Question

A hot-water radiator has a surface temperatue of 80 o C and a surface area of 2 m2 . Treating it as a blackbody, find the net rate at which it will radiate energy to a 20 o C room (see problem in 1.36)

Answer 1

Answer:

925.04 J/s

Explanation:

T = 80 C = 80 + 273 = 353 K

To = 20 c = 20 + 273 = 293 K

A = 2 m^2

Use the formula for Stefan's law

Energy radiated per second

$E = \sigma  A \left ( T^{4}-T_{0}^{4} \right )$

$E = 5.67 \times 10^{-8}\times 2\left ( 353^{4}-293^{4} \right )$

E = 925.04 J/s