Two bodies $A$ and $B$ at temperatures $T_1 \ K$ and $T_2 \ K$ respectively have the same dimensions. Their emissivities are in the ratio $16:1$. At $T_1 = x T_2$,they radiate the same amount of heat per unit area per unit time. The value of $x$ is

The ratio of the energy of emitted radiation of a black body at $27^{\circ}C$ and $927^{\circ}C$ is:

Two black bodies $A$ and $B$ have equal surface areas and are maintained at temperatures $27^{\circ} C$ and $177^{\circ} C$ respectively. What will be the ratio of the thermal energy radiated per second by $A$ to that by $B$?

$A$ black body at a temperature of $227^{\circ} C$ radiates heat energy at the rate of $5 \ cal/cm^{2}-s$. At a temperature of $727^{\circ} C$,the rate of heat radiated per unit area in $cal/cm^{2}-s$ will be:

State and explain the Stefan-Boltzmann law.

$A$ metal ball of surface area $200 \; cm^2$ and temperature $527^{\circ}C$ is surrounded by a vessel at $27^{\circ}C$. If the emissivity of the metal is $0.4$,then the rate of loss of heat from the ball is approximately .......... $J/s$. $(\sigma = 5.67 \times 10^{-8} \; J/(m^2 \cdot s \cdot K^4))$