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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

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

Grade:Upto college level

2 Answers

Saurabh Koranglekar
askIITians Faculty 10341 Points
one year ago
Dear student

The answer is option d

Regards
Vikas TU
14149 Points
one year 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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