Question

The entropy change can be calculated by using the expression, $△S=\frac{Q_{rev}}{T}.$ When water freezes in a glass beaker, choose the correct statement amongst the following:

• A. $△S_{system}$ decreases but $△S_{surroundings}$ remains the same.
• B. $△S_{system}$ increases but $△S_{surroundings}$ decreases.
• C. $△S_{system}$ decreases but $△S_{surroundings}$ increases.
• D. $△S_{system}$ decreases and $△S_{surroundings}$ also decreases.

Answer 1

The correct option is C $△S_{system}$ decreases but $△S_{surroundings}$ increases.

Freezing of water is an exothermic process (heat releases).
As water freezes from liquid to solid, randomness (or entropy) decreases, i.e., $△S_{system}$ decreases. Heat released during the process, is absorbed by the surroundings hence, $△S_{surroundings}$ increases.

Theory:

Second Law of Thermodynamics:

It states that as energy is transferred or transformed, more and more of it is wasted. The Second Law also states that there is a natural tendency of any isolated system to degenerate into a more disordered state.

Note: Entropy of universe increases in the course of a spontaneous change.

$\Delta S_{univ.}$ is greater than 0
$\Delta S_{univ.}=\Delta S_{sys}+\Delta S_{surr}$ > 0
For a spontaneous process in an isolated system, Entropy always increases
Also, When a system is in equilibrium, the entropy is maximum and the change in entropy, $\Delta S=0$.