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Can you think of a way to make a blackbody, a material that absorbs essentially all of the radiant energy falling in it, if you only have a material that reflects half the radiant energy that falls on it?

Short Answer

Expert verified
Answer: A blackbody can be created using a material that reflects half of the radiant energy by layering the material multiple times. Each layer will reflect half of the remaining radiant energy and absorb the other half. With each additional layer, the percentage of energy absorption increases, approaching 100% as the number of layers increases. By using a large number of layers, we can achieve an absorption rate close enough to 100%, creating an approximation of a blackbody.

Step by step solution

01

Understanding the properties of the given material

The given material reflects half of the radiant energy that falls on it. This means that it also absorbs the other half of the energy. In simple terms, when energy falls on the material, 50% is reflected and 50% is absorbed by the material.
02

Layering the material

One approach to increase the absorption of radiant energy is to layer the material. This method will involve placing multiple layers of the given material on top of each other. By doing so, each layer will reflect half of the remaining radiant energy and consequently absorb the other half.
03

Calculating energy absorption after multiple layers

Assume we have n layers of the material. - After the first layer, 50% of energy is absorbed and 50% is reflected. - The second layer now absorbs 50% of the reflected 50%, resulting in the absorption of 50% + 25% = 75% of the initial energy. - The third layer absorbs 50% of the remaining 25%, resulting in the absorption of 75% + 12.5% = 87.5% of the initial energy. With each additional layer, the percentage of energy absorption will increase, approaching 100% as the number of layers increases.
04

Infinite layers of material

Ideally, we need an infinite number of layers to create a perfect blackbody. In practice, using a large number of layers will be sufficient to achieve an absorption rate close enough to 100%. By layering the given material in this way, we can create an approximation of a blackbody, one that absorbs essentially all of the radiant energy falling on it.

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Key Concepts

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

Radiant Energy Absorption
Understanding radiant energy absorption is crucial when exploring how materials interact with energy in the form of light or electromagnetic radiation. This concept deals with how much of the incoming energy is taken up by a substance instead of being transmitted through or reflected away.

When radiant energy, such as sunlight, strikes any surface, some of the energy is absorbed, and the rest may be reflected or transmitted. Different materials have distinct levels of absorption based on their physical and chemical properties. A blackbody is an idealized concept for a material that absorbs all incident radiant energy and reflects none. It is the perfect absorber with an absorption rate of 100%, which means that all of the energy that lands on it is converted into other forms, often heat.

Key Applications in Technology and Science

  • Solar panel design for maximum energy absorption
  • Thermal insulation materials that reduce energy waste
  • Understanding the Earth's heat balance influenced by different surfaces
Energy Reflection and Absorption
The interplay between energy reflection and absorption determines how effective a material is at harnessing energy. Reflection is the process where energy bounces off the surface of a material, while absorption is when the energy is taken up and not re-emitted.

Materials that reflect a higher percentage of radiant energy are often less effective at absorbing it, since the energy must penetrate the material for absorption to occur. Conversely, dark, rough, or matte surfaces tend to absorb more energy because they reflect less. The balance between reflected and absorbed energy is critical in many practical applications, from architectural design to the creation of efficient heating systems.

Optimizing Energy Harnessing

  • Coatings and paints designed to reflect or absorb energy
  • Climate control fabrics that balance comfort with energy efficiency
  • Spacecraft materials that protect from extreme temperature changes
Layering Effect on Energy Absorption
The layering effect on energy absorption is a strategy to enhance the absorbance of a material by adding multiple layers. With each additional layer, a material that partially reflects and absorbs energy can be made more effective, as less energy is reflected back with each successive layer.

Think of each layer acting as a filter. The first layer takes in the first half of the energy, while the second layer deals with the energy that slips through the first. As we add more layers, each subsequent layer has less energy to reflect because the previous layers have already absorbed a portion of it. Therefore, by strategically layering, you can increase the total amount of radiant energy absorbed and come close to creating a near-perfect absorber, or artificial blackbody.

Impact in Real-life Applications

  • Developing thermal blankets that work in multiple environments
  • Improving energy capture in multi-layered solar cells
  • Enhancing audio quality with layered sound absorption materials

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Most popular questions from this chapter

A \(2.0 \cdot 10^{2}\) g piece of copper at a temperature of \(450 \mathrm{~K}\) and a \(1.0 \cdot 10^{2} \mathrm{~g}\) piece of aluminum at a temperature of \(2.0 \cdot 10^{2} \mathrm{~K}\) are dropped into an insulated bucket containing \(5.0 \cdot 10^{2} \mathrm{~g}\) of water at \(280 \mathrm{~K}\). What is the equilibrium temperature of the mixture?

For a class demonstration, your physics instructor pours \(1.00 \mathrm{~kg}\) of steam at \(100.0^{\circ} \mathrm{C}\) over \(4.00 \mathrm{~kg}\) of ice at \(0.00^{\circ} \mathrm{C}\) and allows the system to reach equilibrium. He is then going to measure the temperature of the system. While the system reaches equilibrium, you are given the latent heats of ice and steam and the specific heat of water: \(L_{\text {ice }}=3.33 \cdot 10^{5} \mathrm{~J} / \mathrm{kg}\), \(L_{\text {steam }}=2.26 \cdot 10^{6} \mathrm{~J} / \mathrm{kg}, c_{\text {water }}=4186 \mathrm{~J} /\left(\mathrm{kg}^{\circ} \mathrm{C}\right) .\) You are asked to calculate the final equilibrium temperature of the system. What value do you find?

A metal brick found in an excavation was sent to a testing lab for nondestructive identification. The lab weighed the sample brick and found its mass to be \(3.0 \mathrm{~kg} .\) The brick was heated to a temperature of \(3.0 \cdot 10^{2}{ }^{\circ} \mathrm{C}\) and dropped into an insulated copper calorimeter of mass 1.5 kg containing \(2.0 \mathrm{~kg}\) of water at \(2.0 \cdot 10^{1}{ }^{\circ} \mathrm{C} .\) The final temperature at equilibrium was noted to be \(31.7^{\circ} \mathrm{C}\). By calculating the specific heat of the sample from this data, can you identify the brick's material?

The human body transports heat from the interior tissues, at temperature \(37.0^{\circ} \mathrm{C},\) to the skin surface, at temperature \(27.0^{\circ} \mathrm{C},\) at a rate of \(100 . \mathrm{W}\). If the skin area is \(1.5 \mathrm{~m}^{2}\) and its thickness is \(3.0 \mathrm{~mm}\), what is the effective thermal conductivity, \(\kappa,\) of skin?

Knife blades are often made of hardened carbon steel. The hardening process is a heat treatment in which the blade is first heated to a temperature of \(1346^{\circ} \mathrm{F}\) and then cooled down rapidly by immersing it in a bath of water. To achieve the desired hardness, after heating to \(1346^{\circ} \mathrm{F}\), a blade needs to be brought to a temperature below \(5.00 \cdot 10^{2}{ }^{\circ} \mathrm{F}\). If the blade has a mass of \(0.500 \mathrm{~kg}\) and the water is in an open copper container of mass \(2.000 \mathrm{~kg}\) and sufficiently large volume, what is the minimum quantity of water that needs to be in the container for this hardening process to be successful? Assume the blade is not in direct mechanical (and thus thermal) contact with the container, and neglect cooling through radiation into the air. Assume no water boils but reaches \(100^{\circ} \mathrm{C} .\) The heat capacity of copper around room temperature is \(c_{\text {copper }}=386 \mathrm{~J} /(\mathrm{kg} \mathrm{K}) .\) Use the data in the table below for the heat capacity of carbon steel

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