Suppose there is an isolated system $A$ at time $(-\infty, t_1)$, whose entropy is $S=S_{max}$, i.e. it is at thermodynamical equilibrium.

Between moments $[t_1, t_2)$ the isolation is violated and system's entropy is decrased $S = S_{max} - S_d, \space\space\space 0<S_d <S_{max}$.

Question: the only way that this could happen is energy being added to system?

  • 1
    $\begingroup$ Also, entropy of the universe is maximised in equilibrium. You can shift entropy in and out of the system by dumping into a heat bath. $\endgroup$ – genneth Apr 20 '12 at 14:45

It's also possible that you put system A into a cold room and it has been cooling down...

Or system A is an ideal gas being held at a constant temperature -- in which case its energy is fixed -- and you have lowered its entropy by reducing its volume...

  • $\begingroup$ Putting A into cold room means doing work, i.e. using energy? The same for changing volume? $\endgroup$ – Mooncer Apr 20 '12 at 14:50
  • $\begingroup$ It does not necessarily require energy to move a box into a cold room. For example, maybe the box slides down a ramp into the room. When you shrink a box containing an ideal gas at constant temperature, energy goes out of you and into the constant-temperature-bath surrounding A. Energy is not being added or substracted from A; it is being transferred between two systems, neither of which is A. $\endgroup$ – Steve Byrnes Apr 22 '12 at 2:12

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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