Q. The energy emitted per second by a black body at $27℃$ is $10\ J$. If the temperature of the black body is increased to $327℃$, the energy emitted per second will be
A) $20\ J$
B) $40\ J$
C) $80\ J$
D) $160\ J$
Explanation
(d) $(Q)_{Black\ body} = A\sigma T^{4}t \Rightarrow Q \propto T^{4}$
$\Rightarrow Q_{2} = Q_{1}\left( \frac{T_{2}}{T_{1}} \right)^{4} = 10\left( \frac{273 + 327}{273 + 27} \right)^{4} = 10\left( \frac{600}{300} \right)^{4} = 160J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.