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 and $△S$(surrondings) also decreases
• B. $△S$(system) decreases but $△S$(surrondings) remains the same
• C. $△S$(system) decreases but $△S$(surrondings) increases
• D. $△S$(system) increases but $△S$(surrondings) remains the same

Answer 1

The correct option is C $△S$(system) decreases but $△S$(surrondings) increases

Solution:

$△S=\frac{q_{rev}}{T}$

Freezing of water is an exothermic process, so system releases the heat to the surrounding;
$△S_{system}=\frac{-q_{rev}}{T}$

(- sign indicates loss of heat by the system)

$△S_{surrounding}=\frac{q_{rev}}{T}$

(+ sign indicates gain of heat by the surrounding from system)

The entropy of the system will decrease and that of surrounding will increase.

Hence, option (c) is correct