Q. The top of insulated cylindrical container is covered by a disc having emissivity 0.6 and thickness $1\ cm$. The temperature is maintained by circulating oil as shown in figure. If temperature of upper surface of disc is $127℃$ and temperature of surrounding is $27℃$, then the radiation loss to the surroundings will be (Take $\sigma = \frac{17}{3} \times 10^{- 8}W/m^{2}K^{4}$)
A) $595\ J/m^{2} \times s$
B) $595\ cal/m^{2} \times s$
C) $991.0\ J/m^{2} \times s$
D) $440\ J/m^{2} \times s$
Explanation
(a) Rate of heat loss per unit area due to radiation $i.e.$ emissive power $e = \varepsilon\sigma(T^{4} - T_{0}^{4})$
$= 0.6 \times \frac{17}{3} \times 10^{- 8} \times \lbrack(400)^{4} - (300)^{4}\rbrack$
$= 3.4 \times 10^{- 8} \times \left( 175 \times 10^{8} \right) = 3.4 \times 175 = 595\ J/m^{2} \times s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.