Q. The original temperature of a black body is $727℃$. The temperature at which this black body must be raised so as to double the total radiant energy, is
A) $971\ K$
B) $1190\ K$
C) $2001\ K$
D) $1458\ K$
Explanation
(b) $\frac{Q_{2}}{Q_{1}} = \left( \frac{T_{2}}{T_{1}} \right)^{4} \Rightarrow \frac{2}{1} = \left( \frac{T_{2}}{T_{1}} \right)^{4}$
$\Rightarrow T_{2}^{4} = 2 \times T_{1}^{4} = 2 \times (273 + 727)^{4} \Rightarrow T_{2} = 1190K$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.