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Chapter 12: Problem 22

Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at \(2500 \mathrm{~K}\), a heated metal at \(1500 \mathrm{~K}\), human skin at \(305 \mathrm{~K}\). and a cryogenically cooled metal surface at \(60 \mathrm{~K}\). Estimate the fraction of the solar emission that is in the following spectral regions: the ultraviolet, the visible, and the infrared.

Short Answer

Expert verified
Wavelengths: Sun ~500 nm, Tungsten ~1160 nm, Metal ~1930 nm, Skin ~9500 nm, Cooled Metal ~48300 nm. For solar emission: 44% visible, 50% infrared, 6% ultraviolet.

Step by step solution

01

Understanding Wien's Displacement Law

Wien's Displacement Law states that the wavelength at which the emission of a blackbody is maximized is inversely proportional to the temperature of the blackbody. The formula is given by \( \lambda_{max} = \frac{b}{T} \), where \( \lambda_{max} \) is the wavelength of maximum emission, \( b \) is Wien's displacement constant \( (2.897 \times 10^{-3} \, m \cdot K) \), and \( T \) is the absolute temperature of the blackbody in Kelvin.
02

Calculate Wavelength for the Sun

Assuming the surface temperature of the sun to be about \( 5800 \mathrm{~K} \), use Wien's Law to find the peak wavelength: \( \lambda_{max} = \frac{2.897 \times 10^{-3}}{5800} \approx 500 \mathrm{~nm} \). This indicates that the sun's peak emission is in the visible spectrum.
03

Calculate Wavelength for Tungsten Filament at 2500 K

Use Wien's Law: \( \lambda_{max} = \frac{2.897 \times 10^{-3}}{2500} \approx 1160 \mathrm{~nm} \). The peak emission falls in the infrared spectrum.
04

Calculate Wavelength for Heated Metal at 1500 K

Apply Wien's Law: \( \lambda_{max} = \frac{2.897 \times 10^{-3}}{1500} \approx 1930 \mathrm{~nm} \). The peak emission is also in the infrared spectrum.
05

Calculate Wavelength for Human Skin at 305 K

Using Wien's Law: \( \lambda_{max} = \frac{2.897 \times 10^{-3}}{305} \approx 9500 \mathrm{~nm} \). This peak wavelength is well within the infrared spectrum.
06

Calculate Wavelength for Cooled Metal Surface at 60 K

Applying Wien's Law: \( \lambda_{max} = \frac{2.897 \times 10^{-3}}{60} \approx 48300 \mathrm{~nm} \). This peak wavelength lies in the far infrared to microwave region.
07

Estimate Spectral Distribution of Solar Emission

The sun emits across a range of wavelengths. Most is in the visible light range (about 400-700 nm), a significant fraction is in infrared, and a smaller fraction in ultraviolet (<400 nm). The peak being around 500 nm means approximately 44% is in visible, about 50% in infrared, and 6% in ultraviolet.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Blackbody Radiation
Blackbody radiation is an essential concept in understanding how objects emit energy. A blackbody is an idealized physical object that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence.

When in thermal equilibrium, such an object re-emits energy as a spectrum of light. This emitted spectrum depends solely on the object's temperature and not on its shape or composition.

This phenomenon helps us make sense of various natural and artificial sources of light. By applying concepts like Wien's Displacement Law, which calculates the wavelength at which this radiation is strongest, scientists can determine the dominant light frequency of various bodies, providing insight into their physical properties.
  • Absorbs all radiation
  • Emits based on temperature
  • Wien's Law helps calculate peak wavelength
Solar Emission
Solar emission refers to the wide range of electromagnetic radiation emitted by the sun. The sun acts as a massive blackbody radiator with a surface temperature estimated around 5800 Kelvin. This leads to a peak emission situated in the visible spectrum, approximately at 500 nm.

The sun's spectrum encompasses ultraviolet, visible, and infrared ranges. The majority of this energy is emitted in the visible spectrum, which is why we see sunlight as mainly white light. However, a significant part also spans into infrared frequencies and a smaller percentage into ultraviolet frequencies.
  • Peak emission at visible spectrum (~500 nm)
  • Emits UV, visible, and IR radiation
  • Sun seen as massive blackbody radiator
Infrared Spectrum
The infrared spectrum is a part of the electromagnetic spectrum that lies just beyond the visible light range. It ranges from roughly 700 nm to 1 mm in wavelength. This spectrum is crucial in many fields, from astronomy to climate science, as it helps understand the heat emitted by objects.

