Obtain the equation for the rate of emission of heat for a perfect black body by using the Stefan-Boltzmann law.

(N/A) According to the Stefan-Boltzmann law,the power radiated by a perfect black body at temperature $T$ is given by $P = A \sigma T^4$,where $A$ is the surface area,$\sigma$ is the Stefan-Boltzmann constant,and $T$ is the absolute temperature.
For a body at temperature $T_1$ placed in surroundings at temperature $T_2$ (where $T_1 > T_2$),the body emits energy at a rate of $\frac{dQ_1}{dt} = A \sigma T_1^4$ and absorbs energy from the surroundings at a rate of $\frac{dQ_2}{dt} = A \sigma T_2^4$.
The net rate of emission of heat energy is the difference between the rate of emission and the rate of absorption:
$\frac{dQ}{dt} = \frac{dQ_1}{dt} - \frac{dQ_2}{dt}$
Substituting the expressions:
$\frac{dQ}{dt} = A \sigma T_1^4 - A \sigma T_2^4$
Therefore,the net rate of heat emission is:
$\frac{dQ}{dt} = A \sigma (T_1^4 - T_2^4)$