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A black body radiates 20J of energy at 227oC temperature.If temperature of the black body is changed to 727oC, then energy radiated will be
  • a)
    120 J
  • b)
    240 J
  • c)
    320 J
  • d)
    360 J
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
A black body radiates 20J of energy at 227oC temperature.If temperatur...
Explanation:

Stefan-Boltzmann law states that the energy radiated per unit time by a black body is directly proportional to the fourth power of its absolute temperature.

Mathematically, E ∝ T^4

Where E is the energy radiated and T is the absolute temperature.

Let's assume that the energy radiated by the black body at 227oC is E1 and the energy radiated at 727oC is E2.

So, we can write:

E1 ∝ T1^4

E2 ∝ T2^4

We know that the energy radiated at 227oC is 20 J.

So, E1 = 20 J and T1 = (227 + 273) K = 500 K

Now, we need to find the energy radiated at 727oC.

T2 = (727 + 273) K = 1000 K

Using the Stefan-Boltzmann law, we can write:

E2/E1 = (T2/T1)^4

E2/20 = (1000/500)^4

E2 = 20 x (2)^4

E2 = 20 x 16

E2 = 320 J

Therefore, the energy radiated by the black body at 727oC temperature will be 320 J. Therefore, option (c) is the correct answer.
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A black body radiates 20J of energy at 227oC temperature.If temperature of the black body is changed to 727oC, then energy radiated will bea)120 Jb)240 Jc)320 Jd)360 JCorrect answer is option 'C'. Can you explain this answer?
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