In everyday life, infrared radiation is most commonly associated with heat. For example, heated objects like a tungsten filament or even the human body primarily emit in the infrared region. This is because their temperatures correspond to peak wavelengths in this part of the spectrum.
  • Wavelengths from 700 nm to 1 mm
  • Related to heat emission
  • Significant in science and technology
Ultraviolet Spectrum
The ultraviolet (UV) spectrum lies beyond the visible light range on the shorter wavelength side, spanning from 10 nm to 400 nm. It is a region of great interest and utility, seen in medical applications and various technological fields.

Despite being a small fraction of solar emission, UV radiation significantly impacts life on Earth. It is responsible for phenomena like sunburns and contributes to the synthesis of vitamin D in living organisms.

UV radiation is further divided into UVA, UVB, and UVC, each with distinct effects and penetration abilities. UV stands as a testament to the complexity of solar emission, affecting biological systems and influencing technological innovations.
  • Wavelengths from 10 nm to 400 nm
  • Visible effects on health and environment
  • Further divided into UVA, UVB, and UVC

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Most popular questions from this chapter

An opaque surface, \(2 \mathrm{~m}\) by \(2 \mathrm{~m}\), is maintained at \(400 \mathrm{~K}\) and is simultaneously exponed to solar irradiation with \(G=1200 \mathrm{~W} / \mathrm{m}^{2}\). The surface is diffuse and its spectral absorptivity is \(\alpha_{n}=0,0.8,0\), and \(0.9\) for \(0 \leq \lambda \leq 0.5 \mu m, 0.5 \mu m<\lambda \leq 1 \mu m .1 \mu m<\lambda \leq\) \(2 \mu \mathrm{m}\), and \(\lambda>2 \mu \mathrm{m}\), respectively. Determine the absorbed irradiation, emissive power, radiosity, and net radiation heat transfer from the surface.

Consider an opaque horizontal plate that is well insulated on its back side. The irradiation on the plate is \(2500 \mathrm{~W} / \mathrm{m}^{2}\). of which \(500 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The plate is at \(227^{\circ} \mathrm{C}\) and has an emissive power of \(1200 \mathrm{~W} / \mathrm{m}^{2}\). Air at \(127^{\circ} \mathrm{C}\) flows over the plate with a heat transfer convection coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Detemine the emissivity, absorptivity, and madiosity of the plate. What is the net heat transfer rate per unit area?

A sphere \(\left(k=185 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \alpha=7.25 \times 10^{-3} \mathrm{~m}^{2} / \mathrm{s}\right)\) of 30 -mm diameter whose surface is diffuse and gray with an emissivity of \(0.8\) is placed in a large oven whose walls are of uniform temperature at \(600 \mathrm{~K}\). The temperature of the air in the oven is \(400 \mathrm{~K}\), and the convection heat transfer coefficient between the sphere and the oven air is \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the net heat transfer to the sphere when its temperature is \(300 \mathrm{~K}\). (b) What will be the steady-state temperature of the sphere? (c) How long will it take for the sphere, initially at \(300 \mathrm{~K}\), to come within \(20 \mathrm{~K}\) of the steady-state temperature? (d) For emissivities of \(0.2,0.4\), and \(0.8\), plot the elapsed time of part (c) as a function of the convection coefficient for \(10 \leq h \leq 25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A sphere is suspended in air in a dark room and mair tained at a uniform incandescent temperature. Whet firs viewed with the naked eye, the sphere appears s be brighter around the rim. After several hours, however, it appears to be brighter in the center. Of what type material would you reason the sphere is nude? Give plasible reasons for the nonunifornity of brightness of the sphere and for the changing appes ance with time.

The 50 -mm peephole of a large furnace orerating a \(450^{\circ} \mathrm{C}\) is covered with a material having \(r=0.8\) and \(\rho=0\) for irradiation originating from the furnace. Tle material has an cmissivity of \(0.8\) and is opaque to imsdiation from a source at room temperature. The outer surface of the cover is exposed to surroundings and ambient air at \(27^{\circ} \mathrm{C}\) with a convection heat transet coefficient of \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Assuming that convection effects on the inner surface of the cover are negligitle. calculate the heat loss by the furnace and the tempen: ture of the cover.
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