The temperature of a black body increases from $327^{\circ}C$ to $927^{\circ}C$. If the initial energy possessed is $2 \ kJ$,what is its final energy in $kJ$?

Two spherical black bodies have radii $r_1$ and $r_2$. Their surface temperatures are $T_1$ and $T_2$. If they radiate the same power,then $\frac{r_2}{r_1}$ is:

Two bodies $P$ and $Q$ have thermal emissivities of $\varepsilon_P$ and $\varepsilon_Q$ respectively. Surface areas of these bodies are same and the total radiant power emitted is also at the same rate. If the temperature of $P$ is $\theta_P$ Kelvin,then the temperature of $Q$ i.e.,$\theta_Q$ is:

$A$ sphere is at temperature $600 \text{ K}$. In an external environment of $200 \text{ K}$, its cooling rate is $R$. When the temperature of the sphere falls to $400 \text{ K}$, then the cooling rate $R'$ will become:

$A$ black rectangular surface of area '$A$' emits energy '$E$' per second at $27^{\circ}C$. If length and breadth are reduced to $(1/3)^{rd}$ of their initial values and temperature is raised to $327^{\circ}C$,then the energy emitted per second becomes:

When the temperature of a black body increases, it is observed that the wavelength corresponding to maximum energy changes from $0.26 \mu m$ to $0.13 \mu m$. The ratio of the emissive powers of the body at the respective temperatures is