$A$ cube of side $1 \ m$ has a temperature of $127^{\circ}C$ and an emissivity of $\frac{1}{5.67}$. If the surrounding temperature is $27^{\circ}C$,the rate of loss of heat by radiation is ...... $kW$.

$A$ black body at temperature $127^{\circ} C$ radiates heat at the rate of $5 \ cal / cm^2 \ s$. At a temperature $927^{\circ} C$,its rate of emission in units of $cal / cm^2 \ s$ will be

The emissivity (power radiated per unit area) of a perfect black body is increased to $16$ times by increasing its temperature. If the initial temperature is $T$, then the final temperature of that black body is: (in $\,T$)

If the temperature of a black body is increased from $27^{\circ}C$ to $327^{\circ}C$,the radiation emitted increases by a factor of:

$A$ black body radiates energy at the rate of $E \text{ W/m}^2$ at a high temperature $T \text{ K}$. When the temperature is reduced to $\left(\frac{T}{2}\right) \text{ K}$, the radiant energy is

$A$ rectangular surface of a black body at $127^{\circ}C$ with dimensions $8 \ cm \times 4 \ cm$ emits energy at a rate of $E$. If the length and width are halved and the temperature is increased to $327^{\circ}C$,find the new rate of energy emission.