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Question

If the temperature of a black body increases from 7C to 287C, then the rate of emission of radiation energy is:

A
8 times
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B
16 times
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C
2 times
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D
4 times
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Solution

The correct option is B 16 times
For black body radiation
E=σT4
[E is energy radiated per unit time per unit area, T is temperature of the body]
E2E1=(T2T1)4E2E1=(273+287273+7)4
=(560280)4=161E2=16E1

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