Question

The rate of cooling at $600K$,if surrounding temperature is $300K$ is H. The rate of cooling at $900K$ is:

• A. $\frac{16}{3}H$
• B. $2H$
• C. $3H$
• D. $\frac{2}{3}H$

Answer 1

Answer: (A)

$\frac{16}{3}H$

By Stefan's law, Rate of cooling,

$H\propto (T^4-T_0^4)$

$\therefore \frac{H^{\prime }}{H}=\frac{(900{)}^4-(300{)}^4}{(600{)}^4-(300{)}^4}$

$=\frac{9^4-3^4}{6^4-3^4}=\frac{3^4(3^4-1)}{3^4(2^4-1)}$

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