Question

A black body, initially at temperature T, cools to temperature (T/2) in time Dt in surrounding which is near absolute zero. It will cool further to a temperature (T/4) in additional time

Answer 1

Answer:

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

Explanation:

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

$\frac{dQ}{dt} = \sigma e A T^4$

now we will have

$-ms\frac{dT}{dt} = \sigma eAT^4$

so we have

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

$-\frac{1}{3}(\frac{1}{T^3} - \frac{8}{T^3}) = \frac{\sigma e A}{ms} \Delta t$

$\frac{7}{3T^3} = \frac{\sigma e A}{ms} \Delta t$

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

so we have

$-\frac{1}{3}(\frac{8}{T^3} - \frac{64}{T^3}) = \frac{\sigma e A}{ms} \Delta t'$

$\frac{56}{3T^3} = \frac{\sigma e A}{ms} \Delta t'$

now divide above two equations

$\frac{1}{8} = \frac{\Delta t}{\Delta t'}$

so we have

$\Delta t' = 8 \Delta t$

#Learn

Topic : Thermal Radiation

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