# Radiation - finding emissivity of a sphere

• JoeyBob
In summary, the conversation discusses the use of an equation to calculate the net radiation of a sphere based on its surface area and temperature. The question of whether or not to include the surrounding air as a black body is brought up, but it is determined to be nonsensical as the air is not a black body and does not significantly contribute to the radiation received by the sphere. The concept of subtracting the absorbed radiation from the net radiation is also discussed, with the understanding that the absorbed radiation depends on the e value and not the temperature of the surrounding black body.
JoeyBob
Homework Statement
see attached
Relevant Equations
dQ/dt = Ae(5.67*10^-8)*T^4
So using the above equation, e=dQ/dt / (A*5.67E-8*303.8^4)

The surface area of a sphere is 4(pi)r^2 and I get 136.8478 m^2. dQ/dt would be the net radiation (I think? Its in the correct units), 1074W.

Plugging everything in I get 0.01625, but the answer is 0.0524.

Now as I was writing this I figured out how to do it by trial and error but am a bit confused. So apparently the surrounding air is decreasing the net radiation? Is that because its of a lower temperature? So if the air was hotter would I add it instead of subtracting it? Below is the equation I used to solve.

dQ/dt = SA * constant * e * (T of sphere) ^4 - SA * constant * e * (T of air) ^4

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The question setter seems to expect you to treat the surrounding air as a black body, so subtract the radiation the sphere absorbs from that. The fraction it absorbs will be governed by the same e value.
But it's nonsense. It takes a pretty thick layer of air to radiate much at all, and it certainly is not a black body (fortunately). Even outdoors, most of the received radiation would be from other solid bodies, such as the ground.

JoeyBob
haruspex said:
The question setter seems to expect you to treat the surrounding air as a black body, so subtract the radiation the sphere absorbs from that. The fraction it absorbs will be governed by the same e value.
But it's nonsense. It takes a pretty thick layer of air to radiate much at all, and it certainly is not a black body (fortunately). Even outdoors, most of the received radiation would be from other solid bodies, such as the ground.
Am I understanding correctly that I subtract the absorbed radiation because its of a lower temperature? In other words, would I add it instead if the surrounding black body had a higher temperature than the sphere?

JoeyBob said:
Am I understanding correctly that I subtract the absorbed radiation because its of a lower temperature? In other words, would I add it instead if the surrounding black body had a higher temperature than the sphere?
No, you are told the net radiation from the sphere, so that's radiation out minus radiation in. And 'in' should only be what is absorbed, so depends on e.

JoeyBob

## 1. What is radiation and how does it relate to finding the emissivity of a sphere?

Radiation is the emission or transmission of energy in the form of waves or particles. In the context of finding the emissivity of a sphere, radiation refers to the amount of energy emitted by the sphere compared to a perfect blackbody at the same temperature. The emissivity of a sphere is a measure of how well it can emit or absorb radiation.

## 2. How is the emissivity of a sphere calculated?

The emissivity of a sphere can be calculated using the Stefan-Boltzmann law, which states that the total energy emitted by a blackbody is proportional to the fourth power of its absolute temperature. The equation for calculating emissivity is: ε = E/EBB, where E is the energy emitted by the sphere and EBB is the energy emitted by a blackbody at the same temperature.

## 3. What factors affect the emissivity of a sphere?

The emissivity of a sphere is affected by several factors, including its material properties, surface roughness, and temperature. Materials with high thermal conductivity and low reflectivity tend to have higher emissivity. A rough surface also increases emissivity, as it allows for more energy to be emitted. Additionally, the emissivity of a sphere increases with temperature.

## 4. How is the emissivity of a sphere measured?

The emissivity of a sphere can be measured using a spectrometer or an infrared camera. These instruments measure the amount of radiation emitted by the sphere and compare it to the radiation emitted by a blackbody at the same temperature. The ratio of these two values gives the emissivity of the sphere.

## 5. What are some practical applications of knowing the emissivity of a sphere?

Knowing the emissivity of a sphere is important in various fields, such as materials science, industrial processes, and astronomy. It can be used to accurately measure the temperature of objects, monitor heat transfer, and design more efficient heating and cooling systems. In astronomy, the emissivity of celestial bodies can provide insights into their composition and physical properties.

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