C. 4000 D. 8000