Q. The temperature of a piece of iron is $27℃$ and it is radiating energy at the rate of $Q\ kWm^{- 2}$. If its temperature is raised to $151℃$, the rate of radiation of energy will become approximately
A) $2Q\ kWm^{- 2}$
B) $4Q\ kWm^{- 2}$
C) $6Q\ kWm^{- 2}$
D) $8Q\ kWm^{- 2}$
Explanation
(b) $E_{2} = E_{1}\frac{T_{2}^{4}}{T_{1}^{4}} = Q \times \frac{(273 + 151)^{4}}{(273 + 27)^{4}} = \left( \frac{424}{300} \right)^{4} = 3.99Q = 4Q$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.