Ask Question
6 May, 07:25

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

+1
Answers (1)
  1. 6 May, 07:50
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “Find the emitted power per square meter of peak intensity for a 3000 k object that emits thermal radiation. ...” in 📗 Physics if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers