The rate of radiation of a black body at $0^{\circ} C$ is $E \text{ J}s^{-1}$. The rate of radiation of the black body at $273^{\circ} C$ will be

The energy spectrum of a black body has maximum energy at a wavelength of $\lambda_0$. Now, the temperature of the black body is increased such that the maximum energy is obtained at a wavelength of $3\lambda_0/4$. By what factor will the power radiated by the black body increase?

The temperature of a body is increased from $27\,^{\circ}C$ to $127\,^{\circ}C$. The radiation emitted by it increases by a factor of:

$A$ black body is at a temperature $300 K$. It emits energy at a rate,which is proportional to

Find the rate of radiation emitted per unit area per unit time by a black body at a temperature of $10^{3} \ K$ (in $J \ m^{-2} \ s^{-1}$).

The operating temperature and emissivity of a tungsten filament are $2000 \ K$ and $0.3$,respectively. The surface area of the filament is $A \ cm^2$. If the power of the lamp is $25 \ W$,find $A$. (Given: $\sigma = 5.67 \times 10^{-8} \ W/m^2K^4$)