Q. An object is at a temperature of $400℃$. At what temperature would it radiate energy twice as fast? The temperature of the surroundings may be assumed to be negligible
A) $200℃$
B) $200\ K$
C) $800℃$
D) $800\ K$
Explanation
(d) $\frac{E_{2}}{E_{1}} = \left( \frac{T_{2}}{T_{1}} \right)^{4} \Rightarrow \frac{2}{1} = \left( \frac{400 + 273}{T} \right)^{4} = \left( \frac{673}{T} \right)^{4}$
$\Rightarrow T = 2^{1/4} \times 673 = 800\ K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.