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 $127^o C$ and temperature of surrounding is $27^o C$, then the radiation loss to the surroundings will be (Take $\sigma = \frac{{17}}{3} \times {10^{ - 8}}W/{m^2}{K^4})$
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(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 \,[{(400)^4} - {(300)^4}]$

$ = 3.4 \times {10^{ - 8}} \times (175 \times {10^8}) = 3.4 \times 175 = 595\,J/{m^2} \times \sec $

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