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• radiasi kalor
Ra = 2 Rb
Ta = 27 °C = 300 K
Tb = 327 °C = 600 K
Pa = 10 W
Pb = __?
Daya radiasi
P = e σ T⁴ A
P = e σ T⁴ (πR²)
perbandingan
Pb / Pa
= [e σ Tb⁴ (π R b²)] / [e σ Ta⁴ (π R a²)]
Pb / Pa = (Tb / Ta)⁴ • (Rb / Ra)²
Pb / Pa = 2⁴ • (½)²
Pb / Pa = 16 • ¼
Pb / 10 = 4
Pb = 40 W ← jwb