Question

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

• A. $754^{\circ }C$
• B. $527^{\circ }C$
• C. $254^{\circ }C$
• D. $508^{\circ }C$

Answer 1

Answer: (B)

$\frac{E_2}{E_1}=(\frac{T_2}{T_1}{)}^4$
$\frac{T_2}{T_1}={\left(\frac{E_2}{E_1}\right)}^{1/4}$
$={\left[\frac{16.0\times 10^6}{1.0\times 10^6}\right]}^{1/4}=2$
$T_2=2T_1=2\times T_1$
$=2\times 400$
$=800K=527^{\circ }C$