The entropy change can be calculated by using the expression \Delta S = (q_{rev}/T).
When water freezes in a glass beaker, choose the correct statement amongst the following :
(i) \Delta S (system) decreases, but \Delta S (surroundings) remains the same.
(ii) \Delta S (system) increases but \Delta S (surroundings) decreases.
(iii) \Delta S (system) decreases, but \Delta S (surroundings) increases.
(iv) \Delta S (system) decreases, and \Delta S (surroundings) also decreases.

Answers (1)

The answer is option (iii) \Delta S (system) decreases, but \Delta S (surroundings) increases.

Explanation: As we all know in the case of an exothermic process, heat is released, which leads to a decrease in the surrounding entropy and as a result, the system entropy increases. As freezing is an exothermic process, we choose this option.

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