The temperature of an isolated black body falls from T1 to T2 in time ‘t’. Let ‘c’ be a constant, then……

Q: The temperature of an isolated black body falls from T1 to T2 in time ‘t’. Let ‘c’ be a constant, then……

(a) $ \displaystyle t = c (\frac{1}{T_2}-\frac{1}{T_1} ) $

(b) $ \displaystyle t = c (\frac{1}{{T_2}^2}-\frac{1}{{T_1}^2} ) $

(c) $ \displaystyle t = c (\frac{1}{{T_2}^3}-\frac{1}{{T_1}^3} ) $

(d) $ \displaystyle t = c (\frac{1}{{T_2}^4}-\frac{1}{{T_1}^4} ) $

Ans: (c)

Sol: $ \displaystyle ms\frac{dT}{dt} = -\sigma A(T^4 – 0) $

$ \displaystyle ms\frac{dT}{dt} = -\sigma A T^4 $

$ \displaystyle ms\int_{T_1}^{T_2} \frac{dT}{T^4} = -\sigma A \int_{0}^{t}dt $

$ \displaystyle \frac{ms}{3} (\frac{1}{T_2^3 – \frac{1}{T_1^3}}) = \sigma A t $

$ \displaystyle t = \frac{ms}{3\sigma A} (\frac{1}{T_2^3 – \frac{1}{T_1^3}})$

$ \displaystyle t = c (\frac{1}{T_2^3 – \frac{1}{T_1^3}})$

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