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
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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