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A perfect black body emits radiation at temperature 300 K. If it radiates 16 times this amount, then its temperature will be
- a)300 K
- b)600 K
- c)1200 K
- d)4800 K
Correct answer is option 'B'. Can you explain this answer?
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A perfect black body emits radiation at temperature 300 K. If it radia...
Black Body Radiation and Stefan's Law
Black body radiation is the radiation emitted by a perfect black body at a given temperature. Stefan's Law states that the total energy radiated per unit time per unit area by a black body is proportional to the fourth power of its absolute temperature. Mathematically, it can be written as:
E = σT^4
where E is the energy radiated per unit time per unit area (also known as the radiant emittance), σ is the Stefan-Boltzmann constant, and T is the absolute temperature of the black body.
Solution
Given: A perfect black body emits radiation at temperature 300 K. If it radiates 16 times this amount, then its temperature will be.
Let the initial energy radiated by the black body be E1. Then, according to Stefan's Law:
E1 = σT1^4
where T1 = 300 K.
The black body now radiates 16 times this amount, which means the new energy radiated is 16E1. According to Stefan's Law:
16E1 = σT2^4
where T2 is the new temperature of the black body.
Dividing the second equation by the first, we get:
16E1/E1 = σT2^4/σT1^4
16 = (T2/T1)^4
Taking the fourth root of both sides, we get:
(T2/T1) = 2
T2 = 2T1
Substituting the value of T1 = 300 K, we get:
T2 = 600 K
Therefore, the temperature of the black body after radiating 16 times the initial amount is 600 K.
Answer: b) 600 K
Black body radiation is the radiation emitted by a perfect black body at a given temperature. Stefan's Law states that the total energy radiated per unit time per unit area by a black body is proportional to the fourth power of its absolute temperature. Mathematically, it can be written as:
E = σT^4
where E is the energy radiated per unit time per unit area (also known as the radiant emittance), σ is the Stefan-Boltzmann constant, and T is the absolute temperature of the black body.
Solution
Given: A perfect black body emits radiation at temperature 300 K. If it radiates 16 times this amount, then its temperature will be.
Let the initial energy radiated by the black body be E1. Then, according to Stefan's Law:
E1 = σT1^4
where T1 = 300 K.
The black body now radiates 16 times this amount, which means the new energy radiated is 16E1. According to Stefan's Law:
16E1 = σT2^4
where T2 is the new temperature of the black body.
Dividing the second equation by the first, we get:
16E1/E1 = σT2^4/σT1^4
16 = (T2/T1)^4
Taking the fourth root of both sides, we get:
(T2/T1) = 2
T2 = 2T1
Substituting the value of T1 = 300 K, we get:
T2 = 600 K
Therefore, the temperature of the black body after radiating 16 times the initial amount is 600 K.
Answer: b) 600 K
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A perfect black body emits radiation at temperature 300 K. If it radiates 16 times this amount, then its temperature will bea)300 Kb)600 Kc)1200 Kd)4800 KCorrect answer is option 'B'. Can you explain this answer?
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