$A$ black body is heated from $27^oC$ to $127^oC$. The ratio of their energies of radiations emitted will be

$A$ black body,at a temperature of $227\,^{\circ}C$,radiates heat at a rate of $7\, cal\, cm^{-2} \,s^{-1}$. At a temperature of $727\,^{\circ}C$,the rate of heat radiated in the same units will be ..... units.

The actual temperature of a black body is $727^{\circ}C$. At what temperature in $K$ will the black body emit twice the radiation?

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

The temperature of a body is increased from $T_1 = 127^{\circ}C$ to $T_2 = 227^{\circ}C$. The ambient temperature is $T_0 = 27^{\circ}C$. The energies emitted per second by the body at $T_1$ and $T_2$ are $E_1$ and $E_2$ respectively. The ratio of $\frac{E_2}{E_1}$ is:

$A$ body of area $1\, cm^2$ is heated to a temperature $1000\, K$. The amount of energy radiated by the body in $1\, second$ is .......... $Joule$ (Stefan's constant $\sigma = 5.67 \times 10^{-8}\, W\, m^{-2}K^{-4}$)