Question

Energy is being emitted from the surface of a black body at ${127}^{o}C$ 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. ${727}^{o}C$
• B. ${527}^{o}C$
• C. ${508}^{o}C$
• D. ${254}^{o}C$

Answer 1

The correct option is B ${527}^{o}C$

Here, $T_1={127}^{o}C=400K$

$E_2=16\times {10}^{6}J/s{m}^2$

$E_1=1\times {10}^{6}J/s{m}^2$

Using the relation,

$\frac{E_2}{E_1}={\left(\frac{T_2}{T_1}\right)}^4$

$\frac{T_2}{T_1}=\frac{E_2}{E_1}{\left(\frac{16.0\times {10}^6}{1\times {10}^6}\right)}^{1/4}=2$

$T_2=2\times T_1=2\times 400=800K$

$T_2={527}^{o}C$