Calculate the entropy change of 75.2 g of water that freezes into ice at 273.15 K. ΔHfus for water is 6.01 kJ/mol.?
- Steve4PhysicsLv 72 years agoBest Answer
Molar mass of water M = 2x1 + 16 = 18g
Amount of water: n = m/M = 75.2/18 = 4.178mol
Heat transferred when water freezes = -nΔHfus = -4.178mol x 6.01kJ/mol = -25.11kJ
(Negative as heat is removed, not added.)
Entropy change (ΔS) at temperature T is defined as:
ΔS = ΔQ/T = -25.11kJ / 273.15K = -0.0919 kJ/K = -91.9J/K