# Second Law of Thermodynamics

1. Feb 17, 2008

### Oerg

Hi

I am going through an extreme headache trying to udnerstand the carnot engine and the second law of thermodynamics. To prove that the carnot engine (refreigerator engine) is the most efficient, my text included a prototype engine with a carnot refrigetrator together. So, the author says that the work done in the heat engine drives the carnot engine and we assume that the prototype engine is mroe efficient than the carnot engine. Then he says that because the heat engine is more efficient than the carnot one, heat is expected to flow from the cold to the hot reservoir. ( I agree so far). Then because of thee second law of thermodynamics, this cannot be the case because the net work done is zero. Therefore, the carnot engine must be mroe rfficient than the protoype one.

This is where i got lost. What is so special about the carnot engine that we cannot place a prototype refrigerator in its place?

Second, a carnot engine has alternate adiabatic and isothermal compressions/expansions and is a reversible engine. What is the significance of this in the example above? What is the connection between these adiabatic and isother compresisons expansions that makes the engine reversible?

2. Feb 17, 2008

### Lojzek

The special property of Carnot engine/refrigerator is that when combined, engine can power the refrigerator with the exact power that the refrigerator needs to pump heat back from cold to hot reseirvor.
This system is on the border of being a perpetuum mobile of second type (which can either pump heat to higher temperature or change heat from a single temperature reservoir into work).
The second law of thermodinamics says that perpetuum mobile of 2. type does not exist. However if you would improve the system Carnot engine+Carnot refrigerator with a slightly better engine of refrigerator, you would get perpetuum mobile.
So a better engine/refrigerator can not exist.

3. Feb 17, 2008

### Oerg

i still dun understand. How is this special property related to a carnot heat/refrigerator engine being reversible or having alternate adiabatic and isotherm compresisons and expansions?

4. Feb 17, 2008

### Lojzek

Alternate adiabatic and isotherm compressions/expansions are descriptions how the Carnot engine works. The mentioned "special property" of this engine/refrigerator can only be derived by integrating work dA=-p*dV and heat dQ=dU-dA=m*cv*dT-dA along the cycle of the engine/refrigerator (assuming it uses ideal gas). After you calculate the total work, Q1 and Q2 (heat exchanged from warm and hot reseirvoir) for both the engine and refrigerator, you can verify that one produced as much work as another one uses (if it transfers as much heat as the other one, but in different direction).

The efficiencies are independent on the temperatures of reservoirs, mass of gas and the lowest/highest volume of the cylinder, so the Carnot system always has the "special property", no matter how you chose engine's parameters. You alway get efficiency
A/Qhigh= (Thigh-Tlow)/Thigh for C. engine and
A/(Qhigh)=(Thigh-Tlow)/Thigh for C. refrigerator, so there is possible to balance both work and heat so that the system uses no work and transfers no heat.

Last edited: Feb 17, 2008
5. Feb 17, 2008

### Oerg

geeeez i cant think right now, maye I should sleep first and work on this again

6. Feb 17, 2008

### Mapes

1. We have two reservoirs at different temperatures $T_2>T_1$ (an energy difference) and we want to get some work out of the situation.

2. The Second Law says that total entropy $S_\mathrm{universe}$ cannot decrease; the most efficient scenario is one where $S_\mathrm{universe}$ stays the same. This type of process is called reversible, and reversible heating within the universe also transfers entropy: $\Delta S=\frac{Q}{T}$.

3. Work is the transfer of energy with no accompanying transfer of entropy.

4. Therefore we cannot convert the heat flowing from reservoir $T_2$ to reservoir $T_1$ entirely into work! That would result in $S_\mathrm{universe}$ decreasing. We can extract some work, but we must heat reservoir $T_1$ by some amount. How much? Enough energy to transfer the entropy we removed from reservoir $T_2$.

5. You can calculate that if you use reservoir $T_2$ to heat your engine to gain energy $Q$, you must offload the amount $Q\left(\frac{T_1}{T_2}\right)$ to $T_1$ by heating. You get to extract the remainder as work. Energy is conserved, and because you did everything reversibly, total entropy is conserved.

