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

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