Question

A sphere at temperature $600K$ is placed in environment of temperature $200K$. Its cooling rate is $H.$ If the temperature is reduced to $400K$, 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 $600K$ to $200K$, energy radiated is
$H=\sigma \left[(600{)}^4-(200{)}^4\right]$
Let energy radiated be $H$ when cooled from $400K$ to $200K$, then
$H^{\prime }=\sigma \left[(600{)}^4-(400{)}^4\right]$
$\therefore \frac{H}{H^{\prime }}=\frac{\left[(600{)}^4-(200{)}^4\right]}{\left[(600{)}^4-(400{)}^4\right]}$
Using $a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)$, we have
$\frac{H}{H^{\prime }}=\frac{\left[(600{)}^2-(200{)}^2\right]}{\left[(600{)}^2-(400{)}^2\right]}\times \frac{\left[(600{)}^2+(200{)}^2\right]}{\left[(600{)}^2+(400{)}^2\right]}$
$\frac{H}{H^{\prime }}=\frac{32}{12}\times \frac{40}{20}=\frac{16}{3}$
$\Rightarrow H^{\prime }=\frac{3}{16}H$
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.