# Why does entropy explain why a colder object won't spontaneously give off energy to a hotter one?

I'm very new to my studies of entropy, but in my view entropy is the amount of disorder in a system, which I know is unimaginatively true and not really demonstrative of any true understanding I might have on the subject. It has to do with the number of ways a system has of arranging itself, which is why a cup of cracked, scattered ice has less entropy than a neat glass of water as the molecules in the water have more ways of arranging themselves in the glass while still being water while the ice, with its solid structure has pretty fixed positions and arrangements of its molecules.

However, I still can't tie in my understanding to explain why a cold object won't spontaneously give off energy to a hotter one. I know the answer is tl;dr "because entropy" but my understanding of it doesn't tie this in. I know the hotter object has more disorder, as the system have a higher multiplicity i.e. more range of properties to exhibit (like more arrangements of varying speeds as the cap for speed for each particle has raised) but I don't see how that means it can't spontaneously gain energy from the object with less.

• "I don't see how that means it can't spontaneously gain energy from the object with less." - it can but have you considered all the ways that it would versus all the ways that it goes the other way and asked what is (by far) the most likely outcome? Oct 28 '17 at 2:25
• Just to be precise: it also depends on your precise notion of object. Also in absence of mechanical transformations, if chemical reactions happen inside the object it may give off energy to a hotter object...What the 2nd principle says (or should say to be equivalent to Kelvin's formulation) is that heat does not spontaneously moves form a cooler resevoir to a hotter resevoir. Oct 28 '17 at 8:53

Short answer: that's actually the (or, a) definition of temperature. By definition, object A is at higher temperature than object B if A spontaneously transfers heat to object B when they are placed in thermal contact. So you can leave entropy out of it entirely.

Long answer: Here's how entropy gets involved. Forget about cracked and solid ice or anything macroscopic for a while. Think about some really simple system like, say, electrons in an atom. Each electron can be in one of several quantum states, and the energy of the atom $U$ depends on how many electrons are in each state.

For any amount of energy the atom (or whatever) can have, there may be several different ways to arrange electrons to give it that energy. The number of ways for a system to have a certain energy is called the multiplicity, denoted $\Omega$, and it is a function of energy. Entropy $S$ is (proportional to) the logarithm of $\Omega$, and is also a function of energy.

Now, imagine two systems that can exchange bits of energy between each other. Say system A has 20 energy and system B has 10 energy, so the multiplicity is $\Omega_A(20)\times\Omega_B(10)$. What's likely to happen with this energy? Well, consider this:

• If system A gives a unit of energy to system B, the multiplicity becomes $\Omega_A(19)\times\Omega_B(11)$, and the entropy is $S_A(19) + S_B(11)$
• Or if system B gives a unit of energy to system A, the multiplicity becomes $\Omega_A(21)\times\Omega_B(9)$, and the entropy is $S_A(21) + S_B(9)$
• Or maybe nothing happens, and the multiplicity is still $\Omega_A(20)\times\Omega_B(10)$ and the entropy is still $S_A(20) + S_B(10)$.

Assuming that each individual final state is equally likely, whichever one of these multiplicities - or entropies - is larger, that's the outcome that's going to be more likely. Or to put it another way, the energy transfer that increases the total entropy more is more likely to happen. (That's the second law of thermodynamics, by the way.)

In most realistic situations, one option is overwhelmingly more likely than the others. For example, if the first option (A loses energy to B) is the most likely, that means (among other things) that $$S(U_A + \Delta U) + S(U_B - \Delta U) < S(U_A) + S(U_B)$$ but doing a bit of calculus on that gets you \begin{align} S(U_A + \Delta U) - S(U_A) &< S(U_B) - S(U_B - \Delta U) \\ \lim_{\Delta U\to 0}\frac{S(U_A + \Delta U) - S(U_A)}{\Delta U} &< \lim_{\Delta U\to 0}\frac{S(U_B) - S(U_B - \Delta U)}{\Delta U} \\ \frac{\partial S_A}{\partial U_A} &< \frac{\partial S_B}{\partial U_B} \end{align} Similarly, if the second option (B loses energy to A) is most likely, you can find that $$\frac{\partial S_A}{\partial U_A} > \frac{\partial S_B}{\partial U_B}$$ And if the third option (no energy changes hands) is most likely, you can show (either by elimination or some similar calculus) that $$\frac{\partial S_A}{\partial U_A} = \frac{\partial S_B}{\partial U_B}$$ Clearly, this quantity $\partial S/\partial U$ is related to whether a system gains or loses energy to another system. Systems with smaller values of $\partial S/\partial U$ will tend to lose energy to systems with larger values. But wait! That should remind you of the definition of temperature from the top of my post. And indeed, we define temperature as the reciprocal of this quantity: $$\frac{1}{T} = \frac{\partial S}{\partial U}$$ In other words, the reciprocal of temperature represents the "capacity" of one unit of energy to increase the entropy of a system. If a system is hot, then $1/T$ is small, and that means each unit of energy is "contributing" relatively little entropy. Given the tendency of entropy to increase, a hot system will tend to pass off its energy to other systems where each unit will have more of an effect on the overall entropy.

For entropy I find that it is easier to acquire intuition when going to the statistical definition of entropy.

Given W the multiplicity of possibilities of a given state ( the number of ways you can produce that state) is connected with the entropy directly,

S=k Ln(W) , where k is the Boltzman constant.

Temperature is connected with the kinetic energy available in the system. For a simple example in gases it is defined as

High kinetic energies generate a lot more microstates to be counted in entropy than low kinetic energy microstates, and the probability of equilibrium of energy transfer goes from high average kinetic energies to low average energies, as in the image above. The probability of the right side to transfer all its kinetic energy to the left is very low. Equilibrium will be reached at a lower temperature.(average kinetic energy).

So in the process of reaching equilibrium the right side will transfer some kinetic energy by statistical scatters , but it will be overwhelmed by the higher kinetic energy side.

Suppose that only radiation can be transferred. The Black Body(BB) radiation of the left side will be much higher than the BB radiation of the right side so even though the low temperature will radiate to the high, the number of photons going from left to right and generating microstates will be much higher than the number of photons from right to left.

Hope this helps.

I guess the focus needs to be put on the cold object. If we place a cold object in a hot environment, intuitively, we would say the cold object will not give heat out to its hot environment. Using the concept of entropy and $ds = \frac{\delta Q}T$, we find, if $\delta Q$ is negative, entropy will decrease, which violate the law. In terms of disorder, by giving heat out, the cold object will get even colder and thus more ordered, which is not right as well.