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The human skin is "selective" when it comes to the absorption of the solar radiation that strikes it perpendicularly. The skin absorbs only 50 percent of the incident radiation with wavelengths between \(\lambda_{1}=0.517 \mu \mathrm{m}\) and \(\lambda_{2}=1.552 \mu \mathrm{m}\). The radiation with wavelengths shorter than \(\lambda_{1}\) and longer than \(\lambda_{2}\) is fully absorbed. The solar surface may be modeled as a blackbody with effective surface temperature of \(5800 \mathrm{~K}\). Calculate the fraction of the incident solar radiation that is absorbed by the human skin.

Short Answer

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Question: Calculate the fraction of the incident solar radiation absorbed by the human skin, given that the skin absorbs 100% of the radiation with wavelengths shorter than 0.517 µm and longer than 1.552 µm, and 50% of the radiation with wavelengths between 0.517 µm and 1.552 µm. The solar surface temperature is 5800 K. Solution: To find the fraction of the absorbed solar radiation, first calculate the energy emitted by the blackbody in the given wavelength range using Planck's radiation formula. Divide this range into three parts: wavelengths shorter than 0.517 µm, wavelengths between 0.517 µm and 1.552 µm, and wavelengths longer than 1.552 µm. Then, calculate the energy absorbed by the human skin by adding the energy absorbed in each wavelength range. Finally, divide the absorbed energy by the total energy emitted by the blackbody, and find the fraction of the absorbed energy.

Step by step solution

01

Calculate the energy emitted per unit wavelength range by the blackbody

For a blackbody, the energy emitted per unit wavelength range (\(E_\lambda\)) is given by Planck's radiation formula: \(E_\lambda (\lambda, T) = \frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1}\), where \(h\) is Planck's constant, \(c\) is the speed of light, \(k\) is Boltzmann's constant, and \(T\) is the temperature of the blackbody.
02

Calculate the total energy emitted by the blackbody in the wavelength range of interest

To find the total energy emitted by the blackbody in the wavelength range that includes \(\lambda_1\) and \(\lambda_2\), we have to integrate the energy emitted per unit wavelength range with respect to wavelength: \(\int_{0}^{\infty} E_\lambda (\lambda, T) d\lambda = \int_{0}^{\infty} \frac{2\pi hc^2}{\lambda^5} \frac{1}{e^{\frac{hc}{\lambda k T}} - 1} d\lambda\) However, we need to divide the integral into three parts: wavelengths shorter than \(\lambda_1\), wavelengths between \(\lambda_1\) and \(\lambda_2\), and wavelengths longer than \(\lambda_2\). 1) \(E_{\text{short}} = \int_{0}^{\lambda_1} E_\lambda (\lambda, T) d\lambda\)\ 2) \(E_{\text{middle}} = \int_{\lambda_1}^{\lambda_2} E_\lambda (\lambda, T) d\lambda\)\ 3) \(E_{\text{long}} = \int_{\lambda_2}^{\infty} E_\lambda (\lambda, T) d\lambda\)
03

Calculate the energy absorbed by the human skin

Now let's find out the energy that is absorbed by the human skin. As mentioned earlier, 100% of the radiation with wavelengths shorter than \(\lambda_1\) and longer than \(\lambda_2\) is absorbed, while only 50% of the radiation with wavelengths between \(\lambda_1\) and \(\lambda_2\) is absorbed. Therefore, the total absorbed energy is: \(E_{\text{absorbed}} = E_{\text{short}} + 0.5 \times E_{\text{middle}} + E_{\text{long}}\)
04

Calculate the fraction of the absorbed energy

Lastly, we need to find the fraction of the absorbed energy with respect to the total energy emitted by the blackbody. This can be calculated as: \(\text{Fraction of absorbed energy} = \frac{E_{\text{absorbed}}}{E_{\text{total}}}\),\ where \(E_{\text{total}} = E_{\text{short}} + E_{\text{middle}} + E_{\text{long}}\). By calculating the integrals and plugging in all the known values, we can find the fraction of the absorbed energy by the human skin.

