0
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • 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, 2012 at 14:45

1 Answer 1

Reset to default
2
$\begingroup$

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...

Improve this answer
$\endgroup$
2
  • $\begingroup$ Putting A into cold room means doing work, i.e. using energy? The same for changing volume? $\endgroup$
    – Mooncer
    Apr 20, 2012 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, 2012 at 2:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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