# second law of thermodynamics

I'm a high school rookie learning thermodynamics right now by myself. I got really confused that the second law of thermodynamics states that the entropy of the universe is always increasing. If the entropy is not positive, the process is taking place spontaneously. But when I look at the AP reference book, which says if the entropy of a reaction is negative, it's taking place in the opposite direction spontaneously, I got confused. Is it a hypothetical process? Can the entropy be negative and still happen in the opposite direction? Or does it require work?

• You understood your book all wrong. It's not the entropy of the reaction that matters, it's another thing entirely called Gibbs free energy (that includes entropy). See @maxpesa's answer. – André Chalella Nov 11 '14 at 20:12

the entropy of the universe is always increasing

True. Let's call this the total entropy. (Well, almost true, since the entropy of the universe remains constant for a reversible process).

When a hot stone is dropped in cold water, it's entropy decreases (it gets colder), but the water increases it's entropy at the same time (it receives heat and gets slightly warmer). At any times, when an entropy decrease is happening, an entropy increase is happening somewhere else. The funky thing is that this entropy increase always is numerically larger than the decrease.

You can calculate entropy change for a part of such a system with:

$$\Delta S=\int_1^2 \frac{1}{T} \mathrm{d}Q$$

When you do that for all parts and sum the entropy changes, it is always possitive $\sum \Delta S>0$ (again apart for the ideal case of a purely reversible process, where $\sum \Delta S=0$).

If anyone says entropy decrease it is because they are not talking about the whole system.

An example

Mix water at $T_{a1}=20\mathrm{^o C}$ with the same amount of water at $T_{b1}=80\mathrm{^o C}$.

The water-mix now with double the mass finds an inbetween equillibrium temperature at $T_2=50\mathrm{^o C}$.

• Entropy change for the cold water: $\Delta S_a=\int_1^2 \frac{1}{T} \mathrm{d}Q_a=\int_{T_{a1}}^{T_2} \frac{1}{T} (mc\,\mathrm{d}T)=mc\int_{T_{a1}}^{T_2} \frac{1}{T}\mathrm{d}T$

• Entropy change for the warm water: $\Delta S_b=\int_1^2 \frac{1}{T} \mathrm{d}Q_b=\int_{T_{b1}}^{T_2} \frac{1}{T} (mc\,\mathrm{d}T)=mc\int_{T_{b1}}^{T_2} \frac{1}{T}\mathrm{d}T$

$m$ and $c$ are the same for the two, but the integrals are (numerically) different because of $T_{a1}\neq T_{b1}$. So there must be a change in entropy. That this change is positive is clear if we write them out and solve the integrals:

$$\Delta S_a=mc(\ln T_2-\ln T_{a1})=mc\ln \frac{T_2}{T_{a1}} \text{ and}$$ $$\Delta S_b=mc(\ln T_2-\ln T_{b1})=mc\ln \frac{T_2}{T_{b1}}$$

And the total entropy change will be: $$\sum \Delta S=\Delta S_a+\Delta S_b=mc(\ln \frac{T_2}{T_{a1}}+\ln \frac{T_2}{T_{b1}})=mc\ln (\frac{T_2^2}{T_{a1}T_{b1}})=mc\ln \left(\frac{T_2^2}{(T_2-T_{diff})(T_2+T_{diff})}\right)=mc\ln \left(\frac{T_2^2}{T_2^2-T_{diff}^2}\right)$$

This is some extra rearranging just to prove that the sum is never negative $\sum \Delta S > 0$ because the fraction never is: $\frac{T_2^2}{T_2^2-T_{diff}^2}> 0$ (since $T_{diff}>0$).

So this is a small proof that heat transfer is an example of an irreversible process that will cause the total entropy to increase.

• I know this is somewhat of a stupid question but how is water the surrounding and stone the system....can we reverse them also? – Hydrous Caperilla Apr 2 '18 at 6:45
• @HydrousCaperilla It doesn't matter. Those are just words, or labels if you will, that we use to distinguish two things. In your case you will focus on the stone. Everything else would then be "the surroundings". Reversing them is nothing more than a wording change; they changes none of the physics. – Steeven Apr 2 '18 at 13:02

Nor entropy, neither enthalpy alone describe the spontaneity of a reaction (and temperature is also important): it is described by Gibb's free energy, which states $\Delta\ G = \Delta\ H - T *\Delta\ S$

• If $\Delta\ G$ is positive, the process is endoergonic so not spontaneous;
• if $\Delta\ G$ is negative, the process is esooergonic so spontaneous.

If you are talking about (chemical) reactions, you should consider this.