Question

A sphere at temperature $600\, K$ is placed in environment of temperature $200 \,K$. Its cooling rate is $H .$ If the temperature is reduced to $400\, K$, then the cooling in same environment will be :

• A. $ \frac{H}{16} $
• B. $ \left( \frac{9}{27} \right)H $
• C. $ \left( \frac{16}{3} \right)H $
• D. $ \left( \frac{3}{16} \right)H $

Answer 1

Answer: (D)

From Stefan's law, the total radiant energy emitted per second per unit surface area of a black body is proportional to the fourth power of the absolute temperature of the body.
That is $E=\sigma T^{4}$
Where $\sigma$ is Stefan's constant.
When sphere cools from $600\, K$ to $200 \,K$, energy radiated is
$H=\sigma\left[(600)^{4}-(200)^{4}\right]$
Let energy radiated be $H$ when cooled from $400\, K$ to $200 \,K$, then
$H'=\sigma\left[(600)^{4}-(400)^{4}\right] $
$\therefore \frac{H}{H^{\prime}}=\frac{\left[(600)^{4}-(200)^{4}\right]}{\left[(600)^{4}-(400)^{4}\right]}$
Using $a^{4}-b^{4}=\left(a^{2}-b^{2}\right)\left(a^{2}+b^{2}\right)$, we have
$\frac{H}{H'}=\frac{\left[(600)^{2}-(200)^{2}\right]}{\left[(600)^{2}-(400)^{2}\right]} \times \frac{\left[(600)^{2}+(200)^{2}\right]}{\left[(600)^{2}+(400)^{2}\right]} $
$\frac{H}{H'}=\frac{32}{12} \times \frac{40}{20}=\frac{16}{3} $
$\Rightarrow H'=\frac{3}{16} H$
Note: It is important to note that the rate of radiant energy emitted does not depend upon the shape, size etc. of body. It depends only on temperature.