Question

The rate of emission of radiation of ablack body at temperature ${27}^{o}C$ is $E_1$ . If its temperature is increased to ${327}^{o}C$ the rate of emission of radiation is $E_2.$ The relation between $E_1$ and $E_2$ is:

• A. $E_2=24E_1$
• B. $E_2=16E_1$
• C. $E_2=8E_1$
• D. $E_2=4E_1$

Answer 1

The correct option is B $E_2=16E_1$

In black body radiation

$\frac{d\theta }{dt}=\left(4\pi r^2\right)\sigma T^4$

If at $T={27}^{o}C=300K,\frac{d\theta }{dt}=E_1$

Then,

$E_1=\left(4\pi R^2\right)\sigma (300{)}^4$

If at $T={327}^{o}C=600k,\frac{d\theta }{dt}=E_2$

$E_2=\left(4\pi R^2\right)\sigma \left(2{)}^4\right(300{)}^4$

So, $\frac{E_2}{E_1}=(2{)}^4\frac{\left(4\pi R^2\sigma \right(300{)}^4}{4\pi R^2\sigma (300{)}^4}$

$E_2=(2{)}^4{E}_1$

$\Rightarrow E_{2}10E_1$

Option $B$ is correct