Exercise 1 | Liquids and Solids

Calculate the enthalpy change for converting 10.0 g of water at 25.0°C to steam at 135.0°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

25.0°C → 100.0°C: water

100.0°C → 135.0°C: steam

 

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


ΔHvap (H2O) = 40.67 kJ.mol-1

10.0 g of water ⇒ 5.56 x 10-1 mol of water

So, for this reaction ΔHvap (H2O) (in J) = 40.67 x 0.556 = 22.6 kJ


ΔH = ΔHvap (H2O) (in J) + qP (water) (in J) + qP (steam) (in J)

ΔH = ΔHvap (H2O) + mΔT cS (water) + mΔT cS (steam)

ΔH = 22.6 x 103 + 10.0 x (373-298) x 4.18 + 10.0 x (408-373) x 1.84

ΔH = 26.4 x 103 J = 26.4 kJ