Question

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

• A. $727{}^{\circ }C$
• B. $527{}^{\circ }C$
• C. $508{}^{\circ }C$
• D. $254{}^{\circ }C$

Answer 1

Answer: (B)

Here, $T_1=127{}^{\circ }C=400K$
$E_2=16\times {10}^{6}J/s{m}^2$
$E_1=1\times {10}^{6}J/s{m}^2$
Using the relation. $\frac{E_2}{E_1}={\left(\frac{T_2}{T_1}\right)}^4$
$\frac{T_2}{T_1}=\frac{E_2}{E_1}={\left(\frac{16.0\times {10}^6}{1\times {10}^6}\right)}^{1/4}=2$
$T_2=2\times T_1=2\times 400=800K$
$T_2=527{}^{\circ }C$