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, the cooling is 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)

$\left(\frac{3}{16}\right)H$

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 400 K to 200 K then.

$H^{\prime }=\sigma \left[\right(600{)}^4-\left(400{)}^4\right]$

$\frac{H}{H^{\prime }}=\frac{\left[\right(600{)}^4-\left(200{)}^4\right]}{\left[\right(600{)}^4-\left(400{)}^4\right]}$

Using $a^4-b^4=(a^2-b^2)(a^2+b^2)$ we have

$\frac{H}{H^{\prime }}=\frac{\left[\right(600{)}^4-\left(200{)}^4\right]}{\left[\right(600{)}^4-\left(400{)}^4\right]}\times \frac{\left[\right(600{)}^4+\left(200{)}^4\right]}{\left[\right(600{)}^4+\left(400{)}^4\right]}$

$\frac{H}{H^{\prime }}=\frac{32}{12}\times \frac{40}{20}=\frac{16}{3}$

$H^{\prime }=\frac{3}{16}H$