Question

10 g of ice at 0° are added to 20 g water at 90° in a thermally insulated flask of Negligible heat capacity. The heat of fusion of ice is 6 KJ/mol, change in entropy for system ..? Cp = 75.42 KJ/mol





Urgently Required The Answer...​

Answer 1

10 g of ice at 0° are added to 20 g water at 90° in a thermally insulated flask of Negligible heat capacity. The heat of fusion of ice is 6 KJ/mol, change in entropy for system ..? Cp = 75.42 KJ/mol

Explanation:

The heat gained by ice is equal to the heat lost by water

Energy that is required to melt the ice is 10*6Kj/mol = 60000joules

10 g H₂O (s), 273 K H₂O(l) 273 K H₂O(l), T K , Two entropy changes, ∆S1, ∆S2

90 g H₂O(l) 363 K H₂O(l), T K, ∆S3

Heat balance equation, nLf + Cp(T-273) =

nCp(363-T); T = 306.5

∆S1 = nLf/T = 12.2 JK-1

∆S2 = n(Cp ln T2/T1) = 4.85 JK-1

∆S3 = n(Cp ln T2/T1) = -14.31 JK-1

∆S = sum = 2.75 JK-1, will correspond only to the system