# 2nd law of thermodynamics why?

1. Jul 23, 2011

### gkangelexa

Second Law of Thermodynamics says that "it is impossible to extract an amount of heat QH from a hot reservoir and use it all to do work W"

"Some amount of heat QC must be exhausted to a cold reservoir. "

my questions is.... why?

2. Jul 23, 2011

### Pythagorean

Short answer: energy must be conserved in the midst of a dynamical asymmetry.

Long answer:

Wiki references Clausius' paper elsewhere, on the author's page:

Clausius, R. (1850), "Über die bewegende Kraft der Wärme, Part I, Part II", Annalen der Physik 79: 368–397, 500–524. See English Translation: On the Moving Force of Heat, and the Laws regarding the Nature of Heat itself which are deducible therefrom. Phil. Mag. (1851), 2, 1–21, 102–119.

3. Jul 23, 2011

### arka.12

if you allow to extract heat from a reservoir without any work done then the most fundamental principle of energy conservation will not be valid........................

4. Jul 23, 2011

### Staff: Mentor

If you have a hydroelectric dam and you extract all of the available energy out of the water, it won't have any energy left to leave the turbine!

5. Jul 23, 2011

### gsal

I like to visualize heat flowing downhill, downhill in the temperature scale, that is. In other words, heat flows naturally from higher temperatures to lower temperatures.

If you want heat from a hot reservoir to do some work for you, you need to put a cold reservoir at the bottom of the hill to make it flow; then, during the climb down, such heat can produce some work for you while flowing through some kind of engine, but some heat needs to come out at the bottom of the hill and go into the cold reservoir...if there is no cold reservoir, then there is no bottom of the hill, in the first place...and all the heat that may be going into the engine is simply warming it up and not producing work...you need the engine to already be all warmed up and not needing any heat for itself in order for it to be able to convert some incoming heat into work.

did I loose you? I almost lost myself!

Maybe an analogy could be a hydro plant at a damn...there is upstream and there is downstream...the water would flow naturally between these two points at different heights...you would like to get some work out of the water, so you put a turbine in the middle...but if there is no such thing as the lower height downstream point, water wouldn't flow in the first place. Granted, the same amount of water exists at both ends, but the one below has a lot less potential energy.

does this help?

6. Jul 23, 2011

### Truecrimson

The precise statement is
Note the word "sole result" here. It is possible to convert heat into work completely, such as in an isothermal expansion of an ideal gas. But the process produces additional change in the universe. (In this case, it is the volume of the gas itself that has changed.)

Without loss of generality, we can consider the Carnot engine, which is an idealization by the way. When the engine absorbs heat by an amount $Q_H$, its entropy increases by $\frac{Q_H}{T_H}$. If the engine does not dump some heat, and hence entropy, into a cold reservoir, it cannot return to its original state. This should more or less answer your question.

However, as to why the second law, as stated, is true, you have to know the microscopic interpretation of the second law in statistical mechanics. There, you will see that the second law does not hold absolutely, but only statistically; i.e. it is overwhelmingly unlikely to violate the second law.

7. Jul 23, 2011

### atyy

The statement you quoted is the Kelvin statement of the second law of thermodynamics.

There is an equivalent statement called the Clausius statement, which says heat cannot flow from cold to hot unless work is done.

If you accept the Clausius statement as being more obvious, then it explains the Kelvin statement.

OTOH, if you take it as a law (inexplainable but true), then the Kelvin statement explains the Clausius statement.

8. Jul 23, 2011

### 256bits

My postulate is:
If the cold sink was at absolute zero, the Thot-Tcold = Thot, efficiency would be 100%, and all heat could be turned into work.

9. Jul 23, 2011

### gkangelexa

Why must it dump "entropy" in order to return to its original state?

10. Jul 23, 2011

### Truecrimson

Because entropy is a state function, a property of a system like internal energy and volume.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook