# If heated molecules are more disordered, why doesnt it happen spontaneously?

currently, the way I understand entropy is disorder and that it always increases. I therefore have difficulty to understand why don't object heat spontaneously(i.e take in energy from their surrounding-causing the surroundings to be colder). I am basing my problem on that when molecules are heated they gain energy and move about more, hence more disordered. Therefore, shouldn't heat be taken in by an object as opposed to dissipated?

• It's not bad to understand entropy as disorder but that's not completely accurate. What is accurate is that ENTROPY IS IGNORANCE and you can now have a physical intuition on why the tendency to balance the temperatures of the components of the system, i.e. to have more uncertainty on the state of the system, is equivalent to increasing the entropy. Hope this helps! Commented Jul 13, 2018 at 9:34
• if a group of molecules is put in the presence of a heat source, it will happen spontaneously :) Commented Jul 13, 2018 at 9:38

Consider it from an energy standpoint. If the object were truly heating up "spontaneously" without an external energy source then you are implying that there is no flow of energy from the surroundings into the object yet the internal energy is increasing. This then means that the energy of the universe is increasing thereby violating the 1st law of thermodynamics.

Entropy increasing is what causes spontaneous flow of energy though. And that is why when you put a warm object and a cold one together, the warm one transfers energy to the cold one until they reach equilibrium. That is the maximum entropy state.

Consider a simple example of interacting systems, an Einstein solid that represents the bonds between molecules as harmonic oscillators. Then we want to explore how exchanging energy between them changes the entropy.

Consider that for one ES with $q$ units of energy and $N$ oscillators the multiplicity function is:

$$\Omega = \frac{(q + N - 1)!}{q!(N-1)!}$$

And recall that $S = k \ln\Omega$.

And for interacting ES we have to multiply the multiplicities together so we get:

$$\Omega_{total} = \Omega_A \Omega_B = \frac{(q_A + N_A - 1)!}{q_A!(N_A-1)!} \frac{(q_B + N_B - 1)!}{q_B!(N_B-1)!}$$

The overall number of possible microstates of the system is:

$$\Omega_{overall} = \frac{(q_A + N_A + q_B + N_B - 1)!}{(q_A + q_B)!(N_A + N_B-1)!}$$

To find the probability of a given distribution of energy units, can be calculated simply by the expression:

$$\mathrm{P} = \frac{\Omega_{total}}{\Omega_{overall}}$$

It is most fruitful from here to take on a specific example since simplifying the expressions is rather ugly.

Consider the case of $N_A = N_B = 10$ and $q_A + q_B = 20$.

We can then compute that:

$$\Omega_{overall} \approx 6.9 \times 10^{10}$$

Then express $q_B = 20 - q_A$ and we can express the probability in terms of only the variable $q_A$. We get the expression:

$$\mathrm{P} = \frac{1}{\Omega_{overall}} \frac{(q_A + 9)!}{q_A!9!} \frac{(29 - q_A)!}{(20 - q_A)!9!}$$

Though we could compute the derivative and optimize the function, it is not a very simple problem analytically. The easiest thing to do is to numerically compute a bar chart representing the distribution of the probabilities. When we do this we get the below figure which shows that the probability is maximized when the energy is the same in both solids.

We have optimized the multiplicity here and since entropy can be computed from multiplicity directly and is monatomic, we have also maximized the entropy. Further investigations can show that this holds in other cases and you can consult an undergraduate book on thermal physics for more of the details.

• I do not mean spontaneous in a sense of getting energy from nowhere, rather I mean that energy should flow from the colder to the warmer as that will lead to more disorder in the hotter object Commented Jul 13, 2018 at 9:00
• But you have to look at the disorder as a whole. When calculating the entropy of the system energy flowing from colder to warmer would decrease the overall entropy. An isolated system evolves to maximize the entropy of the system, not any individual object. Commented Jul 13, 2018 at 9:03
• So lets consider the two objects as a system. What you are implying is that there exists more entropy in two medium temperature objects as opposed to one hot and one cold? Commented Jul 13, 2018 at 9:07
• @user1234 Yes, that's correct. In fact, (I'm simplifying here) temperature in statistical physics is defined as the derivative of energy wrt entropy, so in your situation an object with higher temperature loses less entropy than an object with low temperature gains when the heat moves to the colder object.
– JiK
Commented Jul 13, 2018 at 9:29
• I added more to my answer that will hopefully clarify. It is not possible to cover all the details here but I hope the example motivates you for the reason and that if you are interested you will consult a thermodynamics textbook for more details. Schroeder Intro to Thermal Physics is one example of a book with many good examples. Commented Jul 13, 2018 at 9:29