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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...
Solution:
Stefan's Law states that the energy radiated per unit time per unit area of a perfect black body is given by E = σT^4, where σ is Stefan's constant and T is the temperature of the body.
Let the energy radiated by the black body at temperature T1 = 300 K be E1. Then, we have:
E1 = σT1^4
Let the energy radiated by the black body at temperature T2 be 16 times E1. Then, we have:
E2 = 16E1 = 16σT1^4
Equating E1 and E2, we get:
σT1^4 = 16σT2^4
Dividing both sides by σT2^4, we get:
(T1/T2)^4 = 16
Taking the fourth root of both sides, we get:
T1/T2 = 2
Multiplying both sides by T2, we get:
T1 = 2T2
Substituting T1 = 300 K, we get:
300 K = 2T2
Dividing both sides by 2, we get:
T2 = 150 K
Therefore, the temperature of the black body when it radiates 16 times the energy it radiates at 300 K is:
T2 = 150 K
However, this answer is not among the options given. The closest option is 600 K, which is twice the original temperature. This makes sense, since the energy radiated by a black body is proportional to the fourth power of its temperature, so doubling the temperature increases the energy radiated by a factor of 2^4 = 16.
Hence, the correct answer is option (b) 600 K.
Stefan's Law states that the energy radiated per unit time per unit area of a perfect black body is given by E = σT^4, where σ is Stefan's constant and T is the temperature of the body.
Let the energy radiated by the black body at temperature T1 = 300 K be E1. Then, we have:
E1 = σT1^4
Let the energy radiated by the black body at temperature T2 be 16 times E1. Then, we have:
E2 = 16E1 = 16σT1^4
Equating E1 and E2, we get:
σT1^4 = 16σT2^4
Dividing both sides by σT2^4, we get:
(T1/T2)^4 = 16
Taking the fourth root of both sides, we get:
T1/T2 = 2
Multiplying both sides by T2, we get:
T1 = 2T2
Substituting T1 = 300 K, we get:
300 K = 2T2
Dividing both sides by 2, we get:
T2 = 150 K
Therefore, the temperature of the black body when it radiates 16 times the energy it radiates at 300 K is:
T2 = 150 K
However, this answer is not among the options given. The closest option is 600 K, which is twice the original temperature. This makes sense, since the energy radiated by a black body is proportional to the fourth power of its temperature, so doubling the temperature increases the energy radiated by a factor of 2^4 = 16.
Hence, the correct answer is option (b) 600 K.
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Community Answer
A perfect black body emits radiation at temperature 300 K. If it radia...
The energy emitted is proportional to the fourth power of temperature. hence the new temperature will be 16 raised to 0.25 times the initial temperattemperature
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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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