Answer:
[tex]Q=81.56\ W/m^2[/tex]
Explanation:
Given that
[tex]T_1= 210 K[/tex]
[tex]T_2= 150 K[/tex]
Emissivity of surfaces(∈) = 1
We know that heat transfer between two surfaces due to radiation ,when both surfaces are black bodies
[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]
So now by putting the values
[tex]Q=\sigma (T_1^4-T_2^4)\ W/m^2[/tex]
[tex]Q=5.67\times 10^{-8}(210^4-150^4)\ W/m^2[/tex]
[tex]Q=81.56\ W/m^2[/tex]
So rate of heat transfer per unit area
[tex]Q=81.56\ W/m^2[/tex]