# Show that book levitation by absorption of heat violates 2nd law

• Chemistry
• zenterix
zenterix
Homework Statement
Suppose someone makes the following claim.

A book with mass ##m=534\text{g}## rose spontaneously from the table on which it was resting to a height of ##h=3.20\text{cm}## above the table and hovered there for several seconds.

Assume the book accomplished this feat by irreversibly absorbing a large amount of heat from its surroundings. Assume it was a very hot day (##T=312\text{K}##) and that the surroundings are sufficiently large that their temperature was not affected by the book.
Relevant Equations
Compute the heat ##Q## that needs to be absorbed by the book.

Does this necessarily violate the 1st law of thermodynamics?

Compute the entropy change associated with the process.

Does this necessarily violate the 2nd law of thermodynamics?
Let's consider the book to be our system.

The book spontaneously absorbs heat from the surroundings and somehow converts this to gravitational potential energy.

Assuming gravitational potential energy is zero at the table top, the potential energy at ##3.2\text{cm}## above the table is ##mgh=0.1676\text{J}##.

The 1st law tells us that ##U## is extensive and conserved and that ##dU=dQ+dW##.

There is no work happening so ##dU=dQ## and ##\Delta U=Q=0.1676\text{J}##.

There is no violation of the 1st law: the energy absorbed by the book is taken from the surroundings and the total energy of system plus surroundings is constant.

My question is about the calculations involving entropy.

$$dU=TdS-pdV+\mu dN$$

$$=TdS$$

$$=dQ$$

since there is no volume change in the system or change in the number of particles in the system.

$$dS=\frac{dQ}{T}$$

$$\Delta S=\frac{Q}{T}$$

However, apparently, we can only use these relationships for reversible processes which is not the case for the system process.

Note that the surroundings have constant temperature by assumption.

Only the surroundings are large enough so that the heat exchange is reversible.

For the surroundings,

$$\Delta S=-\frac{Q}{T}=-0.000537\mathrm{\frac{J}{M}}$$

In this expression the heat ##-Q## and the temperature ##T## refer to the surroundings, not to the system.

I'd like to understand the above snippet a bit better.

Now, if we accept the snippet then it means that we have a spontaneous process at constant ##V## and ##N## for which ##\Delta S<0## and this violates the 2nd law of thermodynamics.

So here are some questions I have

1) Where does ##\Delta S=-\frac{Q}{T}## come from?

I think it comes from simply considering the differential of entropy for the surroundings

$$dU_{surr}=T_{surr}dS_{surr}=dQ_{surr}=-dQ$$

$$dS_{surr}=-\frac{dQ}{T_{surr}}$$

$$\implies \Delta S_{surr}=\frac{-Q}{T_{surr}}$$

2) Why does ##dS=\frac{dQ}{T}## only apply to reversible processes?

3) What does it mean that the surroundings are large enough that heat exchange can be reversible?

Here is my current understanding

1) I've seen reversible processes in the context of a system containing an ideal gas. In calculations, it allows us to use the ideal gas law.

2) In my thermodynamics course, the function ##S(U,V,N)## was postulated to exist at the very beginning of the course.

From this function we computed the differential form and solved for ##dS## thus implicating the existence of a function ##U(S,V,N)##.

We then defined ##T, P##, and ##\mu## in terms of partial derivatives of ##U##, though we can also do this with partials of ##S##.

A reversible process was introduced as a process that remains at equilibrium with the surrounding environment as the state change occurs. Both temperature and pressure are the same for system and surroundings at all times during the reversible process.

In the case of the problem in the OP, have some sudden transfer of heat to the book. I think we can say that the book is at constant pressure and volume. I don't think we can say this about temperature. These are guesses.

Now, I remember reading a book last year about thermodynamics that now that I think of it seems to clear things up a bit. Here is what I conclude based on that reading.

Since the process did not occur in thermodynamic equilibrium there is no equation of state to describe it. Therefore, we can't actually say much about any of the thermodynamic coordinates for the book.

The surroundings, however, are in thermodynamic equilibrium. This depends on our assumption that the temperature of the surroundings doesn't change.

For the surroundings, therefore, we can use known equations of state.

In summary, here are my answers to my three questions

1) Correct.

2) Only reversible processes have equations of state and differential forms (like ##dS=\frac{dQ}{T}##) are based on equations of state.

3) The assumption that the temperature of the surroundings doesn't change because of the book is justified by it being very large.

