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 cooloing 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)

From Stefans 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 Stefans constant. When sphere cools from 600 K to 200 K, energy 400 K to 200 K then.
$ H=\sigma [{{(600)}^{4}}-{{(400)}^{4}}] $
$ \frac{H}{H}=\frac{[{{(600)}^{4}}-{{(200)}^{4}}]}{[{{(600)}^{4}}-{{(400)}^{4}}]} $
Using $ {{a}^{4}}-{{b}^{4}}=({{a}^{2}}-{{b}^{2}})({{a}^{2}}+{{b}^{2}}), $
we have $ \frac{H}{H}=\frac{[{{(600)}^{2}}-{{(200)}^{2}}]}{[{{(600)}^{2}}-{{(400)}^{2}}]}\times \frac{[{{(600)}^{2}}+{{(200)}^{2}}]}{[{{(600)}^{2}}+{{(400)}^{2}}]} $
$ \frac{H}{H}=\frac{32}{12}\times \frac{40}{20}=\frac{16}{3} $
$ H=\frac{3}{16}H $