# Heat Engines 100% efficiency

1. Jul 21, 2010

### gkangelexa

My question involves heat engines. I understand that a heat engine typically uses energy provided in the form of heat to do work. According to the 2nd law of thermodynamics, however, not all heat energy can be converted into work energy, meaning that heat engines are not perfect. At least some of the energy must be passed on to heat a low-temperature (cold) energy sink. "A heat engine with 100% efficiency is thermodynamically impossible."

Why is this so? Why can't do we need the cold reservoir? Why can't all the heat energy become work energy?

Thanks!

2. Jul 21, 2010

### K^2

Are you asking why converting heat energy to work would decrease entropy and violate second law of thermodynamics?

Or are you asking why there is such a thing as second law of thermodynamics?

These are two entirely separate questions.

3. Jul 21, 2010

### Staff: Mentor

Without a temperature difference, heat can't flow at all, so there isn't any energy to harness.

It may be instructive to look at/think about specific examples such as a hydroelectric dam, a car engine, and a steam engine. All have an area of high energy and an area of low energy - they harnesses the flow of energy from high to low by getting in the way of that energy flow.

4. Jul 21, 2010

### K^2

I don't think he expects heat to flow. He just wants to convert one form of energy to another.

5. Jul 21, 2010

### Staff: Mentor

That's the point: you can only convert it from one form of energy to another if it flows.

Consider the hydoelectric dam: It is an elevated reservoir of water, which has potential energy. How would you convert this potential energy to kinetic or electrical energy without making the water flow?

6. Jul 21, 2010

### K^2

Look you don't have to explain to me that any form of energy conversion/transfer involves work, and therefore, displacement.

But I don't think it's clear to the OP. Chemical energy becomes thermal with no apparent displacement. I think he expects something similar to work in reverse. From heat energy into stored chemical/electrical energy that can be converted directly to mechanical work with no significant losses.

That's why I'm asking if he's not clear on why 2nd law applies, which has to do with heat flows, or why there is such a law, which can also be discussed on microscopic scale.

7. Jul 22, 2010

### gkangelexa

I guess I'm asking both....(and it's a she :-) )

8. Jul 22, 2010

### K^2

I was trying to come up with a simple explanation. By 3rd or 4th page, I realized, I don't have one. Not a simple one, at any rate. So maybe I'm missing something myself.

Basically, keep in mind that heat is composed of several types of mechanical energy intrinsic to the substance. There is kinetic energy of molecules, there is potential energy of the fields keeping them together, etc.

If you take the ideal gas, as a simplest example, put it into a cylinder with a piston, and allow gas to expand to infinity with no external pressure and perfectly insulated cylinder, you can get 100% of gas' internal energy in mechanical work. But your final temperature and pressure are both zero. That's simply not attainable in real world.

More realistically, say you take a cylinder, ignite something inside, releasing amount of heat Q, and let the cylinder expand until pressure is once more equal to external. Lets say putting in Q of heat took temperature from T0 to T2, and expansion took temperature down to T1. Energy you put in is is Q=CV(T2-T0). Work you get out W=CV(T2-T0). I derived all of this in the long explanation with all the physics, but it really doesn't matter. What you have is coefficient of efficiency CoE = W/Q = (T2-T1)/(T2-T0) < 1. So it's not a problem of extracting energy efficiently. It's the problem extracting all of it, because whatever your "exhaust" is going to be, it's going to be carrying off some of the energy, unless you can bring its temperature to absolute zero. And to cool something to absolute zero, you need something already at absolute zero.

9. Jul 22, 2010

### gkangelexa

wow thanks.. i think that sorta makes sense.. i'll work some problems and ask you again if i have more questions

10. Jul 22, 2010

### QuantumPion

Try thinking about it backwards, it makes more sense intuitively.

It is impossible for a hot object to spontaneously get hotter or a cold object to spontaneously get colder than its surroundings, without work being done or some chemical reaction taking place.

Therefore, it is impossible to make a perfect refrigerator which takes heat from a cold reservoir and puts it in a hot reservoir without work being input.

Now, reverse it! While it is possible to make a heat engine which takes heat from a hot reservoir and puts it in a cold reservoir without doing work (a space heater), it is impossible to make a heat engine which takes heat from a hot reservoir and puts it in to work without a cold reservoir. This would be the opposite of a perfect refrigerator.

And if it were physically possible to make a perfect refrigerator, that would mean it would be physically possible for my tea mug to spontaneously get hotter while sitting on my desk without me having to make a trip to to the microwave.