Question #131523

A hot water radiator at 310k temperature radiates thermal radiates like a black body. Its total surface area is 1.6 m^2. Find the thermal power radiated by it.

Expert's answer

Stefan’s Law of Radiation (black body):

E=σAT4E = σAT^4

A is the surface area, m2

T is the surface temperature of the black body, K

Stefan’s constant σ=5.670×10−8W(m2⋅K4)\sigma = 5.670 \times 10^{−8} \frac {W}{( m^2 ⋅ K^4 )}

A=1.6m2A = 1.6 m^2

T=310KT = 310 K

E=5.670×10−8×1.6×3104=5.670×10−8×1.6×9.235×109=837.8WE = 5.670 \times 10^{−8}\times1.6\times 310^4 = 5.670 \times 10^{−8}\times 1.6\times 9.235 \times 10^{9} = 837.8 W

Answer: 837.8 W

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