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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.

Verified Answer :

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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