Posted by sweety on .
A sphere of radius 0.583 m, temperature 39.4°C, and emissivity 0.983 is located in an environment of temperature 85.1°C. At what rate does the sphere (a) emit and (b) absorb thermal radiation? (c) What is the sphere's net rate of energy exchange?
thermal physics -
The area of the sphere is:
A = 4 pi R^2
where R is the radius
The emitted power is given by:
P_em = emissivity*A*sigma T1^4
where T1 is the absolute temperature of the sphere
The absorbed power is given by:
P_ab = emissivity*A*sigma T2^4
where T2 is the absolute temperature of the environment.
sigma is given by:
sigma = 2 pi^5 k^4/(15 h^3 c^2)
Here k is Boltzmann's constant, h is Planck's constant and c is the speed of light. sigma is approximately given by:
sigma = 5.67*10^(-8) Watt/(K^4 m^2)