Question

Energy is emitted from a hole in an electric
furnace at the rate of 20 W, when the
temperature of the furnace is 727 °C. What is
the area of the hole? (Take Stefan's constant
o to be 5.7 x 10-8 J s'm²K4)​

Answer 1

Answer:

350877cm^2=A

Explanation :

The rate of heat transfer by emitted radiation is determined by the Stefan-Boltzmann law of radiation: Q=σeAT4 Q = σ e A T 4 , where σ = 5.67 × 10−8 J/s · m2 · K4 is the Stefan-Boltzmann constant, A is the surface area of the object, and T is its absolute temperature in kelvin.

Q = σ e A T^4

20=5.7 x 10-8×A×(723+273)

20=5.7 x 10-8×A×(1000)

350877cm^2=A

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