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Chapter 12: Problem 38

A 3-mm-thick glass window transmits 90 percent of the radiation between \(\lambda=0.3\) and \(3.0 \mu \mathrm{m}\) and is essentially opaque for radiation at other wavelengths. Determine the rate of radiation transmitted through a \(2-\mathrm{m} \times 2-\mathrm{m}\) glass window from blackbody sources at (a) \(5800 \mathrm{~K}\) and \((b) 1000 \mathrm{~K}\).

Short Answer

Expert verified
Question: Calculate and compare the rate of radiation transmitted through a glass window (2m x 2m) from two blackbody sources at temperatures 5800 K and 1000 K. The glass window transmits 90 percent of the radiation between wavelengths 0.3 to 3.0 µm, while the remaining wavelengths are opaque.

Step by step solution

01

Understand the problem and constants

First, let's list down the given information and the constants needed for this problem: - Window thickness = 3 mm - Transmittable radiation range: 0.3 to 3.0 µm (90% transmittance) - Window size = 2 m x 2 m - Two blackbody sources: 5800 K and 1000 K - Planck's constant (h) = \(6.626 \times 10^{-34} \text{Js}\) - Boltzmann constant (k) = \(1.381 \times 10^{-23} \text{JK}^{-1}\) - Speed of light (c) = \(3 \times 10^8 \text{m/s}\) Step 2: Planck's Law formula
02

Planck's Law formula

We'll use Planck's Law to calculate the intensity of radiation for each wavelength. The formula is: \(I(\lambda, T) = \dfrac{2 \pi hc^2}{\lambda^5} \dfrac{1}{e^{\frac{hc}{\lambda kT}} - 1}\) Here, I(λ, T) represents the intensity for a given wavelength λ and temperature T. Step 3: Integration over the transmittable wavelength range
03

Integration over the transmittable wavelength range

To find the total radiation transmitted through the window between the range of 0.3 to 3.0 µm, we need to integrate the intensity function: \(Q(T) = \int_{0.3 \times 10^{-6}}^{3 \times 10^{-6}} I(\lambda, T)d\lambda\) Step 4: Calculate transmitted radiation for both sources
04

Calculate transmitted radiation for both sources

We need to find the transmitted radiation for both sources (a) 5800 K and (b) 1000 K: \(Q_{5800} = \int_{0.3 \times 10^{-6}}^{3 \times 10^{-6}} I(\lambda, 5800)d\lambda\) \(Q_{1000} = \int_{0.3 \times 10^{-6}}^{3 \times 10^{-6}} I(\lambda, 1000)d\lambda\) Step 5: Consider the window transmittance and size
05

Consider the window transmittance and size

Since the window transmits 90% of the radiation between the given wavelength range, we need to find 90% of the radiation calculated in step 4. Furthermore, we need to multiply this by the window size (area) to find the rate of radiation transmitted through the window: \(Q_{final\,5800} = 0.9 \times Q_{5800} \times 2 \times 2\) \(Q_{final\, 1000} = 0.9 \times Q_{1000} \times 2 \times 2\) Step 6: Present final answers
06

Present final answers

Finally, we present the answer for the rate of radiation transmitted through the window from the blackbody sources at (a) 5800 K and (b) 1000 K: \(Q_{final\,5800} = \text{Rate of radiation transmitted through the window for a blackbody source at 5800 K}\) and \(Q_{final\, 1000} = \text{Rate of radiation transmitted through the window for a blackbody source at 1000 K}\) By following these steps, the student can calculate the rate of radiation transmitted through a glass window from blackbody sources at different temperatures.

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Most popular questions from this chapter

A furnace that has a \(40-\mathrm{cm} \times 40-\mathrm{cm}\) glass window can be considered to be a blackbody at \(1200 \mathrm{~K}\). If the transmissivity of the glass is \(0.7\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and zero for radiation at wavelengths greater than \(3 \mu \mathrm{m}\), determine the fraction and the rate of radiation coming from the furnace and transmitted through the window.

Consider a surface at \(-5^{\circ} \mathrm{C}\) in an environment at \(25^{\circ} \mathrm{C}\). The maximum rate of heat that can be emitted from this surface by radiation is (a) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(155 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(293 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(354 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(567 \mathrm{~W} / \mathrm{m}^{2}\)

What is the solar constant? How is it used to determine the effective surface temperature of the sun? How would the value of the solar constant change if the distance between the earth and the sun doubled?

What is a blackbody? Does a blackbody actually exist?

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 \(3 \mathrm{~m}\) ?
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