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)
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Answered by
0
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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Answered by
0
Answer:
The temperature of an isolated black body falls from
to
in time
. Let
be a constant
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