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

925.04 J/s

Step-by-step 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 * 10^(-8)* 2\left ( 353^(4)-293^(4) \right )`

E = 925.04 J/s