Question

The rate of emission of radiation of a black body at ${273}^{\circ }C$ is $E$, then the rate of emission of radiation of this body at $0^{\circ }C$ will be

• A. $\frac{E}{16}$
• B. $\frac{E}{4}$
• C. $\frac{E}{8}$
• D. $0$

Answer 1

The correct option is A $\frac{E}{16}$

From Stefan Boltzmann law
power radiated at temperature $T$ is
$E_T=eA{T}^4$
When $T={273}^{\circ }C$
$E_{273}=eA(273+273{)}^4=E$ (given)
When $T=0^{\circ }C$
$E_0=eA(273+0{)}^4$
$\frac{E_0}{E}={\left(\frac{273}{546}\right)}^4=\frac{1}{16}$
$E_0=\frac{E}{16}$