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°C and temperature of surrounding is 27°C, then the radiation loss to the surroundings will be $(Take \sigma = \frac{17}{3} \times 10^{-8} \mathrm{W/m^{2}K^{4}})$
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Rate of heat loss per unit area due to radiation i.e. emissive power $e = \varepsilon \sigma \left( T^{4} - T_{0}^{4} \right)$
$= 0.6 \times \frac{17}{3} \times 10^{-8} \times [(400)^4 - (300)^4]$
$= 3.4 \times 10^{-8} \times (175 \times 10^{8}) = 3.4 \times 175 = 595]/m^{2} \times sec$
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