A body at a temperature of $728^{o}$ C and has surface area $5 c m$ , radiates $300 J$ of energy each minute. The emissivity is
(Given Boltzmann constant $= 5.67 \times 10^{- 8} W m^{2} K^{4}$ )

e = 0.18

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e = 0.02

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e = 0.2

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e = 0.15

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Solution

The correct option is A $e = 0.18$
According to the Stefan-Boltzmann law,
$P = e σ A T^{4}$
$⟹ e = \frac{P}{A σ T^{4}}$
$= \frac{300 / 60}{0.0005 \times 5.67 \times 10^{- 8} \times (728 + 273)^{4}}$ $= 0.18$

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A body at a temperature of 728oC and has surface area 5cm, radiates 300J of energy each minute. The emissivity is(Given Boltzmann constant =5.67×10−8Wm2K4)

The correct option is A e=0.18According to the Stefan-Boltzmann law,P=eσAT4⟹e=PAσT4=300/600.0005×5.67×10−8×(728+273)4 =0.18

A body at a temperature of $728^{o}$ C and has surface area $5 c m$ , radiates $300 J$ of energy each minute. The emissivity is
(Given Boltzmann constant $= 5.67 \times 10^{- 8} W m^{2} K^{4}$ )

The correct option is A $e = 0.18$
According to the Stefan-Boltzmann law,
$P = e σ A T^{4}$
$⟹ e = \frac{P}{A σ T^{4}}$
$= \frac{300 / 60}{0.0005 \times 5.67 \times 10^{- 8} \times (728 + 273)^{4}}$ $= 0.18$