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:A. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` remains the sameB. `Delta S_(("system"))` increases but `Delta S_(("surroundings"))` decreasesC. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` increasesD. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` also decrease

The entropy change can be calculated by using the expression `DeltaS-(

1.	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:A. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` remains the sameB. `Delta S_(("system"))` increases but `Delta S_(("surroundings"))` decreasesC. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` increasesD. `Delta S_(("system"))` decreases but `Delta S_(("surroundings"))` also decrease
Answer» Correct Answer - C 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

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