Energy is emitted from a hole in an electric furnace at the rate of 20 W, when the temperature of the furnace is 727 ºC. What is the area of the hole? (Take Stefan’s constant σ to be 5.7 ×10-8 J s-1 m-2 K-4)
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Given: Q/t = 20W, T=273+727=1000 K, σ = 5.7×10-8 Js-1/m2.K-4
Q/t = σAT4
∴ The area of the hole,
A = (Q/t)/(σT4 ) = 20(5.7×10-8 ×10004 ) = 20×10-4/5.7 m2
= 3.509×10-4 m2
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