Question

What amount of heat is required to convert 1.0 g of water at 67.0°C to 1.0 g of steam at 100.0°C? (heat capacity of water = 4.184 J/g • C; ∆Hvap = 40.7 kJ/mol)

Answer 1

Q= Q₁ + Q₂

Q₁ = cm(T₂ - T₁) =4.184 *1.0*(100-67)= 138.072 J

Q₂= ∆H_vap*n

n = m/M = 1.0 / 18.02 = 0.055 mol

Q₂= 40700J /mol * 0.055 mol = 2239 J

Q = 138.072 J + 2239 J = 2377.072 J = 2.377 kJ

Answer: 2.377 kJ