# 100% Efficient Transfer of Energy

1. Sep 11, 2010

### grounded

Is there ANY way for ANY type of energy to be transformed from one form to another with no losses, a perfect 100% efficiency?

Thank you...

2. Sep 11, 2010

### ZapperZ

Staff Emeritus
Gravitational PE to KE. Drop a ball in low vacuum.

Zz.

3. Sep 11, 2010

### Staff: Mentor

Electrical to heat.

4. Sep 11, 2010

### grounded

I'm not arguing, just trying to understand if the 2nd law of thermodynamics applies only for heat to mechanical or if it applies to all forms off energy tranformation.

It seems logical to me that the amount of energy required to stop the ball from falling should always be more than just the energy needed to hold the weight of the ball.

My logic is that by suspending the ball, it will somewhat be compressed causing some amount of heat loss energy, although really small. ??

Also, I'm not to sure that PE turns into KE, it seems more like PE stops KE from increasing, also KE should increases with distance where PE should decrease with distance.

Any other examples?

5. Sep 11, 2010

### grounded

Wait, I'm wrong.
PE is not the energy required to hold the object from falling, its basically what KE can be at different distances. In that sense they are equal.

6. Sep 11, 2010

Its not possible to get 100% Efficient energy transfer, Only under Ideal condition it is possible, Ideal conditions are assumptions made by us.

In most of energy transfers taking place in the universe, losses take place generally due to friction, or the energy might be lost to atmosphere if its not a perfect closed system.

7. Sep 11, 2010

### ZapperZ

Staff Emeritus
They are? This doesn't make any sense. What if, instead of dropping it, I roll a wheel down an inclined plane. Do you think the KE of that wheel is the same as the dropped ball after it has gone through the same height? If it isn't, then how can PE be nothing more than "what KE can be at different distance" when these two examples produced DIFFERENT KE?

Zz.

8. Sep 11, 2010

### Dr Lots-o'watts

Energy is always transferred from one input type to many output types (the application + "losses") with perfect efficiency (law of conservation). The issue here is to single out one of the output types. It comes down to a matter of how close you get to what you actually want. It comes down to how perfect is heat isolation, how perfect is your crystal, how perfect is your vacuum etc. depending on what types of energy you work with.

9. Sep 11, 2010

### grounded

I must be missing something simple, I thought the PE of the ball at the top of the hill is equal to the KE of the ball at the bottom of the hill? (minus friction)

Isn't the potential amount of energy the ball can have equal to the the energy the ball can obtain by falling or rolling?

10. Sep 11, 2010

### ZapperZ

Staff Emeritus
Not if the ball is rolling! The original PE has been transferred into the KE of the ball and the rotational energy.

Still, this is besides the point. By definition, the potential energy is never defined as what it's KE can be. The potential energy for an electron in a stable state in an atom has nothing to do with what its KE can be when it is liberated from that atom.

Zz.

11. Sep 11, 2010

### Staff: Mentor

Look at the definition of change in entropy, as it appears in most textbooks:

$$dS = \frac{dQ}{T}$$

If dQ = 0, then dS = 0.

12. Sep 11, 2010

### Bob S

If you look at the transfer of energy from one nuclear isomer to another identical nucleus by way of an emitted photon, the energy has to be conserved. Look in particular at Fe-57 isomer used in Mossbauer experiments. The emitted photon energy spread is
~5 *10-9 eV compared to the emitted photon energy of 14.4 *103 eV, so the energy transfer has to be 100% to within 1 part in 1012 to absorbed by another Fe-57 nucleus. The major energy loss is due to nuclear recoil because of the finite nuclear mass. This can be minimized by 'locking" the Mossbauer atoms in a crystal lattice. (Sorry, this is not Classical Physics.)

Bob S