# Does the Second law of thermodynamics violate the First law of thermodyanmics?

Since entropy has the units of Joules per Kelvin, and the net entropy of the entire universe is always increasing. Does it mean that the energy of the universe is always increasing?

Edit1: To be more precise. Consider two systems. A glass of hot water ($$A$$) and rest of the universe $$B$$, at time $$t$$. We have $$[S_A + S_B]_{t + \Delta t} > [S_A + S_B]_{t}$$ and $$[E_A + E_B]_{t + \Delta t}= [E_A + E_B]_{t}$$. Where $$S_i$$ and $$E_i$$ are entropy and energy, respectively.

My question: A transfer of a given amount of energy ($$\Delta E$$) from $$A$$ to $$B$$ means: subtract the energy $$\Delta E$$ in $$A$$ and add it to $$B$$, so the total energy is constant. Why isn't this true for the entropy? Why doesn't the subtraction of a given amount of entropy $$\Delta S$$ from $$A$$, result in exactly the same amount of addition in $$B$$?

• But energy is just Joules. Entropy is not energy. They have different units. Sure, they are related, but there are many instances in physics where quantities with related units are not correlated. – Aaron Stevens Aug 13 at 19:55
• I have updated my answer to respond to the question at the end of your edit. Hope it helps clarify things for you. – Bob D Aug 14 at 14:02
• @Patrick: Don't use edits to comment - wait until you have enough reputation (10 is enough, I believe - and you can get some reputation by making MEANINGFUL edits to other posts!) – Norbert Schuch Aug 29 at 17:21

Just because entropy has units of energy per degree kelvin and is always increasing does not mean energy is always increasing. The simplest example to show this is heat transfer that always flows naturally from a hot object to a cold object.

Let two objects be massive enough relative to the amount of heat transferred so that their temperatures do not change as a result of heat $$Q$$ transferring from the hot object at temperature $$T_{H}$$ to the cold object at temperature $$T_{C}$$ (i.e.,they are thermal reservoirs).

The change in entropy of the hot object is then

$$\Delta S_{hot}=-\frac{Q}{T_H}$$

It is negative because heat is transferred out of the hot object.

The change in entropy of the cold object is then

$$\Delta S_{cold}=+\frac{Q}{T_C}$$

It is positive because heat is transferred to the cold object.

The total entropy change is then

$$\Delta S_{tot}=+\frac{Q}{T_C}-\frac{Q}{T_H}$$

Since $$T_{H}>T_{C}$$, the total entropy change is $$\Delta S>0$$.

So the entropy has increased, but there has been no increase in energy. That is because the energy lost by the hot object in the form of heat exactly equals the energy gained by the cold object in the form of heat. Energy is conserved.

thanks. But what is it that actually increases the entropy?

Irreversible processes increase entropy, and all real processes are irreversible.

In the case of heat, it transfers spontaneously from a hot object to a cold object and is an irreversible process. It is irreversible because the same heat will not spontaneously go in the reverse direction, that is, from the cold object to the hot object. That has never been observed. If it did, in this example we would have $$\Delta S_{tot}<0$$, in violation of the second law. And yet, the reverse process does not violate the first law of thermodynamics. That is, energy would still be conserved if the reverse process occurred.

Why doesn't the subtraction of a given amount of entropy Δ𝑆 from 𝐴, result in exactly the same amount of addition in 𝐵?

It is because entropy is defined as heat (energy) transfer divided by the temperature at which the transfer occurs, or

$$dS=\frac{dQ_{rev}}{T}$$

and not energy alone.

The heat transferred out of your hot glass of water (A) occurs at a higher temperature than the heat transferred into into the environment (the rest of the universe, B) which, by necessity, has to be at a lower temperature.

From the definition of entropy, that means the increase in entropy of the environment is greater than the decrease in entropy of the glass of water, so that, overall $$\Delta S_{universe}>0$$.

Hope this helps.

• thanks. But what is it that actually increases the entropy? – Patrick Aug 13 at 20:04
• @Patrick I have answered your question in a revision to my answer. Hope it helps. – Bob D Aug 13 at 20:31

This is not how physics works. Units tell us nothing about how a quantity behaves. There are a million examples but here is one: kinetic energy has units of Joule but is not conserved (only the total energy is).

But to answer the question directly:

The first and second laws are two independent statements and one does not violate the other. Take a closed rigid box that is insulated. We can divide the box into two parts and distribute its mass and energy in any arbitrary way we want (one half hot the other cold or high pressure in one part and low in the other). If we remove the wall between the two parts and allow the system to reach equilibrium then:

• the total energy in the box will the same as before (first law)
• the total entropy will be higher than before (second law)

The increase of entropy indicates that the system has moved towards equilibrium. In this sense we may think of entropy as a counter that counts how may steps the system has taken as it approaches equilibrium.

• Kindly refer to my edits. – Patrick Aug 14 at 7:38
• @Patrick In your edited question you are asking (I am paraphrasing) "If energy is conserved, why isn't entropy also conserved?" You must give some argument to connect the two parts of the question. Otherwise it sounds as if you are asking "if energy is conserved, why isn't everything conserved?" – Themis Aug 15 at 15:03