Are Newton's law of Cooling and Stefan's law related?

Question

Many of Indian school textbooks claim a proof of Newton's law of cooling from Stefan's law of black-body radiation.

As far as I am aware of, Newton's law is based on cooling from convection currents and Stefan's law on radiation. There is not supposed to be any relation between them.

Question: Is there any relation between them, and can Newton's law of cooling be derived from Stefan's law?

I found many answers and resources on Google, but an answer referring to a well-established paper, book or resource of same kind will be highly appreciated.

UPDATE

There's a question in the syllabus they teach us such as "Derive Newton's law of Cooling from Stefan's Law." And here's one of the links which show the solution. They are approximating Stefan's law to Newton's law(mathematically) by considering $T-T_0$ very small. They even claim that Newton's law is applicable for even small temperature differences whereas in reality Newton's law is applicable for all temperature ranges.

Please help. Is there any strong reference which can help shut down this misconception. I understand why are they different but a reference might help much.

Answer 1

Newton's Law of Cooling is fundamentally an empirical relation for the rate of heat transfer into a body in the limit of a small temperature difference between the body and its surroundings. Given any arbitrary heat transfer law,

$\dot{Q} = f(T)$,

a corresponding first order law of cooling can be deriving by performing a Taylor expansion around the equilibrium temperature $T_0$ as follows:

$\dot{Q} = f'(T_0) \cdot (T - T_0) + \mathcal{O}(T-T_0)^2$

The mechanism of heat transfer here can be arbitrary, since Newton's Law of Cooling holds for any such mechanism in the limit where $T$ does not differ too much from $T_0$. Indeed, in the heat transfer literature, you will find that heat transfer coefficients are reported for systems involving static conduction, convection, radiation, or any combination of these mechanisms.

From the equation above, we can see by inspection that given a heat transfer law $f(T)$, the heat transfer coefficient $h$ in Newton's Law of Cooling is given by

$h = f'(T_0)$

Thus, in the specific example of the Stefan-Boltzmann law, we have

\begin{align*} \dot{Q} &= \sigma_B \,(T^4-T_0^4)\\&= 4\, \sigma_B\,T_0^3 \,\left(T - T_0 \right ) + \mathcal{O} \left(T-T_0\right)^2 \\ h &= 4 \, \sigma_B \, T_0^3 \end{align*}

Your confusion arises from wrongly thinking of Newton's Law of Cooling as a fundamental law of heat transfer, where in fact it is simply an approximation that makes solving heat transfer problems much easier in the limit of small temperature differences. So Newton's Law of Cooling is not strictly valid for all temperatures, or put in a different way, the heat transfer coefficient $h$ in the law will take on different values at different temperatures. For simple systems like the one above, $h$ can be derived from first principles, but in practice it must be estimated from experimental data.

Answer 2

Edit: I'm no longer confident in the correctness of my answer. The response given by Arvind Kannan seems to me superior to my own.

You are correct - the Stefan-Boltzmann law and Newton's Law of Cooling are unrelated. The former deals only with radiation heat exchange whereas the latter with conduction.

This can be considered mathematically as well. The Stefan-Boltzmann law states that heat is transferred at a rate proportional to the fourth power of temperature:

$\frac{dT}{dt} = -k(T^4 - T^4_0)$

whereas Newton's Law of Cooling involves a first power rate of heat transfer:

$\frac{dT}{dt} = -k(T - T_0)$

This causes the two laws to be fundamentally different and unrelated.

Answer 3

Newton's law of cooling is a special case of Stefan's law. $$ \begin{aligned} \frac{d Q}{d t} &=e \sigma A\left(T^{4}-T_{0}^{4}\right) \\ &=e \sigma A\left(T^{2}+T_{0}^{2}\right)\left(T^{2}-T_{0}\right) \\ &=e \sigma A 2 T_{0}^{2}\left(T+T_{0}\right)\left(T-T_{0}\right) \\ m s \frac{d T}{d t} &=e \sigma A 4 T_{0}^{3}\left(T-T_{0}\right) \\ \frac{d T}{d t} &=\frac{e \sigma A 4 T_{0}^{3}}{m s}\left(T-T_{0}\right) \\ \frac{d T}{d t} &=k\left(T-T_{0}\right) \end{aligned} $$