6. The Carnot cycle is the process described in (5). To answer your question about cycle components, the adiabatic (and reversible) steps are the ones where we extract work (recall that adiabatic reversible means isentropic: $Q=0$). The isothermal (and reversible) steps are when reservoir $T_2$ heated the engine, or when the engine heated reservoir $T_1$.

7. The Carnot engine is an idealization. No real heating step can ever be reversible, since we need a temperature gradient for heat to flow in the real world.

7. Feb 18, 2008

### Oerg

Thanks for your reply. I still dont understand the method of proof by contradiction that the carnot engine is the most efficient. In the text book, the proof did not make use of anything that is associated with a carnot refrigerator. In other words, I still do not see the difference if we were to use a prototype refrigerator instead of a carnot one.

8. Feb 18, 2008

### Mapes

It's from point (7) above. Any real refrigerator requires a temperature gradient for heat to flow. But you'll find that this gradient decreases efficiency because it creates entropy.* If we let the gradient tend to zero, we have the Carnot refrigerator, the most efficient possible engine.

*Try to prove this to yourself, it may make things clearer.

9. Feb 19, 2008

### Oerg

ok, I have realised that for the contradiction method to work, one of the important factors is that the carnot engine's efficiency is only dependent on the temperature difference between the 2 reservoirs of energy. Whereas the efficiency of other engines are not constant as long as there is a change of work supplied to the engine(refrigerator). i realised this must be the case. Am I right?

10. Feb 19, 2008

### Mapes

Sounds good to me.

11. Feb 19, 2008

### Oerg

hey thanks for the help. I understand now.

12. Feb 19, 2008

### GT1

Why there is no transfer of entropy when work is done ?

13. Feb 19, 2008

### Oerg

This is because the change in entropy is defined as the change in the transfer of heat at a specified temperature. The more general form of the second law of thermodynamics state that the change in entropy is always positive for a closed system.

Last edited: Feb 19, 2008
14. Feb 19, 2008

### Mapes

To do work on a system is to change the energy levels of the system microstates without changing their population. (Heating a system has the opposite effect.) Entropy, being a measure of the number of possible microstates only, does not change when energy is added through work.

Mathematically, we see this in the energy equation

$$dU=T\,dS-p\,dV+\gamma\,dA+F\,dl+E\,dQ+\cdots$$

for a closed system, where the entropy term (associated with heating) is distinct from any generalized displacements (volume, area, length, charge, etc.) associated with work.

15. Feb 20, 2008

### GT1

I still don't understand physically why work doesn't change the entropy- the number of degrees of freedom that the system has ,and why adding heat to the system can change the number of degrees of freedom that the system has.
I see it from equations, but i don't understand the physical source of it.

16. Oct 25, 2008

### Sohan

I don't really know whether this would make it simpler .....
Consider a combination of a heat engine and a refrigerator (which work's according to the carnot cycle) working between the same high and low temp reservoirs.Here we assume that the heat engine is more efficient than the ref.
let the ref extract an amount of heat QL from the lower temp reservoir and release an amount of heat QH to the higher temp reservoir(for which a work QH-QL has to be done)
Now let the heat engine extract the same amount of heat that was released by the ref from the high temp reservoir..i.e QH and rejects a an amount of heat QL1 to the low temp reservoir.hence work equal to QH-QL1 can be obtained from this heat engine.
Now since we have assumed that the heat engine is more efficient than the ref(working on carnot cycle)it follows that QH-QL1>QH-QL.....hence the work output from the heat engine can be used to drive the refrigerator.... but now if you look at it the net heat tranfer from the high temp reservoir is zero and hence it can be neglected and heat QH could be imagined as being directly supplied to the heat engine from the ref. So this system consisting of a heat engine and the ref is producing a net work by exchanging heat from a single reservoir.(low temp reservoir) The heat extracted from the low temp reservoir is equal to QL-QL1 ..and by some simple math it follows that the net work is also equal to QL-QL1.
This clearly violates the kelvin planck statement of the second law,hence impossible and we can conclude that our initial assumption that the heat engine is more efficient than ref(carnot engine,also known as reversible engine) is wrong...

Last edited: Oct 25, 2008