The thermal radiative flux from a surface of emissivity = 0.4 is 22.68 kW/m2. The approximate surface temperature (K) is (Stefan-Boltzmann constant = 5.67 × 10-8 W/m2.K4) ?
Correct answer: A
Explanation
The formula is $q = ext{emissivity} imes ext{Stefan-Boltzmann constant} imes T^4$. Solving for T gives approximately 1000 K.