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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.

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प्रश्न

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 σ to be 5.7 × 10-8 Js-1 m-2K-4.)

संख्यात्मक
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उत्तर

`"Q"/"t" = 20`W, T = 273 + 727 = 1000 K

σ = 5.7 × 10-8 Js-1 m-2K-4.

`"Q"/"t" = sigma"AT"^4`

∴ The area of the hole,

A = `("Q"//"t")/(sigma"T"^4) = 20/((5.7 xx 10^-8)(10^3)^4)` m2

`= (20xx10^-4)/5.7 = 3.509 xx 10^-4` m2 

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पाठ 3: Kinetic Theory of Gases and Radiation - Exercises [पृष्ठ ७४]

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बालभारती Physics [English] Standard 12 Maharashtra State Board
पाठ 3 Kinetic Theory of Gases and Radiation
Exercises | Q 20 | पृष्ठ ७४

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