The second Law of Thermodynamics states that entropy always increases in the universe: things become more disorganised.

This means, that if I have a hot coffee in a cold cup, then the heat will transfer/distribute evenly over the two, leading to a more disorganised state, hence corresponding to a higher entropy of the whole system coffee+cup.

But if the hot coffee becomes hotter and the cup colder then the order will be higher, meaning a lower entropy for the system.

Is the above true?

  • $\begingroup$ What is the question? $\endgroup$
    – Sofia
    Commented Jan 29, 2015 at 22:31
  • $\begingroup$ Welcome to the site Sana. I edited your question trying to make it more fit for this site. Feel free to revert the edit if you think it does not reflect your original intent. $\endgroup$
    – glS
    Commented Jan 29, 2015 at 22:39
  • $\begingroup$ Normally when a cup of hot liquid cools, the heat is mostly going out into the larger environment around it--the air, the table etc.--not just into the material of the cup. Do you want to specifically assume that this doesn't happen, and that the cup and liquid are totally isolated (and inside a mirrored sphere so that even thermal radiation can't escape)? Or do you want an answer that allows for the entropy of the larger environment to increase? Either way, you're going to have some components of the system that increase in entropy and others that decrease, and the increase'll be larger. $\endgroup$
    – Hypnosifl
    Commented Jan 29, 2015 at 22:51
  • $\begingroup$ I want to assume that we are talking about an isolated system, consisting of just the coffee and the cup :) $\endgroup$
    – Sanuks
    Commented Jan 29, 2015 at 22:54
  • $\begingroup$ Comment: Entropy is not disorder. Entropy is a measure of whether energy is localized (low entropy) or dispersed (high entropy). In the case of your hot coffee in a cool cup, energy is localized in the coffee. The entropy is low at the start. The energy is maximally dispersed when the coffee+cup system comes to thermal equilibrium. The entropy is high at the end. $\endgroup$ Commented Jan 30, 2015 at 16:35

1 Answer 1


You are correct that the entropy of the coffee will decrease while the entropy of the cup increases. However this will not decrease the total entropy of the system. Rather, heat will continue to flow between the two objects until entropy can no longer increase.

Let $\Delta S$ mean the change in total entropy as the energy of contents of the two components change by $\Delta Q_\text{cup,coffee}$. For infinitesimal energy transfer, we should have $$\Delta S = \beta_\text{cup} \Delta Q_\text{cup} + \beta_\text{coffee} \Delta Q_\text{coffee}$$ for some quantity $\beta$, that is not necessarily the same for both components. Let us see what this quantity is.

Since energy is conserved, we have $\Delta Q_\text{coffee}= - \Delta Q_{cup}$ and we can write the change in entropy as $$\Delta S = \Delta Q(\beta_\text{cup} - \beta_\text{coffee}).$$ Here I have taken $\Delta Q = \Delta Q_\text{cup}$ to be positive -- it's the heat gained by the cup. Thus the entropy change is consistent with thermodynamics if $$\beta_\text{cup} \ge \beta_\text{coffee}.$$ The condition for equilibrium is $$0 = \Delta S \Leftrightarrow \beta_\text{cup} = \beta_\text{coffee}.$$ We see that $\beta$ is a quantity that is equal when two systems that can exchange heat are equilibrium -- it must be a function of the temperature. Since heat flows from objects with smaller $\beta$ to objects with greater $\beta$, we define temperature through the relation $\beta = 1/T$.

  • 1
    $\begingroup$ Also might be worth making explicit the point that if $\beta_{cup}$ is not equal to $\beta_{coffee}$, and if the two $\Delta Q$'s have equal absolute value but one is negative and the other is positive, then $\Delta S$ will always be positive if the larger $\beta$ (corresponding to a smaller $T$) is paired with the positive $\Delta Q$ (heat is going into the object with the lower temperature) and the smaller $\beta$ is paired with the negative $\Delta Q$ (heat is leaving the object with the higher temperature). $\endgroup$
    – Hypnosifl
    Commented Jan 29, 2015 at 23:32

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