And to further conclude the actual problem, we showed that if the book were to levitate as claimed then the entropy of the surroundings would decrease. This contradicts the 2nd law of thermodynamics.

How would you include the work to lift the book in a reversible way?

It is not clear to me what work would be in the context of this book levitation.

As I stated initially, it seems the book absorbs heat and then in some way unbeknownst to me transforms this into gravitational potential energy. I have no idea how the displacement of the book occurs.

There doesn't seem to be any ##pV## work happening either.

Perhaps you could give me some hints as to what you are getting at.

Last edited by a moderator:
what is the change of U if you lift a book of mass m in earths gravitational acceleration g to height h?

##mgh=0.1676\text{J}## as present in the OP.

Last edited by a moderator:
So dW=mg dh

There are two phases - lifting, levitation. Work done during the lifting is mgh. There is no work done during the levitation, as the book doesn't move (force × displacement = 0).

Perhaps we understand the word levitate differently - for me it means it doesn't move at all. It has to be lifted first, but than it just stays in a position.

DrDu
I think the solution to the problem is as always to consider a reversible process between the same initial and final states. Lifting a book reversibly can be done with Q=0 and W=mgh. So Delta S=0 and Delta U =mgh.
In the strange process we are considering, Delta S of the book is still 0, but the entropy of the surrounding has changed by -Q/T as correctly argued by zenterix. If you use the reversible process to bring the book back to the table you can transform the potential energy into work. So we would have a process in which nothing else has happened than the conversion of heat to mechanical work, which is a clear violation of the second law.

DrDu said:
I think the solution to the problem is as always to consider a reversible process between the same initial and final states. Lifting a book reversibly can be done with Q=0 and W=mgh. So Delta S=0 and Delta U =mgh.
In the strange process we are considering, Delta S of the book is still 0, but the entropy of the surrounding has changed by -Q/T as correctly argued by zenterix. If you use the reversible process to bring the book back to the table you can transform the potential energy into work. So we would have a process in which nothing else has happened than the conversion of heat to mechanical work, which is a clear violation of the second law.
I understood your reasoning up until the last point.

"Nothing else happened than the conversion of heat to mechanical work, which is a clear violation of the 2nd law".

Lifting the book reversibly means we move it very slowly with infinitesimal work $dW=mgdh$. The total work is $W=mgh=\Delta U$. The change in entropy is 0 since there was no heat involved.

If we now reversibly bring the book back down to the table, it means we are very slowly bringing it down such that there is no acceleration and $W=-mgh$.

Again, no change in entropy.

So, considering the same cyclic state change but with an irreversible process, the change in entropy for the book must also be zero.

On the other hand, the surroundings lost some heat and so entropy went down.

So where was the 2nd law violated?

1) $\Delta S<0$ for the composite system, which is impossible by the 2nd law.

2) The book being lifted and then levitating required heat. The process of being lifted seems to be spontaneous (irreversible) and so $\Delta S$ should be larger than 0.

One formulation of the second law states, that it is impossible to have a cyclic process in the course of which heat is completely transformed into work. If you couple the spontaneous levitation with a reversible process only generating work, this obviously violates the second law. We also calculated that the entropy of the system + surrounding decreased, which violates the second law, too.
Letting a book drop on the table generates heat and is clearly an irreversible process with Delta S> 0 (of the surrounding). The spontaneous levitation is the time reversed process which clearly violates both the irreversibility and the second law.

Lord Jestocost
DrDu said:
The spontaneous levitation is the time reversed process which clearly violates both the irreversibility and the second law.
One has to understand that the second law of thermodynamics is a "secondary law". Sir Arthur Stanley Eddington in “THE NATURE OF THE PHYSICAL WORLD” (Cambridge, At the University Press (1929)):

"Primary and Secondary Law. I have called the laws controlling the behaviour of single individuals "primary laws”, implying that the second law of thermodynamics, although a recognised law of Nature, is in some sense a secondary law. This distinction can now be placed on a regular footing. Some things never happen in the physical world because they are impossible; others because they are too improbable. The laws which forbid the first are the primary laws; the laws which forbid the second are the secondary laws…..

….But for all its completeness primary law does not answer every question about Nature which we might reasonably wish to put. Can a universe evolve backwards, i.e. develop in the opposite way to our own system? Primary law, being indifferent to a time direction, replies, "Yes, it is not impossible". Secondary law replies, "No, it is too improbable". The answers are not really in conflict; but the first, though true, rather misses the point." [Bold by LJ]

sbrothy and weirdoguy

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