Exercise 4 | Liquids and Solids

Calculate the enthalpy change for converting 2.00 mol of ice at -50.0°C to water at 70.0°C.


Data: cS (ice) = 2.09 J.g-1.K-1; cS (water) = 4.18 J.g-1.K-1; cS (steam) = 1.84 J.g-1.K-1

ΔHfus (H2O) = 6.01 kJ.mol-1; ΔHvap (H2O) = 40.67 kJ.mol-1

-50.0°C → 0°C: ice

0°C → 70.0°C: water


cS = qPm T   with qP = energy added as heat (in J)


ΔHfus (H2O) = 6.01 kJ.mol-1

ΔHfus (H2O) (in J) = ΔHfus (H2O) (in J.mol-1) x nH2O

ΔHfus (H2O) (in J) = 6.01 x 103 x 2 = 1.20 x 104 J


n = mM ⇒ mH2O = nH2O x MH2O = 2.00 x 18.0 = 36.0 g


ΔH = ΔHfus (H2O) (in J) + qP (ice) (in J) + qP (water) (in J)

ΔH = ΔHfus (H2O) + m ΔT cS (ice) + m ΔT cS (water)

ΔH = 1.20 x 104 + 36.0 x (273-223) x 2.09 + 36.0 x (343-273) x 4.18

ΔH = 26.3 x 104 J = 26.3 kJ