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A cylindrical log 15cm in diameter and 68cm long is glowing red hot in a fireplace. If it 's emitting radiation at the rate of 38kW, what is its temperature? the log's emissivity e=1.

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HomertheGenius
T = ( P / e S A ) ^(1/4)
P = 38 kW = 38,000 W
e = 1
S ( sigma ) = 5.67 * 10^(-8)
A = 2 r² π + 2 r π h = 2 * ( 0.075)² * 3.14 + 2 * 0.075 * 3.14 * 0.68
A = 0.355605 m² 
T = ( 38,000 / (1 * 5.67 * 10^(-8) * 0.355605 )) ^(1/4) =
= ( 3.8 * 10^(12) / 2.0162803 ) ^(1/4)=
= 1.172 * 10^3 = 1,172 K

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