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Question Number 58286 by Umar last updated on 21/Apr/19

A small body with temperature θ and  absorbtivity τ is placed in a large   evaluated capacity whose interior   walls are at a temperature θ_w .  when    θ_w −θ    is  small, show that the  rate of heat transfer by radiation is       Q^• = 4θ_w ^3 Aτδ(θ_w −θ).

$$\mathrm{A}\:\mathrm{small}\:\mathrm{body}\:\mathrm{with}\:\mathrm{temperature}\:\theta\:\mathrm{and} \\ $$$$\mathrm{absorbtivity}\:\tau\:\mathrm{is}\:\mathrm{placed}\:\mathrm{in}\:\mathrm{a}\:\mathrm{large}\: \\ $$$$\mathrm{evaluated}\:\mathrm{capacity}\:\mathrm{whose}\:\mathrm{interior}\: \\ $$$$\mathrm{walls}\:\mathrm{are}\:\mathrm{at}\:\mathrm{a}\:\mathrm{temperature}\:\theta_{\mathrm{w}} . \\ $$$$\mathrm{when}\:\:\:\:\theta_{\mathrm{w}} −\theta\:\:\:\:\mathrm{is}\:\:\mathrm{small},\:\mathrm{show}\:\mathrm{that}\:\mathrm{the} \\ $$$$\mathrm{rate}\:\mathrm{of}\:\mathrm{heat}\:\mathrm{transfer}\:\mathrm{by}\:\mathrm{radiation}\:\mathrm{is} \\ $$$$\:\:\:\:\:\overset{\bullet} {\mathrm{Q}}=\:\mathrm{4}\theta_{\mathrm{w}} ^{\mathrm{3}} \mathrm{A}\tau\delta\left(\theta_{\mathrm{w}} −\theta\right). \\ $$

Commented by Umar last updated on 21/Apr/19

help please

$$\mathrm{help}\:\mathrm{please} \\ $$

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