Stefan’s law of radiation states that the rate of change in the temperature of a body at T (t) kelvins in a medium at M (t) kelvins is proportional to ${\mathbf{M}}^{\mathbf{4}}\mathbf{-}{\mathbf{T}}^{\mathbf{4}}$. That is, $\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{K}\left[\mathbf{M}{\left(\mathbf{t}\right)}^{\mathbf{4}}\mathbf{-}\mathbf{T}{\left(\mathbf{t}\right)}^{\mathbf{4}}\right]$ where K is a constant. Let $\mathbf{K}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}{\left(\mathbf{min}\right)}^{\mathbf{-}\mathbf{1}}$ and assume that the medium temperature is constant, M (t) = 293 kelvins. If T (0) = 360 kelvins, use Euler’s method with h = 3.0 min to approximate the temperature of the body after

(a) 30 minutes.

(b) 60 minutes.

Euler’s method is a tool to approximate the solution of an initial value problem at a given point. For a given initial value problem $\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{=}\mathbf{f}\left(\mathbf{x}\mathbf{,}\mathbf{y}\right)\mathbf{,}\mathbf{y}\left({\mathbf{x}}_{\mathbf{0}}\right)\mathbf{=}{\mathbf{y}}_{\mathbf{0}}$, the recursive relation that is referred to as Euler’s scheme is

$$\begin{gathered} {\mathbf{x}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{x}}_{\mathbf{n}}\mathbf{+}\mathbf{h} \\ {\mathbf{y}}_{\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}{\mathbf{y}}_{\mathbf{n}}\mathbf{+}\mathbf{hf}\left({\mathbf{x}}_{\mathbf{n}}\mathbf{,}{\mathbf{y}}_{\mathbf{n}}\right) \end{gathered}$$

Here, h is the step size.

The prescribed initial value problem is

$\frac{\mathbf{dT}}{\mathbf{dt}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\left({\mathbf{293}}^{\mathbf{4}}\mathbf{-}{\mathbf{T}}^{\mathbf{4}}\right)\mathbf{,}\mathbf{T}\left(\mathbf{0}\right)\mathbf{=}\mathbf{360}$

The given step size is h = 3.0.

Here,

$$\begin{gathered} \mathbf{f}\left(\mathbf{t}\mathbf{,}\mathbf{T}\right)\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\left({\mathbf{293}}^{\mathbf{4}}\mathbf{-}{\mathbf{T}}^{\mathbf{4}}\right) \\ {\mathbf{t}}_{\mathbf{0}}\mathbf{=}\mathbf{0} \\ {\mathbf{T}}_{\mathbf{0}}\mathbf{=}\mathbf{360} \end{gathered}$$

Apply the Euler’s scheme as defined in step 1.

Putting n = 0, it results,

$$\begin{gathered} {\mathbf{t}}_{\mathbf{1}}\mathbf{=}{\mathbf{t}}_{\mathbf{0}}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{0}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{3} \\ {\mathbf{T}}_{\mathbf{1}}\mathbf{=}{\mathbf{T}}_{\mathbf{0}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{0}\mathbf{,}\mathbf{360}\right) \\ \mathbf{=}\mathbf{360}\mathbf{+}\mathbf{3}\left[\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\left({\mathbf{293}}^{\mathbf{4}}\mathbf{-}{\mathbf{360}}^{\mathbf{4}}\right)\right] \\ \mathbf{=}\mathbf{357}\mathbf{.}\mathbf{26642833229} \end{gathered}$$

Putting n = 1, it gives,

$$\begin{gathered} {\mathbf{t}}_{\mathbf{2}}\mathbf{=}{\mathbf{t}}_{\mathbf{1}}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{3}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{6} \\ {\mathbf{T}}_{\mathbf{2}}\mathbf{=}{\mathbf{T}}_{\mathbf{1}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{3}\mathbf{,}\mathbf{357}\mathbf{.}\mathbf{26642833229}\right) \\ \mathbf{=}\mathbf{357}\mathbf{.}\mathbf{26642833229}\mathbf{+}\mathbf{3}\left[\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\left({\mathbf{293}}^{\mathbf{4}}\mathbf{-}\mathbf{357}\mathbf{.}{\mathbf{26642833229}}^{\mathbf{4}}\right)\right] \\ \mathbf{=}\mathbf{354}\mathbf{.}\mathbf{679123642067818} \end{gathered}$$

Continuing in this process, the result obtained is listed below.

From the above table, you can see that the approximate value of T (30) is 338.139059077222714. Rounding off to the one-decimal place, the approximate value of T (30) is 338.1.

Hence, after 30 mins, the temperature of the body is approximately 338.1 kelvins.

In the previous part, the temperature of the body has already been calculated. In this step, perform the iteration for n = 30 and onwards.

Putting n = 30 in the iteration scheme, it follows,

$$\begin{gathered} {\mathbf{t}}_{\mathbf{31}}\mathbf{=}{\mathbf{t}}_{\mathbf{30}}\mathbf{+}\mathbf{h} \\ \mathbf{=}\mathbf{30}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{33} \\ {\mathbf{T}}_{\mathbf{31}}\mathbf{=}{\mathbf{T}}_{\mathbf{30}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{30}\mathbf{,}\mathbf{338}\mathbf{.}\mathbf{1}\right) \\ \mathbf{=}\mathbf{338}\mathbf{.}\mathbf{1}\mathbf{+}\mathbf{3}\left[\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\left({\mathbf{293}}^{\mathbf{4}}\mathbf{-}\mathbf{338}\mathbf{.}{\mathbf{1}}^{\mathbf{4}}\right)\right] \\ \mathbf{=}\mathbf{333}\mathbf{.}\mathbf{143528759580873} \end{gathered}$$

Putting n = 31, it results

$$\begin{gathered} {\mathbf{t}}_{\mathbf{32}}\mathbf{=}{\mathbf{t}}_{\mathbf{31}}\mathbf{+}\mathbf{h} \\ \mathbf{=}\mathbf{33}\mathbf{+}\mathbf{3} \\ \mathbf{=}\mathbf{36} \\ {\mathbf{T}}_{\mathbf{32}}\mathbf{=}{\mathbf{T}}_{\mathbf{31}}\mathbf{+}\mathbf{3}\mathbf{f}\left(\mathbf{33}\mathbf{,}\mathbf{333}\mathbf{.}\mathbf{143528759580873}\right) \\ \mathbf{=}\mathbf{333}\mathbf{.}\mathbf{143528759580873}\mathbf{+}\mathbf{3}\left[\mathbf{2}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\left({\mathbf{293}}^{\mathbf{4}}\mathbf{-}\mathbf{333}\mathbf{.}{\mathbf{143528759580873}}^{\mathbf{4}}\right)\right] \\ \mathbf{=}\mathbf{328}\mathbf{.}\mathbf{839175029307695} \end{gathered}$$

Continuing in this process, the result obtained is listed below.

From the above table, you can see that the approximate value of T (60) is 308.480461276162297. Rounding off to the one-decimal place, the approximate value of T (60) is 308.5.

Thus, after 60 mins, the temperature of the body is approximately 308.5 kelvins.