Question

A black body, initially at temperature T, cools to temperature (T/2) in time t in surrounding which is near absolute zero. It will cool further to a temperature (T/4) in additional time-
(A) 8t
(B) 7t
(C) 9t
(D) None

Answer 1

It will cool further to a temperature (T/4) in additional time to 8t.

Explanation:

As we know that rate of heat radiation by the object is given as

dQ/dt = σ e A T^4

Now we will have

- msdT / dt =  σ e A T^4

Therefore we have.

-∫ dT / T^4 = σ e A / ms∫ dt

- 1/3 (1/T^3 - 8 /T^3 ) = σ e A / ms ΔT

7 / 3T^3 =  σ e A / ms ΔT

Now again we need to find the further time to change its temperature from T/2 to T/4.

Thus,

- 1/3 (1/T^3 - 64/T^3) = σ e A / ms Δt'

56 / 3 T^3 = σ e A / ms Δt'

Now divide the above equations.

1/8 = Δt / Δt'

Δt' = 8 Δt

Thus it will cool further to a temperature (T/4) in additional time to 8t.

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