This post is wrong. See this comment.

A high-school physics textbook will tell you that, according to Newton's law of cooling, a hot thing will cool down according to

$\frac{d Q}{d t} \propto T_{\infty} - T$

where $Q$ is heat, $t$ is time, $T$ is temperature, and $T_{\infty}$ is the temperature of the environment. But this is slightly wrong! It turns out that the correct cooling rate, which I'll call Lambert's law, is

$\frac{d Q}{d t} \propto \frac{1}{T} - \frac{1}{T_{\infty}}$ .

(Newton's law approximates this pretty well.) Let's derive this from first principles, and try to test it empirically.

Epistemic status: Logical deductions from a speculative physical model. I don't actually know nonequilibrium thermodynamics.

Rates of change

First, a lemma: I claim that if $X$ is any macroscopic variable, then the rate of change of $X$ is proportional to the derivative of the entropy with respect to $X$ :

$\frac{d X}{d t} \propto \frac{d S}{d X}$

Why? Suppose our system changes in discrete timesteps, and $X$ can only change in steps of size $Δ X$ . And suppose the probability of a transition $X \mapsto X + Δ X$ is proportional to the number of microstates $Ω (X + Δ X)$ . Then you can calculate the expected change in $X$ per timestep, and from there prove the lemma.^[1]

The Lambert cooling law

Now let our system comprise two subsystems. There's a fixed amount of heat split between the subsystems. Apply our lemma, taking $X$ to be $Q_{1}$ , the amount of heat in subsystem 1. We find that^[2]

$\frac{d Q_{1}}{d t} \propto \frac{d S_{1}}{d Q_{1}} - \frac{d S_{2}}{d Q_{2}} = \frac{1}{T_{1}} - \frac{1}{T_{2}}$

Here the inverse of temperature, or coldness, shows up. The coldness of a system is defined to be the derivative of entropy with respect to energy. (The only energy here is heat.)

Now if subsystem 2 is very big, its temperature doesn't change much. So let's rename $Q_{1}$ to $Q$ , $T_{1}$ to $T$ ; and $T_{2}$ to $T_{\infty}$ , because that's the temperature subsystem 1 will have in the limit:

$\frac{d Q}{d t} \propto \frac{1}{T} - \frac{1}{T_{\infty}}$

We can use this to get an expression for $d T / d t$ involving the second derivative of entropy, which is related to the heat capacity:

$\frac{d T}{d t} \propto - \frac{d^{2} S}{d Q^{2}} T^{2} (\frac{1}{T} - \frac{1}{T_{\infty}})$

However, textbooks tend to assume that the heat capacity is temperature-independent, in which case heat is roughly proportional to temperature. So we have

$\frac{d T}{d t} \propto \frac{1}{T} - \frac{1}{T_{\infty}}$

This differential equation has a neat solution after a change of variables. Let

$τ = (\frac{T}{T_{\infty}} - 1) e^{T / T_{\infty} - 1}$

Then the solution is exponential decay:

$τ = τ_{0} e^{- r t}$

where $τ_{0}$ is the value of $τ$ ...

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