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Standard XII

Physics

Stefan-Boltzman's Law

A small hole ...

Question

A small hole in a furnace acts like a black body. Its area is $1$ cm $^{2}$ , and its temperature is the same as that of the interior of the furnace, $1727$ $^{o}$ C. The energy radiated out of the hole per second is nearly :
(Stefans constant = $5.67 \times 10^{- 8} W m^{- 2} K^{- 4}$ )

79 J

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60 J

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91 J

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104 J

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Solution

The correct option is B 91 J
We know that radiation from the body is given as: $E = ϵ σ A T^{4}$
$E = 5.67 \times 10^{- 8} \times 0.0001 \times 2000^{4} = 90.72 \approx 91 w a t t$ (watt=joule/sec) here emissivity is constant.

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A small hole in a furnace acts like a black body. Its area is 1cm2, and its temperature is the same as that of the interior of the furnace, 1727oC. The energy radiated out of the hole per second is nearly : (Stefans constant = 5.67×10−8Wm−2K−4)

The correct option is B 91 JWe know that radiation from the body is given as: E=ϵσAT4E=5.67×10−8×0.0001×20004=90.72≈91watt (watt=joule/sec) here emissivity is constant.

A small hole in a furnace acts like a black body. Its area is $1$ cm $^{2}$ , and its temperature is the same as that of the interior of the furnace, $1727$ $^{o}$ C. The energy radiated out of the hole per second is nearly :
(Stefans constant = $5.67 \times 10^{- 8} W m^{- 2} K^{- 4}$ )

The correct option is B 91 J
We know that radiation from the body is given as: $E = ϵ σ A T^{4}$
$E = 5.67 \times 10^{- 8} \times 0.0001 \times 2000^{4} = 90.72 \approx 91 w a t t$ (watt=joule/sec) here emissivity is constant.