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

Question

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

Dr. Abubacker · Accepted Answer

(c) $\frac{E_{2}}{E_{1}} = \left( \frac{T_{2}}{T_{1}} ight)^{4} \Rightarrow T_{2} = \left( \frac{E_{2}}{E_{1}} ight)^{1/4} 	imes T_{1} = (16)^{1/4} 	imes (273 + 127)$
$\Rightarrow T_{2} = 800\ K = 527℃$

Q. Energy is being emitted from the surface of a black body at $127℃$ 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

Explanation

Reviewed by Dr. Abubacker

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