Question

According to Stefan’s law of radiation the absolute temperature $\mathit{T}$of a body cooling in a medium at constant absolute temperature ${\mathit{T}}_{\mathbf{m}}$is given by

$\frac{\mathbf{d}\mathbf{T}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathit{k}\left(T^4-{T_m}^4\right)\mathbf{,}$

where $\mathit{k}$is a constant. Stefan’s law can be used over a greater temperature range than Newton’s law of cooling.

(a) Solve the differential equation.

(b) Show that when$T-T_m$ is small in comparison to$T_m$ then Newton’s law of cooling approximates Stefan’s law. [Hint: Think binomial series of the right-hand side of the DE.]

Short answer

The solution is$\left(\frac{T-T_m}{T+T_m}\right)-2{\tan }^{-1}\left(\frac{T}{T_m}\right)=4k{T_m}^{3}t+h$

Step-by-step solution

 Definition of differential equation

A differential equation is an equation containing the derivatives or differentials of one or more dependent variables, with respect to one or more independent variables.

 Simplify the differential equation

a)We have a body cooling by radiation in a medium at a temperature $T_m$, and then the temperature of body comes from the differential equation

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

And we have to solve this differential equation as the following technique

We can simplify our differential equation as

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

Since this differential equation shown in equation(1) is separable, then we can solve it as.

Make partial fraction  

Before we find the value of integration shown above, we have to make partial fraction to the fraction $\frac{1}{\left(T-T_m\right)\left(T+T_m\right)\left(T^2+{T_m}^2\right)}$as

$$\begin{gathered} \frac{1}{\left(T-T_m\right)\left(T+T_m\right)\left(T^2+{T_m}^2\right)}=\frac{A}{T-T_m}+\frac{B}{T+T_m}+\frac{CT+D}{T^2+{T_m}^2} \\ =\frac{\left[A\left(T+T_m\right)+B\left(T-T_m\right)\right]\left(T^2-{T_m}^2\right)+\left(CT+D\right)\left(T^2-{T_m}^2\right)}{\left(T-T_m\right)\left(T+T_m\right)\left(T^2-{T_m}^2\right)} \\ =\frac{\left[AT+A{T}_m+BT-B{T}_m\right]\left(T^2+{T_m}^2\right)+C{T}^3+D{T}^2-C{T_m}^{2}T-D{T_m}^2}{\left(T-T_m\right)\left(T+T_m\right)\left(T^2-{T_m}^2\right)} \\ =\frac{\left(A+B+C\right)+\left(A{T}_m-B{T}_m+D\right)T^2+\left(A+B-C\right)T+\left(A{T_m}^3-B{T_m}^3\right)}{\left(T-T_m\right)\left(T+T_m\right)\left(T^2+{T_m}^2\right)} \end{gathered}$$

Then by comparison we can have the equations

$$\begin{gathered} A+B+C=0 \\ A{T}_m-B{T}_m+D=0 \\ A+B-C=0 \\ A{T_m}^3-B{T_m}^3-D{T_m}^2=1 \end{gathered}$$

Then by solving these equations, we find that

$A=\frac{1}{4{T_m}^3},B=-\frac{1}{4{T_m}^3},C=0,D=-\frac{1}{2{T_m}^2}$

Then we have

$\frac{1}{\left(T-T_m\right)\left(T+T_m\right)\left(T^2+{T_m}^2\right)}=\frac{1}{4{T_m}^3}\frac{1}{T-T_m}-\frac{1}{4{T_m}^3}\frac{1}{T+T_m}-\frac{1}{2{T_m}^2}\frac{1}{T^2+{T_m}^2}$

 Solving the equation  

Then we have

$\frac{1}{4{T_m}^3}\int \frac{1}{T-T_m}dT-\frac{1}{4{T_m}^3}\int \frac{1}{T+T_m}dT-\frac{1}{2{T_m}^2}\int \frac{1}{T^2+{T_m}^2}=kt+h_1$

$$\begin{gathered} \frac{1}{4{T_m}^3}\mathrm{In}\left(T-T_m\right)-\frac{1}{4{T_m}^3}\mathrm{In}\left(T+T_m\right)-\frac{1}{2{T_m}^3}{\tan }^{-1}\left(\frac{T}{T_m}\right)=kt+h_1 \\ \frac{1}{4{T_m}^3}\mathrm{In}\left(\frac{T-T_m}{T+T_m}\right)-\frac{1}{2{T_m}^3}{\tan }^{-1}\left(\frac{T}{T_m}\right)=kt+h_1 \\ \mathrm{In}\left(\frac{\mathrm{T}-{\mathrm{T}}_{\mathrm{m}}}{\mathrm{T}+{\mathrm{T}}_{\mathrm{m}}}\right)-2{\tan }^{-1}\left(\frac{\mathrm{T}}{{\mathrm{T}}_{\mathrm{m}}}\right)=4{{\mathrm{T}}_{\mathrm{m}}}^3\mathrm{t}+\mathrm{h} \end{gathered}$$

is the general solution of the given equation.

$\mathrm{Hence}\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{In}\left(\frac{\mathrm{T}-{\mathrm{T}}_{\mathrm{m}}}{\mathrm{T}+{\mathrm{T}}_{\mathrm{m}}}\right)-2{\tan }^{-1}\left(\frac{\mathrm{T}}{{\mathrm{T}}_{\mathrm{m}}}\right)=4{{\mathrm{T}}_{\mathrm{m}}}^3\mathrm{t}+\mathrm{h}$