# Radiation - finding emissivity of a sphere

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

#### Attachments

• question.PNG
14.6 KB · Views: 39

Homework Helper
Gold Member
2022 Award
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
JoeyBob
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?

• 