Question

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 ${127}^{\circ }$ Cand temperature of surrounding is ${27}^{\circ }$ C, then the radiation loss to the surroundings will be (Take $\sigma =\frac{17}{3}\times {10}^{-8}W/m^2{K}^4$)

• A. $595J/m^2\times sec$
• B. $595cal/m^2\times sec$
• C. $991.0J/m^2\times sec$
• D. $440J/m^2\times sec$

Answer 1

The correct option is D $440J/m^2\times sec$

Rate of heat loss per unit area due to radiation i.e. emissive power $e=\epsilon \sigma (T^4-T_0^4)$
$=0.6\times \frac{17}{3}\times {10}^{-8}\times \left[\right(400{)}^4-\left(300{)}^4\right]$
$=3.4\times {10}^{-8}\times (175\times {10}^8)=3.4\times 175=595J/m^2\times sec$