The Zeroth Law of Thermodynamics states that if two systems are in thermodynamic equilibrium with a third system, the two original systems are in thermal equilibrium with each other.

It says that they reach the same temperature after an infinite amount of time. Why can't we quantify this time taken to reach thermal equilibrium?


4 Answers 4


In the simplest model of heat transfer, the rate of heat transfer is proportional to the temperature difference between the two systems. This gives an exponential temperature curve, for which the temperature difference between the two systems approaches zero as closely as we like, but never reaches zero.

This is, however, only an approximate model of reality. It is similar to the simple model of radioactivity in which a fixed proportion of a radioactive substance decays in each time period. In this model the number of decays per second approaches but never reaches zero. But in reality we know that when the number of undecayed atoms is small then the decay rate will deviate from this model and eventually all atoms will decay within a finite time.

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    $\begingroup$ So, just to be clear, like with radioactivity, eventually the two systems really will be at thermal equilibrium, not just arbitrarily close to it? $\endgroup$
    – gardenhead
    Jun 26, 2021 at 21:55
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    $\begingroup$ I’m not sure the radioactivity metaphor works. Radioactivity is a discrete event, but temperature is not. I think an object can have any arbitrarily small template rather difference. $\endgroup$
    – Tim
    Jun 26, 2021 at 23:01
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    $\begingroup$ @gardenhead Yes, in real life. Or at least they will become so close to thermal equilibrium that quantum effects take over. $\endgroup$
    – gandalf61
    Jun 27, 2021 at 7:54
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    $\begingroup$ @Tim Temperature, like radioactivity, is a statistical property. Eventually the notional temperature difference between two objects becomes so small that it is dominated by random quantum fluctuations. $\endgroup$
    – gandalf61
    Jun 27, 2021 at 8:02

The zeroth law of Thermodynamics is speaking about systems already at thermodynamic equilibrium. It does not say anything about the time required to get equilibrium. The equilibration time depends on the efficiency of the exchanges of energy (not only thermal exchanges but also volume variations and particle exchanges, in simple systems) and their relaxation times.

Macroscopic transport theories, like heat conduction or diffusion theory, may provide some generic guide, but they depend on non-thermodynamic quantities (the transport coefficients like thermal conduction coefficient, viscosity coefficients, diffusion, and inter-diffusion coefficients) whose determination requires some detailed model of the underlying atomic processes. Moreover, macroscopic solutions for the time relaxation towards equilibrium, often providing exponentially decaying behaviors, shouldn't be taken too seriously when the difference of thermodynamic state variables becomes comparable with the size of thermodynamic fluctuations always present in finite-size samples.

As a consequence, the time to reach thermodynamic equilibrium is strongly dependent on all the relaxation times in a system. In practice, what really matters is the presence of a well definite macroscopic time scale such that the system properties are stationary. I often like to cite Feynman's definition of thermodynamic equilibrium, given at the beginning of his Statistical Mechanics textbook: it is the state such that "all the "fast" things have happened and all the "slow" things not.


The temperature of a system that is hotter than the surroundings decreases and it's temperature is of the form $$T=A + Be^{-kt} $$

where $A$,$B$ and $k$ are constants and $t$ is time.

It approaches temperature $A$, but never reaches it.

enter image description here

In the image $A$ is $2$, $B$ is $2$ and $k$ is $0.2$

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    $\begingroup$ If the temperature difference is smaller than the temperature fluctuation of the bodies you could say it reached the temperature. $\endgroup$
    – lalala
    Jun 26, 2021 at 11:12
  • $\begingroup$ @lalala Yes, good point, the answer was just trying to explain the sentence in the question. $\endgroup$ Jun 26, 2021 at 11:23
  • $\begingroup$ It may be useful to point out this is an approximate relationship known as Newton's Law of Cooling: "The law is frequently qualified to include the condition that the temperature difference is small and the nature of heat transfer mechanism remains the same. As such, it is equivalent to a statement that the heat transfer coefficient, which mediates between heat losses and temperature differences, is a constant. " $\endgroup$
    – jim
    Jun 26, 2021 at 11:37

Why can't we quantify this time taken to reach thermal equilibrium?

While the time to reach equilibrium is infinite, we can quantify it in a useful manner by looking at a finite measure of the time.

Consider a simple model where the temperature difference $T$ evolves according to $$ \frac{\mathrm{d}{T}}{\mathrm{d}t} = -a T. $$ The constant $a$ must have dimensions of reciprocal time, so we might as well write $\tau = 1/a$, i.e. $$ \frac{\mathrm{d}{T}}{\mathrm{d}t} = -\frac{T}{\tau}, $$ where $\tau$ is the time scale. We get $$ T(t) = T(0) \exp(-t / \tau). $$ Now, if $t / \tau = 4$ then the exponential factor is already less than 2%. So while the theoretical time to reach equilibrium is always infinite, a practical measure would be $t = 4 \tau$.

A similar example would be an RC circuit. The voltage decays exponentially according to $$ V(t) = V(0) \exp \left( -\frac{t}{RC} \right), $$ so we have a time scale of $\tau = RC$ and a "practical" equilibrium time of $4\tau = 4 RC$ (which is a lot more useful than just saying the time to reach equilibrium is infinite).


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