The original temperature of a black body is 727- C- Calculate temperature at which total radiant energy from this black body becomes double-

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Standard XII

Physics

Blackbody

The original ...

Question

The original temperature of a black body is $727^{\circ} C$ . Calculate temperature at which total radiant energy from this black body becomes double:

971 K

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1189 K

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2001 K

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1458 K

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Solution

The correct option is B $1189 K$
$R e d i a n t E n e r g y = σ T^{2}$
$E n e r g y = σ (1000)^{4}$
$E_{2} = 2 E_{1}$
$T h e n$
$σ T_{2}^{4} = 2 \times σ (1000)^{4}$
$T_{2} = 2^{\frac{1}{4}} \times 1000$
$T_{2} = 1189 K$

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The original temperature of a black body is 727∘C. Calculate temperature at which total radiant energy from this black body becomes double:

The correct option is B 1189KRediantEnergy=σT2Energy=σ(1000)4E2=2E1ThenσT42=2×σ(1000)4T2=214×1000T2=1189K

The original temperature of a black body is $727^{\circ} C$ . Calculate temperature at which total radiant energy from this black body becomes double:

The correct option is B $1189 K$
$R e d i a n t E n e r g y = σ T^{2}$
$E n e r g y = σ (1000)^{4}$
$E_{2} = 2 E_{1}$
$T h e n$
$σ T_{2}^{4} = 2 \times σ (1000)^{4}$
$T_{2} = 2^{\frac{1}{4}} \times 1000$
$T_{2} = 1189 K$