If the temperature of a black body increases from 7- C to 287- C- the rate of energy radiation increases by

From Stefan's law, E=σ T^4 ∴ E_1 = σ (273+7)^4 = 280^4.....(1) and E_2 = σ (273+287)^4 = 560^4.....(2) From (1) and (2), E_2/E_1 = ( 560/280 )^4 = 16 Hence, th ...

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Standard XIII

Physics

Stefan-Boltzman's Law

If the temper...

Question

If the temperature of a black body increases from $7^{\circ} C$ to $287^{\circ} C$ , the rate of energy radiation increases by

${(\frac{287}{7})}^{4}$

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Solution

The correct option is B
16

From Stefan's law, $E = σ T^{4}$
$∴ E_{1} = σ (273 + 7)^{4} = 280^{4}$ .....(1)
and $E_{2} = σ (273 + 287)^{4} = 560^{4}$ .....(2)
From (1) and (2), $\frac{E_{2}}{E_{1}} = {(\frac{560}{280})}^{4} = 16$
Hence, the correct choice is (b).

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If the temperature of a black body increases from 7∘C to 287∘C, the rate of energy radiation increases by

The correct option is B 16 From Stefan's law, E=σT4 ∴E1=σ(273+7)4=2804.....(1) and E2=σ(273+287)4=5604.....(2) From (1) and (2), E2E1=(560280)4=16 Hence, the correct choice is (b).

If the temperature of a black body increases from $7^{\circ} C$ to $287^{\circ} C$ , the rate of energy radiation increases by

The correct option is B
16

From Stefan's law, $E = σ T^{4}$
$∴ E_{1} = σ (273 + 7)^{4} = 280^{4}$ .....(1)
and $E_{2} = σ (273 + 287)^{4} = 560^{4}$ .....(2)
From (1) and (2), $\frac{E_{2}}{E_{1}} = {(\frac{560}{280})}^{4} = 16$
Hence, the correct choice is (b).