# Two isolated metal cubes: Negative ΔS? Thermodynamics

1. May 31, 2014

### Nikitin

Hi. So let's say you have two identical cubes with expansion coefficients of approx. zero. These cubes are thermally in contact with each other, and otherwise isolated from the surroundings. One cube has a higher temperature than the other, so heat will flow from the hotter cube and to the cooler.

For argument's sake, let's assume we can make it flow reversibly in such a way that the temperature difference will run a Carnot engine that can do useful work on the surroundings. (You guys are probably already familiar with this common exergy problem.)

But here's my problem: If heat flows reversibly from the hot cube and too the cold, the entropy change of the system will be zero. Further, if we use the generated work to, say, compress a gas, we can make the total entropy change of the universe to be negative.

Is this allowed by the rules of thermodynamics? A process in which the final result is a negative change in entropy for the universe seems sketchy to me...

Last edited: May 31, 2014
2. May 31, 2014

### Nikitin

By the way, does the 2nd law of thermodynamics say "The entropy change for the universe must be positive for any cyclic process" or does it say "The entropy change for the universe must be either positive or zero for any cyclic process"?

If the latter wording is correct; does that mean a carnot engine and reversible processes are not declared impossible by thermodynamics? Rather, they are impossible due to practical considerations?

3. May 31, 2014

### Staff: Mentor

If the gas is compressed adiabatically and reversibly, then its entropy change will be zero. Even though its pressure will be higher, its temperature will also be higher, and the two effects on entropy will exactly cancel. If the compression is non-adiabatic, then the temperature of something else in the surroundings will have to increase.

Chet

4. May 31, 2014

### WannabeNewton

Carnot engines are perfectly allowed by thermodynamics, as are more general reversible processes. A heat engine operates between a hot reservoir of fixed temperature $T_h$ and a cold reservoir of fixed temperature $T_c$. It draws heat $Q_h$ from $T_h$, converts some of it to work $W$, and dumps the rest $Q_c$ into $T_c$. In this process one has the hot reservoir losing entropy $\frac{Q_h}{T_h}$ and the cold reservoir gaining entropy $\frac{Q_c}{T_c}$. The heat engine itself has a net entropy change of zero since it undergoes a cyclic process. The second law of thermodynamics simply states that $\Delta S = \frac{Q_c}{T_c}-\frac{Q_h}{T_h} \geq 0$ for general heat engines so that a process in which $Q_h = W$ becomes impossible, which can be shown to be equivalent to the statement that heat cannot flow spontaneously from a colder body to a hotter body. The statement $\Delta S = \frac{Q_c}{T_c}-\frac{Q_h}{T_h} \geq 0$ becomes for a Carnot engine $\Delta S = \frac{Q_c}{T_c}-\frac{Q_h}{T_h} = 0$ since the entire process is reversible by hypothesis for Carnot engines. This simply leads to the statement that Carnot engines have maximum possible efficiency and moreover that all Carnot engines have the same efficiency $\eta = \frac{T_h - T_c}{T_h}$.

Of course Carnot engines are practically useless since they tacitly assume, on top of employing quasi-static processes, that the Carnot engine is, at the isothermal expansion stage, at a temperature infinitesimally lower than $T_h$ which makes for painstakingly slow heat flow into the Carnot engine and similarly a temperature at the isothermal compression stage that is infinitesimally higher than $T_c$ thus leading to painstakingly slow heat rejection into the cold reservoir.

Last edited: May 31, 2014