Question-03 A rectangular block of surface area A emits energy E per second at 27^\circ C. If length and breadth is reduced to half of initial value and temperature is raised to 327^\circ C then energy emitted per second becomes
(A) 2E
(B) 4E
(C) E
(D) 8E
Answer : (B)
Explanation:
By Stefan’s law, energy radiated per second will be,
E=A\sigma T^4∴ \frac{E_2}{E_1}=\left(\frac{A_2}{A_1}\right)\left(\frac{T_2}{T_1}\right)^4=\left(\frac{\frac{l}{2}\times\frac{b}{2}}{l\times b}\right)\left(\frac{327+273}{27+273}\right)^4
=\left(\frac{1}{4}\right)\left(\frac{600}{300}\right)^4∴ \frac{E_2}{E_1}=\frac{1}{4}\times(16)
∴ E_2=4E ….(E_1=E)