Question

The energy radiated from a black body at a temperature of ${727}^{0}C$ is E. By what factor the radiated energy shall increase if the temperature is raised to ${2227}^{0}C$?

Answer 1

According to Stefan-Boltzmann law, the total energy radiated per unit area of a block body is directly proportional to the fourth power of the black body’s temperature T i.e.

$j=\sigma T^4$

Where, $\sigma$ is constant of proportionality called as Stefan–Boltzmann constant

Case 1.

$T_1={727}^{0}C=1000K$

So, $E=\sigma \times {\left(1000\right)}^4..............\left(1\right)$

Case 2

$T_2={2227}^{0}C=2500K$

$E^{{}^{\prime }}=\sigma \times {\left(2500\right)}^4.....\left(2\right)$

Taking ratio of (1) and (2)

$\frac{E}{E^{{}^{\prime }}}=\frac{\sigma \times {\left(1000\right)}^4}{\sigma \times {\left(2500\right)}^4}$

$\frac{E}{E^{{}^{\prime }}}=\frac{16}{625}$

$\frac{625E}{16}=E^{\prime }$

The energy increases by a factor of $\frac{625E}{16}$.