The entropy change can be calculated by using the expression Δ S = (qrev/T). When water freezes in a glass beaker, choose the correct statement amongst the following :

Question

The entropy change can be calculated by using the expression Δ S = (qrev/T). When water freezes in a glass beaker, choose the correct statement amongst the following : (A) Δ Ssystem decreases but Δ Ssurroundings remains the same (B) Δ Ssystem increases but Δ Ssurroundings decreases (C) Δ Ssystem decreases but Δ Ssurroundings increases (D) Δ Ssystem decreases and Δ Ssurroundings also decreases

Tardigrade Admin · Accepted Answer

As water freezes from liquid to solid, randomness decreases, i.e., Δ Ssystem decreases. Heat released during the process, is absorbed by the surroundings hence, Δ Ssurroundings increases.

Q. The entropy change can be calculated by using the expression $Δ S = \frac{q _{re v}}{T}$ . When water freezes in a glass beaker, choose the correct statement amongst the following :

$Δ S_{sys t e m}$ decreases but $Δ S_{s u rro u n d in g s}$ remains the same

$Δ S_{sys t e m}$ increases but $Δ S_{s u rro u n d in g s}$ decreases

$Δ S_{sys t e m}$ decreases but $Δ S_{s u rro u n d in g s}$ increases

$Δ S_{sys t e m}$ decreases and $Δ S_{s u rro u n d in g s}$ also decreases

As water freezes from liquid to solid, randomness decreases, i.e., $Δ S_{sys t e m}$ decreases. Heat released during the process, is absorbed by the surroundings hence, $Δ S_{s u rro u n d in g s}$ increases.

Q. The entropy change can be calculated by using the expression ΔS=Tqrev​​. When water freezes in a glass beaker, choose the correct statement amongst the following :

ΔSsystem​ decreases but ΔSsurroundings​ remains the same

ΔSsystem​ increases but ΔSsurroundings​ decreases

ΔSsystem​ decreases but ΔSsurroundings​ increases

ΔSsystem​ decreases and ΔSsurroundings​ also decreases