$A$ 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 ...... $kW$.

Write the value and dimensional formula of the Stefan-Boltzmann constant.

Two identical bodies have temperatures $277^{\circ} C$ and $67^{\circ} C$. If the surroundings temperature is $27^{\circ} C$, the ratio of loss of heats of the two bodies during the same interval of time is (approximately) (in $ : 1$)

$A$ black body radiates maximum energy at wavelength $\lambda$ and its emissive power is $E$. Now,due to a change in temperature of that body,it radiates maximum energy at wavelength $\frac{\lambda}{3}$. At that new temperature,the emissive power is: (in $E$)

Radiated energy at $T$ temperature is $E$ for a body of diameter $d$. If the temperature becomes $2T$ and the diameter becomes $d/4$,then the radiated energy will be:

The 'Kangri' is an earthen pot used to stay warm in Kashmir during the winter months. Assume that the 'Kangri' is spherical and of surface area $7 \times 10^{-2} \,m^{2}$. It contains $300 \,g$ of a mixture of coal,wood,and leaves with a calorific value of $30 \,kJ/g$ (and provides heat with $10 \%$ efficiency). The surface temperature of the 'Kangri' is $60^{\circ}C$ and the room temperature is $0^{\circ}C$. Then,a reasonable estimate for the duration $t$ (in $h$) that the 'Kangri' heat will last is (take the 'Kangri' to be a black body).