A sphere, 0.30 m in radius, has a surface emissivity of 0.48 and is at a temperature of 600 K.

This Physical Question Drive me Crazy
A sphere, 0.30 m in radius, has a surface emissivity of 0.48 and is at a temperature of 600 K. The
sphere is surrounded by a concentric spherical shell whose inner surface has a radius of 0.90 m
and an emissivity of 1.00. The temperature of the shell is 400 K. The net heat current radiated,
including direction, in the space between the sphere and the shell, in kW, is closest to:
A) 8.3, outward , B) 18.8, outward, C) 4.0, outward, D) 6.5, inward, E) 10.8, inward

Would you please explain the answer in details and write the solution so I can study it and understand it.
thanks in advance.

From the Stefan-Boltzmann law, a graybody with emissivity ε and temperature T radiates energy at a rate of:

J = ε*σ*T^4

per unit surface area, where σ is the Stefan-Boltzman constant = 5.67*10^-8 W/(m^2 * K^4)

This gives the radiant flux (energy/unit area). The total energy radiated by a body is J*A, where A is the surface area of the body that's radiating.

The inner sphere then radiates at a rate of:

J*A = 0.48*σ*(600K)^4 * (4π*(0.3m)^2) = 3.99kW (outward)

The outer shell radiates at a rate of:

1.00*σ*(400K)^4 * (4π*(0.9m)^2) = 14.78kW (inward)

The net radiation flux is therefore inward (because the shell is radiating more energy/unit time than the inner sphere), and is given by:

14.78kW - 3.99kW = 10.8 kW (inward)

Just one formula is to be used here
Q = A*e*t*σ*T^4
Q = Heat, A=Area that is emitting heat, t=time, σ=Stefen-Boltzman constant=5.67*10^-8 W m−2 K−4, T= Temp of body

So for solid sphere
Q = (4*pi*0.30*0.30)*0.48*1*5.67*10^-8*(600^…
Q = 3.989 KJ = Power as time is taken as 1 sec = 3.389 KW

For shell
Q = (4*pi*0.9*0.9)*1*5.67*10^-8*(400^4)
Q =14.775 KJ = 14.775 KW

so clearly heat is flowing into solid sphere = 14.775-3.989 = 10.786 KW
So option E is correct.