91Ó°ÊÓ

According to Stefan’s law,

$\frac{dT}{dt}=k\left(M^4-T^4\right),$

According to Stefan’s law,

$\frac{dT}{dt}=k\left(M^4-T^4\right),$

Where, k is a positive constant, T degrees Kelvin is the temperature of the body degrees and M degrees Kelvin is the change in the temperature in a medium.

We have to solve this equation and to explain why Newton’s law and Stefan’s law are nearly the same when T is close to M and M is constant.

$$\begin{gathered} \frac{dT}{dt}=k\left(M^4-T^4\right) \\ \frac{dT}{\left(T^4-M^4\right)}=-kdt \\ \frac{1}{T^4-M^4}=\frac{1}{\left(T^2-M^2\right)\left(T^2+M^2\right)} \end{gathered}$$

Now using the partial fractions

Now,

$$\begin{gathered} ∫\left(\frac{1}{2M^2\left(T^2-M^2\right)}-\frac{1}{2M^2\left(T^2+M^2\right)}\right)dT=-∫2k{M}^{2}dt \\ ∫\frac{dT}{2M\left(T-M\right)}-∫\frac{dT}{2M\left(T+M\right)}-∫\left(\frac{dT}{\left(T^2+M^2\right)}\right)=-2k{M}^{2}t+C \\ \frac{1}{2M}\ln \left|T-M\right|-\frac{1}{2M}\ln \left|T+M\right|-\frac{1}{M}\arctan \frac{T}{M}=-2k{M}^{2}t+C \\ \ln \left|\frac{T-M}{T+M}\right|=2\arctan \frac{T}{M}-4k{M}^{3}t+C \\ \left|\frac{T-M}{T+M}\right|=C{e}^{2\arctan \frac{T}{M}-4k{M}^{3}t} \\ \left(T-M\right)=C\left(T+M\right)e^{2\arctan \frac{T}{M}-4k{M}^{3}t} \end{gathered}$$

When is T close to M then

$$\begin{gathered} \frac{1}{M^4-T^4}=4M^3\left(M-T\right) \\ \frac{dT}{dt}=4M^{3}k\left(M-T\right) \\ =k_1\left(M-T\right) \end{gathered}$$

Therefore, the results are $\left(\mathbf{T}\mathbf{-}\mathbf{M}\right)\mathbf{=}\mathbf{C}\left(\mathbf{T}\mathbf{+}\mathbf{M}\right){\mathbf{e}}^{\mathbf{2}\mathbf{arctan}\frac{\mathbf{T}}{\mathbf{M}}\mathbf{-}\mathbf{4}{\mathbf{kM}}^{\mathbf{3}}\mathbf{t}}$and When T is close to M then