Question

Find the emitted power per square meter of peak intensity for a 3000 k object that emits thermal radiation.

Answer 1

Step-by-step explanation:

we can calculate total emitted power per square from Stefan-Boltzmann law

P = e σ AT⁴

where

P= power of body's thermal radiation

A= surface area of the body

σ= Stefan-Boltzmann constant it's value is
`5.67 *10^-8 W/m^2k^4`

T= temperature of the body

e= emissivity of the substance here equal to 1

now by putting all the values in given formula

`P/A= (1) (5.67*10^-8 W/m^2K^4)(3000)^4`

we get the answer

`P/A= 4.5*10^ 6 W/m^2`

Answer 2

According to Stefan-Boltzmann Law, the thermal energy radiated by a radiator per second per unit area is proportional to the fourth power of the absolute temperature. It is given by;
P/A = σ T⁴ j/m²s
Where; P is the power, A is the area in square Meters, T is temperature in kelvin and σ is the Stefan-Boltzmann constant, ( 5.67 × 10^-8 watt/m²K⁴)
Therefore;
Power/square meter = (5.67 × 10^-8) × (3000)⁴
= 4.59 × 10^6 Watts/square meter