Energy is being emitted from the surface of a black body at $127\,^oC$ temperature at the rate of $1.0 \times {10^6}\,J/s - {m^2}$. Temperature of the black body at which the rate of energy emission is $16.0 \times {10^6}\,J/s - {m^2}$ will be ......... $^oC$
Medium
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$T_{1}=127^{\circ} C=400 K$
$E_{2}=16 \times 10^{6} J / s m^{2}$
$E_{1}=1 \times 10^{6} J / \mathrm{sm}^{2}$
Using the relation.
$\frac{E_{2}}{E_{1}}=\left(\frac{T_{2}}{T_{1}}\right)^{4}$
$\frac{T_{2}}{T_{1}}=\frac{E_{2}}{E_{1}}=\left(\frac{16.0 \times 10^{6}}{1 \times 10^{6}}\right)^{1 / 4}=2$
$T_{2}=2 \times T_{1}=2 \times 400=800 K$
$T_{2}=527^{\circ} \mathrm{C}$
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