Q. A black body at a temperature of 227$℃$ radiates heat at the rate of 20 $cal\ m^{- 2}s^{- 1}$. When its temperature rises to 727$℃$, the rate of heat radiated will be
A) $40\ cal\ m^{- 2}s^{- 1}$
B) $160\ cal\ m^{- 2}s^{- 1}$
C) $320\ cal\ m^{- 2}s^{- 1}$
D) $640\ cal\ m^{- 2}s^{- 1}$
Explanation
(c) The Stefan’s law,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E = \sigma T^{4}$
Given, $T_{1} = 227℃ = 227 + 273 = 500$ K
$\ \ \ \ \ \ \ \ \ \ \ \ \ T_{2} = 727℃ = 273 + 727 = 1000$ K
$\therefore$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}}$
$\Rightarrow$ $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2} = \frac{T_{2}^{4}}{T_{2}^{4}}\ E_{1}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2} = \frac{(1000)^{4}}{(500)^{4}} \times 20$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2} = 16 \times 20$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ E_{2} = 320\ cal\ m^{- 2}s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.