Question

The tempearture of an isolated black body falls from T₁ to
T₂ in time t, then t is (Let c be a constant)
(a)t = c (1/T₂ - 1/T₁)
(b)t = c (1/T₂² - 1/T₁²)
(c)t = c (1/T₂³ - 1/T₁³)
(d)t = c (1/T₂⁴ - 1/T₁⁴)

Answer 1

Answer:

By stefan-boltzmann law, rate of energy radiated by black body is given as: q=Aσ(T

4

)⇒msdT/dt=Aσ(T

4

), where T is the temperature of the body.

So, dt=C(dT/T

4

), where C is the constant of proportionality. Integrating above we get, ∫

0

t

dt=C∫

T

1

T

2

dT/T

4

⇒t=(−C/3)(1/T

3

)∣

T

1

T

2

⇒t=c(

T

2

3

1

−

T

1

3

1

) where c=−C/3

Explanation:

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Answer 2

Answer:

The temperature of an isolated black body falls from

to

in time

. Let

be a constant