Question

say core answer only..​

Answer 1

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The loss of thermal energy at the rate of ‘8 × 10⁴ W.m²’ .

Stefan's Constant $\bf{(\sigma)}$ = $\bf\red{5.7\times{10^{-8}}\:W.m^{-2}.K^{-4}}$

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The temperature .

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✈︎ We know that,

$\\ {\color{aqua}\bigstar}\:\bf{\pink{\underbrace{\overbrace{\blue{Energy\: radiates\:per\:unit\:time\:\atop{per\:unit\:Area\:(E)\:}=\:\sigma\:T^4\:}}}}}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{E}{\sigma}\:}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{8\times{10^4}}{5.7\times{10^{-8}}}\:}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{8}{5.7}\times{10^{12}}\:}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{80}{57}\times{10^{12}}\:}$

$\\ \bf{:\implies\:T^4\:=\:1.4035\times{10^{12}}\:}$

$\\ \bf{:\implies\:T\:=\:\sqrt[4]{1.4035\times{10^{12}}}\:}$

$\\ \bf{:\implies\:T\:=\:1.0884\times{10^{3}}\:}$

$\\ \bf\green{:\implies\:T\:=\:1088.4\:K\:}$

$\huge\orange\therefore$ The temperature at which black body losses thermal energy at the rate of 8 × 10⁴ W.m² is ‘1088.4 K’ .

Answer 2

Answer:

$\\ \bf{:\implies\:T^4\:=\:\dfrac{E}{\sigma}\:}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{8\times{10^4}}{5.7\times{10^{-8}}}\:}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{8}{5.7}\times{10^{12}}\:}$

$\\ \bf{:\implies\:T^4\:=\:\dfrac{80}{57}\times{10^{12}}\:}$

$\\ \bf{:\implies\:T^4\:=\:1.4035\times{10^{12}}\:}$

$\\ \bf{:\implies\:T\:=\:\sqrt[4]{1.4035\times{10^{12}}}\:}$

$\\ \bf{:\implies\:T\:=\:1.0884\times{10^{3}}\:}$

$\\ \bf\blue{:\implies\:T\:=\:1088.4\:K\:}$