A black body at 127 $^{o}$ C emits the energy at the rate of 10 $^{6}$ J/m $^{2}$ s. The temperature of a black body at which the rate of energy emission is 16x10 $^{6}$ J/m $^{2}$ s is :

508^{o} C

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273^{o} C

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400^{o} C

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527^{o} C

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Solution

The correct option is D $527^{o} C$
Using Stefan's Law:
$\frac{E_{2}}{E_{1}} = {(\frac{T_{2}}{T_{1}})}^{4} T_{2} = T_{1} \sqrt[4]{\frac{E_{2}}{E_{1}}} T_{2} = (127 + 273) \sqrt[4]{\frac{16 \times 10^{6}}{10^{6}}} = 800 K T_{2} = 527^{\circ} C$

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Similar questions

Q. Energy is being emitted from the surface of black body at

127^{o} C

at the rate of

1.0 \times 10^{6} J / s m^{2}

. The temperature of black body at which the rate of energy is

16.0 \times 10^{6} J / s m^{2}

will be :

Energy is being emitted from the surface of a black body at $127^{o} C$ temperature at the rate of $1.0 \times 10^{6} J / s e c - m^{2}$ . Temperature of the black body at which the rate of energy emission is $16.0 \times 10^{6} J / s e c - m^{2}$ will be