Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\frac{E_2}{E_1}=(\frac{T_2}{T_1}{)}^4\Rightarrow$
$T_2=(\frac{E_2}{E_1}{)}^{\frac{1}{4}}\times T_1=(16{)}^{\frac{1}{4}}\times (273+127)$
$\Rightarrow T_2=800K={527}^{o}C$