Question

Calculate the enthalpy change (in joules) involved in converting 5.00 grams of water at 14.0 °c to steam at 115 °c under a constant pressure of 1 atm. the specific heats of ice, liquid water, and steam are, respectively, 2.03, 4.18, 1.84 j/g-k, and for water δhfusion = 6.01 kj/mole and δhvap = 40.67 kj/mole. hints

Answer 1

Enthalpy change is the term given to the concentration of the heat absorbed or evolved in a reaction carried out at a constant pressure. It is given by the symbol ΔH.

Enthalpy change = m (Cwater × dT + Hvap + Csteam × dT

dHvap = 40.67 kJ/mol = 40.67 × 103/18 = 2259.4 J/g

Enthalpy change = 5.00 [4.18 × (100-14) + 2259.4 + 1.84 × (115-100)]

= 13232.4 J