$A$ rectangular block of surface area $A$ emits energy $E$ per second at $27^{\circ} C$. If length and breadth are reduced to half of their initial values and the temperature is raised to $327^{\circ} C$,then the energy emitted per second becomes:

$A$ black body emits energy at the rate of $1.0 \times 10^{6} \ J/s \cdot m^{2}$ at $127^{\circ}C$. At what temperature will the energy emission rate be $16.0 \times 10^{6} \ J/s \cdot m^{2}$ in $^{\circ}C$?

The spectral energy distribution of a star is maximum at a temperature twice that of the Sun. The total energy radiated by the star is

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?

$A$ black body is at temperature $827^{\circ}\text{C}$. The rate at which it emits energy is proportional to

The top of an insulated cylindrical container is covered by a disc having emissivity $0.6$ and thickness $1\, cm$. The temperature is maintained by circulating oil as shown in the figure. If the temperature of the upper surface of the disc is $127^\circ C$ and the temperature of the surroundings is $27^\circ C$,then the radiation loss to the surroundings will be (Take $\sigma = \frac{17}{3} \times 10^{-8} \, W/m^2 K^4$)