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A perfectly black body emits radiation at temperature T1 K. if it sis to radiate 16 times this power, its temperature T2. will be (a) T2 = 16 T1 (b) T2 = 8 T1 (c) T1 = 4 T1 (d) T2 = 2 T1

sumit kumar , 11 Years ago
Grade Upto college level
anser 2 Answers
Saurabh Koranglekar

Last Activity: 4 Years ago

Dear student

The answer is option d

Regards

Vikas TU

Last Activity: 4 Years ago

The rate of loss of heat by radiation is given by the Stefen-boltzmann law as: Q=σT4A where σ is the stefen-boltzmann constant, A is the area of the radiating body. So Q∝T4. If T is doubled, Q becomes 24=16 times.

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