Question

Calculate the power per unit area (the exitance, W/m2) radiating from a blackbody at 77 K (liquid nitrogen temperature) and at 298 K (room temperature).

Short answer

The value of Planck distribution at 77k and 298k

$M=1.99w/m^2;M=447.13W/m^2\sqrt{2}$

Step-by-step solution

Step1:Define planck distribution:

Planck's law describes the unique and characteristic spectral distribution for electromagnetic radiation in thermodynamic equilibrium, when there is no net flow of matter or energy

Step 2:Calculating the power unit:

The existence is:$M=\sigma .T^4$

For$T=77K:M=5.6698.{10}^{-8}.10W/m^2{K}^4.{\left(77K\right)}^4$

$M=1.99W/m^2$

For$T=298k:M=5.6698.{10}^{-8}W/m^2{k}^4.{\left(298K\right)}^4$

$M=447.13W/m^2$