Question

Calculate the entropy change when 72.00 g of ice, at 273.2 K and 1.000 bar pressure, is melted and then heated to 298.2 K. The enthalpy of fusion of ice is 6.009 kJ/mole and the heat molar heat capacity of water at 1.000 atm is 75.43 J/ K mole

Please help! Can't figure this homework problem out.

Answer 1

Given mass of ice = 72 g

Mass of 1 mole of ice = 18 g

=> number of moles of ice = 72/18 = 4 moles

Total entropy on heating ice at 273.2 K to 298.2 K = entropy change associated with melting of ice + entropy change associated with heating of water

--(1)

AS total = AS melting + AS heating

-----------------(2)

NAHfusion ASfusion = T

--------------------(3)

ASheating = ncin 3)

From equation (2)

$\Delta$S_fusion = 4*6.009/273.2 = 0.08798 kJ = 87.98 J

From equation (3)

$\Delta$S_heating = 4*75.43*ln(298.2/273.2) = 26.41 J

Therefore from equation (1)

$\Delta$S_total = 87.98 + 26.41 = 114.40 J

Therefore, when 72 g of ice at 273.2 K and 1 bar pressure is melted and then heated to 298.2 K, the total entropy change is

$\Delta$S_total = 114.40 J