- #1

zenterix

- 557

- 76

- 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.

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

$$=TdS$$

$$=dQ$$

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.

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?

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?