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 
Count Iblis,
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)