Question

Two identical metal balls at temperature $200^{\circ }C$ and $400^{\circ }C$ kept in air at $27^{\circ }C$. The ratio of net heat loss by these bodies is

• A. 1/4
• B. 1/2
• C. 1/16
• D. $\frac{473^4-300^4}{673^4-300^4}$

Answer 1

Answer: (D)

If temperature of surrounding is considered, then net loss of energy of a body by radiation
$Q=A\epsilon \sigma \left(T^4-T_0^4\right)t$
$\Rightarrow Q\propto \left(T^4-T_0^4\right)$
$\Rightarrow \frac{Q_1}{Q_2}=\frac{T_1^4-T_0^4}{T_2^4-T_0^4}$
$=\frac{(273+200{)}^4-(273+27{)}^4}{(273+400{)}^4-(273+27{)}^4}$
$=\frac{(473{)}^4-(300{)}^4}{(673{)}^4-(300{)}^4}$