Absorption and Emission help

So we can set the equations from parts 1 and 2 equal to each other and solve for the equilibrium temperature:σA(T1^4 - T2^4) = σT^4Solving for T, we get:T = (T1^4 + T2^4) / 2 = (300^4 + 300^4) / 2 = 300KSo the equilibrium temperature of the slab is also 300K.In summary, we can use the Stefan-Boltzmann law and the principle of energy conservation to solve for the rate of heat transfer and the equilibrium temperature of a slab of matter contained between two infinite planes.
  • #1
test2k8
5
0
Consider a slab of matter contained within two infinite planes a distance of s = 1m apart.

absorption coefficent =1m2kg-1
density of slab 1kgm-3
heat capacity 103^3jk-1kg-1

assume the absorption coefficient to be independent of wave length

1.Suppose one face of a slab is illuminated by a source emitting blackbody radiation at temp 300k, compute the rate at which the slab would tend to warm up by absorption of incident radiance.

2. assuming the slab is also at temp of 300k compute the irradiance emitted by the slab. Compute the rate at which the slab would tend to cool by emission.

3. assuming that the incident black body radiation remains unchanged in time, compute the temp at which the slab will come into equilibrium.


given equation:
emissivity at given wavelength = Rλ / Bλ

My attempt to show with a little confusing. I believe we have to use the Kirchhoff's Law to solve this problem. From my understand..

Rλ = Emissivity*Bλ = aλ*Iλ

Rλ = emitted spectral radiance
Bλ = black body
aλ = absorptance
Iλ = incident radiance..

therefore, to find 1)

Iλ = Emissivity*Bλ / aλ??

for 2)...
Since emissivity is given by Rλ / Bλ. Therefore the actual radiance emmitted by the slab is just ..

Rλ = Bλ * Emissivity.

3) Question 3 I'm not sure how to do it..

Can you help?
 
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  • #2


Sure, I'd be happy to help you with this problem. Let's break it down step by step and solve each part separately.

1. To compute the rate at which the slab would warm up, we can use the Stefan-Boltzmann law, which states that the rate of heat transfer by radiation is proportional to the fourth power of the temperature difference between the two objects. In this case, we have a blackbody source at 300K and a slab at an initial temperature of 300K as well. So the rate of heat transfer by radiation can be calculated as:

Q = σA(T1^4 - T2^4)

Where Q is the rate of heat transfer, σ is the Stefan-Boltzmann constant (5.67x10^-8 W/m^2K^4), A is the surface area of the slab (which we can assume to be 1m^2), T1 is the temperature of the blackbody source (300K), and T2 is the initial temperature of the slab (also 300K).

Plugging in these values, we get:

Q = 5.67x10^-8 * 1 * (300^4 - 300^4) = 0

This means that there is no net heat transfer between the blackbody source and the slab, so the slab will not warm up.

2. To compute the irradiance emitted by the slab, we can use the Stefan-Boltzmann law again, but this time with the temperature of the slab (300K) as the only variable:

E = σT^4 = 5.67x10^-8 * 300^4 = 153 W/m^2

This is the rate at which the slab emits radiation. To compute the rate at which the slab cools, we can use the same formula as in part 1, but with the temperature difference between the slab and its surroundings (which we can assume to be at 300K as well):

Q = 5.67x10^-8 * 1 * (300^4 - 300^4) = 0

Again, there is no net heat transfer between the slab and its surroundings, so the slab will not cool down.

3. To find the equilibrium temperature of the slab, we can use the principle of energy conservation. At equilibrium, the rate of heat transfer by absorption must be equal to the rate of heat transfer
 

What is the difference between absorption and emission?

Absorption is when an object or substance takes in and retains energy from a source, while emission is when an object or substance releases energy in the form of light or heat. In other words, absorption is the intake of energy, and emission is the release of energy.

How do absorption and emission relate to the color of objects?

The color of an object is determined by the wavelengths of light that it reflects. Objects that absorb all wavelengths of light appear black, while objects that reflect all wavelengths appear white. Objects that absorb some wavelengths and reflect others will appear as different colors. For example, a red object absorbs all wavelengths of light except for those in the red spectrum, which it reflects, making it appear red.

What factors affect absorption and emission?

The material, temperature, and energy source all affect the absorption and emission of energy. Different materials have different properties that make them better at absorbing or emitting energy. Temperature also plays a role, as higher temperatures typically result in higher energy absorption and emission. The type and intensity of the energy source can also impact absorption and emission.

How are absorption and emission used in everyday life?

Absorption and emission are used in many everyday objects and processes. For example, solar panels absorb sunlight and convert it into electricity, light bulbs emit light when electricity is passed through them, and plants absorb sunlight to perform photosynthesis. These are just a few examples of how absorption and emission are essential for our daily lives.

What are some real-world applications of absorption and emission?

There are countless real-world applications of absorption and emission, ranging from medical imaging techniques to environmental monitoring. Some specific examples include using infrared absorption to detect gas leaks, using UV light emission to disinfect surfaces, and using fluorescence in medical imaging to highlight specific cells or tissues. Additionally, absorption and emission are crucial in understanding and studying the behavior of atoms and molecules in different states of matter.

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