When the temperature of a black body is increased, the value of $\lambda_m$ changes from $0.26 \mu m$ to $0.13 \mu m$. What is the ratio of its emissive power corresponding to these temperatures?

If a black body emits $0.5 \ J$ of energy per second when it is at $27^{\circ} C$,then the amount of energy emitted by it when it is at $627^{\circ} C$ will be (in $J$)

$Assertion :$ Bodies radiate heat at all temperatures.
$Reason :$ Rate of radiation of heat is proportional to the fourth power of absolute temperature.

$A$ very small hole in an electric furnace is used for heating metals. The hole nearly acts as a black body. The area of the hole is $200 \ mm^2$. To keep a metal at $727^{\circ} C$, the heat energy flowing through this hole per second, in joules, is (given $\sigma = 5.67 \times 10^{-8} \ W m^{-2} K^{-4}$):

$A$ metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated $(P)$ by the metal. The sensor has a scale that displays $\log_{2}(P / P_0)$,where $P_0$ is a constant. When the metal surface is at a temperature of $487^{\circ} C$,the sensor shows a value of $1$. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to $2767^{\circ} C$?

The ratio of radiant energies radiated per unit surface area by two bodies is $16 : 1$. The temperature of the hotter body is $1000 \ K$. Then,the temperature of the colder body will be ....... $K$.