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?

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    $\begingroup$ 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


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, 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. $\endgroup$ Apr 22, 2012 at 2:12

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