The entropy change can be calculated by using the expression `DeltaS-(q_(rev))/(T)`. When water freezes in a glass beaker, choose the correct statement amongst the following:

Options

A. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` remains the same
B. `Delta S_(("system"))` increases but `Delta S_(("surroundings"))` decreases
C. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` increases
D. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` also decrease

Answer

C

Explanation

During the process of freezing energy is released, which is absorbed by the surroundings
`:. Delta S_("system") = (-q_(rev))/(T)`,
`Delta S_("surroundings") = (q_(rev))/(T)`
Therefore the entropy of the system decrease and that of surroundings increases