Question

The rectangular surface of area $6\text{cm}\times 2\text{cm}$ of a black body at a temperature of ${127}^{\circ }\text{C}$ emits energy at the rate of $E_1$ per second. If the length and breadth of the surface are each reduced to half of the initial value and the temperature is raised to ${327}^{\circ }\text{C}$, Find the rate of emission of energy.

• A. $\frac{81E_1}{49}$
• B. $\frac{81E_1}{64}$
• C. $\frac{49E_1}{64}$
• D. $\frac{64E_1}{81}$

Answer 1

The correct option is B $\frac{81E_1}{64}$

According to Stefan's law,
Energy radiated per second by the body :
$E=\frac{Q}{t}=A\sigma T^4[e=1]$
From the data given in the question, for a rectangular surface we can write that,
$\frac{E_2}{E_1}=\frac{l_2}{l_1}\times \frac{b_2}{b_1}\times {\left(\frac{T_2}{T_1}\right)}^4=\frac{\left(l_1/2\right)}{l_1}\times \frac{\left(b_1/2\right)}{b_1}\times {\left(\frac{600}{400}\right)}^4$
$\frac{E_2}{E_1}=\frac{1}{2}\times \frac{1}{2}\times {\left(\frac{3}{2}\right)}^4\Rightarrow E_2=\frac{81}{64}{E}_1$
Thus, option (b) is the correct answer.