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 $J$)

Radiation from a black body at the thermodynamic temperature $T_1$ is measured by a small detector at distance $d_1$ from it. When the temperature is increased to $T_2$ and the distance to $d_2$,the power received by the detector is unchanged. What is the ratio $d_2/d_1$?

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

The surface of a black body is at a temperature $727^{\circ} C$ and its cross-section is $1 \,m^2$. Heat radiated from this surface in one minute in joules is (Stefan's constant $=5.7 \times 10^{-8} \,W / m^2 / K^4$ )

The total energy radiated from a black body source is collected for one minute and is used to heat a quantity of water. The temperature of water is found to increase from $20^{\circ}C$ to $20.5^{\circ}C$. If the absolute temperature of the black body is doubled and the experiment is repeated with the same quantity of water at $20^{\circ}C$,the temperature of water will be....... $^{\circ}C$