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Chapter 12: Problem 39

Daylight and candlelight may be approximated as a blackbody at the effective surface temperatures of \(5800 \mathrm{~K}\) and \(1800 \mathrm{~K}\), respectively. Determine the radiation energy (in \(\mathrm{W} / \mathrm{m}^{2}\) ) emitted by both lighting sources (daylight and candlelight) within the visible light region \((0.40\) to \(0.76 \mu \mathrm{m})\).

Short Answer

Expert verified
Question: Determine the radiation energy emitted by daylight and candlelight within the visible light region (0.40 to 0.76 micrometers). Answer: To find the radiation energy emitted by daylight and candlelight within the visible light region, we need to use Planck's Law and Stefan-Boltzmann Law. First, calculate the total energy emitted by daylight and candlelight using the Stefan-Boltzmann Law. Next, calculate the ratio of energy emitted in the visible light region compared to the total energy by finding the integral of Planck's Law over the visible light range divided by total energy from Stefan-Boltzmann Law using numerical methods. Finally, multiply the calculated ratio with the total energy emitted by daylight and candlelight to find the radiation energy within the visible light region.

Step by step solution

01

Identify the Constants

First, we identify the constants provided in the exercise: - Effective surface temperature of daylight (T1): \(5800 \mathrm{~K}\) - Effective surface temperature of candlelight (T2): \(1800 \mathrm{~K}\) - Range of visible light: 0.40 to 0.76 \(\mu m\) (0.40 × 10⁻⁶ to 0.76 × 10⁻⁶ m)
02

Introduce Planck's Law

Planck's Law describes the intensity of radiation emitted by a blackbody at a particular temperature and wavelength. The equation is as follows: $$ I(\lambda, T) = \frac{2\pi h c^2}{\lambda^5} \cdot \frac{1}{e^{\frac{h c}{\lambda k_\mathrm{B} T}} - 1}, $$ where - \(I(\lambda, T)\): Radiation intensity at wavelength \(\lambda\) and temperature \(T\) - \(h\): Planck's constant (\(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s}\)) - \(c\): Speed of light (\(2.998 \times 10^{8} \mathrm{m/s}\)) - \(k_\mathrm{B}\): Boltzmann constant (\(1.381 \times 10^{-23} \mathrm{J/K}\)) - \(\lambda\): Wavelength - \(T\): Temperature
03

Introduce Stefan-Boltzmann Law

The Stefan-Boltzmann Law helps us calculate the total energy emitted by a blackbody across all wavelengths. The equation is as follows: $$ E(T) = \sigma \cdot T^4, $$ where - \(E(T)\): Total energy emitted by a blackbody at temperature \(T\) - \(\sigma\): Stefan-Boltzmann constant (\(5.670 \times 10^{-8} \mathrm{Wm^{-2}K^{-4}}\)) - \(T\): Temperature
04

Calculate Total Energy Emitted by Daylight and Candlelight

Using the Stefan-Boltzmann Law, we calculate the total energy emitted by daylight and candlelight: $$ E_{\text{daylight}} = \sigma \cdot T_1^4 = (5.670 \times 10^{-8}) \cdot (5800)^4 = 6.422 \times 10^{7} \mathrm{W/m^2} $$ $$ E_{\text{candlelight}} = \sigma \cdot T_2^4 = (5.670 \times 10^{-8}) \cdot (1800)^4 = 1.310 \times 10^{6} \mathrm{W/m^2} $$
05

Calculate the Ratio of Energy Emitted in Visible Light Region

To find the radiation energy emitted by daylight and candlelight within the visible light region, we need to know the ratio of energy emitted in the visible light region compared to the total energy. To find this ratio, we need the integral of Planck's Law over the visible light range divided by total energy from Stefan-Boltzmann Law: $$ \text{Ratio} = \frac{\int_{\lambda_1}^{\lambda_2} I(\lambda, T) d\lambda}{E(T)} $$ This integral does not have an analytical solution, so we need to use numerical methods like Simpson's rule, trapezoidal rule, or a relevant numerical library to calculate the integral.
06

Calculate the Radiation Energy in Visible Light Region

Multiply the calculated ratio with the total energy emitted by daylight and candlelight to find the radiation energy within the visible light region: $$ E_{\text{daylight-visible}} = \text{Ratio}_1 \times E_{\text{daylight}} $$ $$ E_{\text{candlelight-visible}} = \text{Ratio}_2 \times E_{\text{candlelight}} $$ Now we have the radiation energy emitted by daylight and candlelight within the visible light region, which are the answers to this exercise.

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Most popular questions from this chapter

Consider two identical bodies, one at \(1000 \mathrm{~K}\) and the other at \(1500 \mathrm{~K}\). Which body emits more radiation in the shorter-wavelength region? Which body emits more radiation at a wavelength of \(20 \mu \mathrm{m}\) ?

A small surface of area \(A=1 \mathrm{~cm}^{2}\) emits radiation as a blackbody at \(1800 \mathrm{~K}\). Determine the rate at which radiation energy is emitted through a band defined by \(0 \leq \phi \leq 2 \pi\) and \(45 \leq \theta \leq 60^{\circ}\), where \(\theta\) is the angle a radiation beam makes with the normal of the surface and \(\phi\) is the azimuth angle.

At a wavelength of \(0.7 \mu \mathrm{m}\), the black body emissive power is equal to \(10^{8} \mathrm{~W} / \mathrm{m}^{3}\). Determine \((a)\) the temperature of the blackbody and \((b)\) the total emissive power at this temperature.

Consider an opaque horizontal plate that is well insulated on the edges and the lower surface. The plate is uniformly irradiated from above while air at \(T_{\infty}=300 \mathrm{~K}\) flows over the surface providing a uniform convection heat transfer coefficient of \(40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Under steady state conditions the surface has a radiosity of \(4000 \mathrm{~W} / \mathrm{m}^{2}\), and the plate temperature is maintained uniformly at \(350 \mathrm{~K}\). If the total absorptivity of the plate is \(0.40\), determine \((a)\) the irradiation on the plate, \((b)\) the total reflectivity of the plate, \((c)\) the emissive power of the plate, and \((d)\) the total emissivity of the plate.

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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