Question

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

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

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

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

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

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

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

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.

Key concepts

Planck's Law

Planck's Law provides a fundamental tool for understanding the intensity of radiation emitted by a blackbody, which is an idealized object that absorbs all incident radiation. The law allows us to calculate how the intensity changes with different wavelengths and temperatures.
Here’s the formula for Planck's Law: \[I(\lambda, T) = \dfrac{2 \pi hc^2}{\lambda^5} \dfrac{1}{e^{\frac{hc}{\lambda kT}} - 1}\] Where:

• \(I(\lambda, T)\) is the intensity of radiation at wavelength \(\lambda\) and temperature \(T\).
• \(h\) represents Planck's constant, \(6.626 \times 10^{-34} \text{Js}\).
• \(c\) is the speed of light, \(3 \times 10^8 \text{m/s}\).
• \(k\) is the Boltzmann constant, \(1.381 \times 10^{-23} \text{JK}^{-1}\).

Planck’s Law is essential for calculating the spectral distribution of emission for any blackbody. This calculation is foundational in the field of thermal radiation, helping us understand how different temperatures affect radiative power.

Blackbody Radiation

Blackbody radiation refers to the emission of electromagnetic radiation by an idealized non-reflective object, known as a blackbody. In practice, blackbodies, which perfectly absorb and emit all radiation frequencies, are theoretical constructs but are useful ideals in physics. The sun and heated objects can be approximated as blackbody emitters.
Blackbody radiation helps us study real-world materials because it represents a baseline against which actual emissions can be compared. It follows that as a blackbody heats up, it emits more radiation and the peak of its emission shifts to shorter wavelengths.
This is formalized in key laws and principles including Planck's Law for spectral energy distribution, the Stefan-Boltzmann Law for total energy radiation, and Wien's Displacement Law which correlates peak wavelength with temperature. Each law facilitates our understanding of thermal radiation and is useful in applications such as calculating the transmitted radiation through a material.

Wavelength Integration

Wavelength integration is the process of summing the radiation intensities over a range of wavelengths to compute total radiative output. This method allows for more reliable calculations of energy transfer through various mediums.
In the case of our original exercise, to determine the total radiation passing through the glass, one must calculate the integral of the Planck’s Law equation over the defined spectral range from \(0.3 \mu m\) to \(3.0 \mu m\). This requires integrating: \[Q(T) = \int_{0.3 \times 10^{-6}}^{3 \times 10^{-6}} I(\lambda, T)d\lambda\] This integral effectively gives us the energy emitted at all wavelengths within the transmitting range of the glass. It represents a comprehensive evaluation of the energy, informing estimates that can be factored by additional parameters like transmittance and surface area for real-world applications.

Transmittance

Transmittance is the measure of how much light passes through a material. It is a percentage of incoming light at specific wavelengths that exit on the other side. Transmittance is crucial when considering materials' transparency and absorption qualities, especially in evaluating radiation.
In the specific problem, the glass window transmits 90% of the radiation within the given wavelength range, making it semi-transparent only in that region and opaque outside it. Practically, this means the glass will allow 90% of all radiation in these wavelengths emitted from blackbody sources at specific temperatures to pass through, while blocking the rest.
This parameter is vital for assessing the efficiency and effectiveness of the material in any light-based or radiation-intensive application. In contexts such as solar panels or greenhouse effects, understanding and leveraging transmittance can lead to more controlled environments and optimized energy use.

Thermal Radiation

Thermal radiation is the emission of electromagnetic waves generated by the thermal motion of charged particles in matter. It is a natural process that depends on the material's temperature, where higher temperatures typically result in more radiation over a broader range of wavelengths.
Thermal radiation plays a significant role in heat transfer, fundamentally influencing the thermal management strategies in engineering, climate control, and electronic devices.
It’s the principle underlying the Planck's Law, which leads to predicting the spectral composition and intensity of radiation from substances like blackbodies. Understanding thermal radiation allows engineers and scientists to design systems that either enhance or mitigate heat transfer for energy efficiency or environmental control. For instance, it's imperative in architectural designs to choose materials with appropriate thermal emissivities to ensure energy-efficient heating and cooling.