The rate of radiation of a black body at 0 $^{0}$ C is E watt. Then the rate of radiation of this black body at 273 $^{0}$ C will be :

8 E

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16 E

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4 E

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Solution

The correct option is D 16 E
$E$ $α$ $T^{4}$
Given: $T_{1} = 0^{0} C = 273^{0} K$
Given: $T_{2} = 273^{0} C = 546^{0} K$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}} = (\frac{273}{546})^{4} = \frac{1}{16}$
$\Rightarrow E_{2} = 16 E_{1}$
$\Rightarrow E_{2} = 16 E$

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The rate of radiation of a black body at 0 0C is E watt. Then the rate of radiation of this black body at 273 0C will be :

The correct option is D 16 EE α T4Given: T1=00C=2730KGiven:T2=2730C=5460KE1E2=T41T42=(273546)4=116⇒E2=16E1⇒E2=16E

The rate of radiation of a black body at 0 $^{0}$ C is E watt. Then the rate of radiation of this black body at 273 $^{0}$ C will be :

The correct option is D 16 E
$E$ $α$ $T^{4}$
Given: $T_{1} = 0^{0} C = 273^{0} K$
Given: $T_{2} = 273^{0} C = 546^{0} K$
$\frac{E_{1}}{E_{2}} = \frac{T_{1}^{4}}{T_{2}^{4}} = (\frac{273}{546})^{4} = \frac{1}{16}$
$\Rightarrow E_{2} = 16 E_{1}$
$\Rightarrow E_{2} = 16 E$