Complete beginner on the physics of heat absorption / loss, so bear with me. When I make a hot beverage, it starts out hot, but then gradually cools down as it loses heat. However, it eventually stops cooling down, or else it would reach the impossible temperature of absolute zero. My question is why do objects stop losing heat?

My best guess is that it enters into a type of chemical equilibrium with the atmosphere. Am I thinking along the correct lines here?

  • 1
    $\begingroup$ At some point the heat coming in equals the heat going out. $\endgroup$ – Hot Licks Jan 1 '18 at 15:48

When two objects/systems have the same temperature, they can be said to be in thermal equilibrium, and this means that there will be no net heat transfer between them. I cannot further emphasize how important the word net here is. This means that there is heat transfer, but heat transferred from system A to B equals to heat transferred from system B to A, so they remain at the same temperatures.

chemical equilibrium with the atmosphere.

It is thermal equilibrium here, not chemical. Hence to conclude, an object never stops losing heat, but will stop losing net heat when it achieves thermal equilibrium with the atmosphere.

  • $\begingroup$ So the concept of thermal equilibrium is the same as chemical equilibrium, in that both parties are losing/giving materials to each other at the same rate. The only difference is that thermal equilibrium involves heat, whereas chemical equilibrium involves concentration of chemicals? $\endgroup$ – Inertial Ignorance Jan 1 '18 at 9:04
  • $\begingroup$ Not only at the same rate, but same amount too. Chemical equilibrium is something different altogether. Chemical equilibrium is the state when both reactants have no tendency to change with time. $\endgroup$ – QuIcKmAtHs Jan 1 '18 at 9:10

I'd like to point something out that the other answers are glossing over (understandably; this is somewhat of a small detail).

As mentioned; heat transfer is proportional to temperature difference. As the object cools; the temperature decreases and less heat is transferred.

So consider the process as it plays out. As the object approaches the temperature of the surroundings, it begins to cool less. This process actually creates an exponential decay; where the hot cup temperature approaches the value of the surroundings temperature; but only reaches it after a very long time.

(Note: with the usual large scale engineering heat transfer models, it would reach that temperature after infinite time, but there may be higher order/microscopic effects which may make that not apply)

So the reason your object doesn't go to absolute zero is because the cooling doesn't even reach room temperature very fast. It may cover 99% of the distance relatively quickly; but the cooling speed will continue to slow down as it approaches this equilibrium.

  • $\begingroup$ Interesting -- what's the reason for the object never actually reaching the temperature of its surroundings? $\endgroup$ – Inertial Ignorance Jan 3 '18 at 9:33
  • $\begingroup$ @InertialIgnorance Because the heat loss keeps getting smaller and smaller as the temperatures are closer. When you look at the problem on continuous scales. it creates a situation where you get fractionally closer as time goes on; but the extent that you get closer decreases as you get closer, so you never actually reach the final value. Again; this doesn't necessarily apply on the microscopic scale; as things don't behave perfectly continuous at that scale. $\endgroup$ – JMac Jan 3 '18 at 13:42

heat transfer is driven by temperature differences. once your beverage cools down to a temperature equal to that of its surroundings, heat transfer stops.

  • $\begingroup$ The other poster said that heat transfer doesn't stop when both parties are at an equal temperature. When you say heat transfer stops, do you mean net heat transfer stops? $\endgroup$ – Inertial Ignorance Jan 1 '18 at 9:07
  • $\begingroup$ yes. equal amounts of heat going both ways. net transfer = zero. $\endgroup$ – niels nielsen Jan 1 '18 at 9:09
  • $\begingroup$ @nielsnielsen Perhaps the wording could be "net heat transfer stops". It's trivial; but it fits in more with what actually happens in equilibrium. $\endgroup$ – JMac Jan 1 '18 at 17:09
  • $\begingroup$ I agree. this is another case where I went for simpler rather than more complete. $\endgroup$ – niels nielsen Jan 1 '18 at 19:06

Cooling is loss of heat, and that heat goes elsewhere (your hand around the cup, the warm air rising above the beverage...). Other objects lose heat too, any of which can go into your beverage. Heat is never destroyed (by the second law of thermodynamics), so it is unsurprising that your beverage doesn't approach absolute zero. Other objects lose heat too, any of which can go into your beverage.

The temperature balance will eventually be reached with any objects that have a thermal connection to the beverage, and that (by definition, or 'the zeroth law' of thermodynamics) means zero temperature difference. It does not imply zero temperature, rather zero expected heat exchange (gain or loss).


The basic reason for this is probability. The situation where two objects have the same temperature is the most likely state of the system, subject to the constraint that the energy of the total system (for example your room) is constant. This is the macroscopic state with the maximal number of microstates. Let us call this number the multiplicity $\Omega$.

This number is the product of the multiplicity of the cup and the multiplicity of the rest of the room: $\Omega = \Omega_{cup} \times \Omega_r$. The multiplicities depend very strongly on the internal energy $E$. When the cup is hot, energy is transferred to the room because the fractional increase in $\Omega_r$ is greater than the fractional decrease of the multiplicity of the cup. This makes that the total $\Omega$ increases.

Eventually, thermal equilibrium will be reached, the most probable macrostate, the state with the highest multiplicity. Then the fractional changes of $\Omega$ of the cup and of the room are equal and opposite when a small amount of energy is transferred between them. For example when the multiplicity of the cup goes up by one promille, the multiplicity of the room goes down by one promille, leaving the product unchanged at its maximum.

This is what temperature is. When two systems have the same temperature, they have the same relative change of multiplicity with energy $\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E}$. At room temperature this is about 4 % per milli-eV.

For the cognoscenti: with the definition of entropy $S = k \ln\Omega$ and $\frac{dS}{dE} = \frac{k}{\Omega} \frac{d\Omega}{dE} = \frac{1}{T}$ we recognize the fractional change of $\Omega$ with energy as the thermodynamic beta (coldness) $\beta = \frac{1}{kT} =\frac{1}{\Omega}\frac{{\rm d}\Omega}{{\rm d}E}$.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.