0
$\begingroup$

Second law of thermodynamics states that we cannot take heat from a hot source and make it 100% useful work. Is this because the heat is high disordered energy (ie molecules moving in a disorder manner) so we cannot harvest all these movements, because to make work we need movements to one direction only? Is this legit? If not how can we explain the reason behind the validity of 2nd law in its molecular basis? Also, how Carnot proves 2nd law? How are these two connected?

$\endgroup$
2
$\begingroup$

The second law of thermodynamics simply says that heat doesn't move from cold to hot on its own. That is the fundamental definition of temperature. It's this simplicity and independence of any microscopic physics that makes thermodynamics so useful. You should embrace this simplicity instead of trying to find (non-existent) proofs for it from other principles. This is (probably) the most simple form the description of thermodynamics can take.

You can compare this to the analogous case of Newton's second law, which is the definition of inertial mass, the definition of force and which establishes a fundamental relationship between both and acceleration. We don't have to know anything about what mass is or what causes force to use Newton, but a large and useful body of physics can be derived from it.

Similarly, we don't need to know anything about molecules, order, disorder and statistical mechanics to use the second law of thermodynamics. Carnot's results and the absolute thermodynamic temperature scale can be derived from it (and the other laws) easily.

On the other hand, we can't derive the laws of thermodynamics from anything else without further assumptions, and the connection between statistical mechanics and thermodynamics requires the ergodic hypothesis, which is invalid for some rather trivial systems and which is hard to prove for others. Compared to the simplicity of the first, second and third laws, that's a huge can of worms and not that much more useful.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.