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Key Concepts

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

blackbody radiation
Blackbody radiation refers to the type of electromagnetic radiation (light) that is emitted by any object that is called a "blackbody." A blackbody is an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. However, in this exercise, we use the blackbody concept to model the sun's effective surface temperature of 5800 K. This means that the sun is treated as though it emits radiation uniformly in all wavelengths, which is a crucial assumption for calculating the radiation absorbed by the human skin. This radiation has a spectrum and distribution defined by its temperature and peaks at a specific wavelength depending on its temperature. In our scenario, this assumption helps us estimate how much of the sun's energy in the form of solar radiation interacts and gets absorbed by human skin.
Planck's radiation formula
Planck's radiation formula is fundamental to understanding blackbody radiation. It provides the spectral distribution of radiation emitted by a blackbody at a certain temperature. The equation is given by: \[ E_\lambda(\lambda, T) = \frac{2\pi hc^2}{\lambda^5} \cdot \frac{1}{e^{\frac{hc}{\lambda k T}} - 1} \]Where:
  • \(E_\lambda\) is the energy per unit wavelength.
  • \(h\) is Planck's constant.
  • \(c\) is the speed of light.
  • \(k\) is Boltzmann's constant.
  • \(\lambda\) is the wavelength.
  • \(T\) is the absolute temperature of the blackbody.
This formula describes how the intensity of radiation depends on both the wavelength and temperature. The ability of this formula to predict the radiation distribution across different wavelengths allows us to simulate solar radiation and understand how much energy reaches and gets absorbed by the human skin.
wavelength integration
Integration across wavelengths is essential to determine the total energy emitted over a specified range of wavelengths. In the given problem, we perform integration over three different wavelength ranges to assess the amount of energy emitted and subsequently absorbed by the skin. These ranges are:
  • Wavelengths shorter than \(\lambda_1\).
  • Wavelengths between \(\lambda_1\) and \(\lambda_2\).
  • Wavelengths longer than \(\lambda_2\).
Each of these integrates to a different value due to the varying contributions of each range as described by Planck's formula. Understanding the absorption of solar radiation requires these integrations, as they account for the variations in energy across different parts of the electromagnetic spectrum. Integration helps determine \(E_{\text{short}}\), \(E_{\text{middle}}\), and \(E_{\text{long}}\), which are subsequently used to calculate how much of this energy is absorbed by the skin. Knowing the energy contributions from these ranges allows us to compute the fraction of absorbed solar radiation accurately.

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Most popular questions from this chapter

A 1-m-diameter spherical cavity is maintained at a uniform temperature of \(600 \mathrm{~K}\). Now a 5 -mm-diameter hole is drilled. Determine the maximum rate of radiation energy streaming through the hole. What would your answer be if the diameter of the cavity were \(3 \mathrm{~m}\) ?

Solar radiation is incident on the front surface of a thin plate with direct and diffuse components of 300 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), respectively. The direct radiation makes a \(30^{\circ}\) angle with the normal of the surface. The plate surfaces have a solar absorptivity of \(0.63\) and an emissivity of \(0.93\). The air temperature is \(5^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The effective sky temperature for the front surface is \(-33^{\circ} \mathrm{C}\) while the surrounding surfaces are at \(5^{\circ} \mathrm{C}\) for the back surface. Determine the equilibrium temperature of the plate.

Consider an opaque plate that is well insulated on the edges and it is heated at the bottom with an electric heater. The plate has an emissivity of \(0.67\), and is situated in an ambient surrounding temperature of \(7^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To maintain a surface temperature of \(80^{\circ} \mathrm{C}\), the electric heater supplies \(1000 \mathrm{~W} / \mathrm{m}^{2}\) of uniform heat flux to the plate. Determine the radiosity of the plate under these conditions.

Why do skiers get sunburned so easily?

A small circular surface of area \(A_{1}=2 \mathrm{~cm}^{2}\) located at the center of a 2-m-diameter sphere emits radiation as a blackbody at \(T_{1}=1000 \mathrm{~K}\). Determine the rate at which radiation energy is streaming through a \(D_{2}=1\)-cm-diameter hole located \((a)\) on top of the sphere directly above \(A_{1}\) and \((b)\) on the side of sphere such that the line that connects the centers of \(A_{1}\) and \(A_{2}\) makes \(45^{\circ}\) with surface \(A_{1}\).

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