What amount of heat, in kJ, is required to convert 1.00 g of water at 67.0 °C to 1.00 g of steam at 100.0 °C? (specific heat capacity of water = 4.184 J/g • °C; ∆Hvap = 40.7 kJ/mol)

Answer 1

Step 1: Raise temp of 1.00 g water @ 67º to 1 g water at 100º

q = mC∆T = (1.00 g)(4.184 J/gº)(33º) = 138.1 J = 0.1381 kJ

Step 2: Convert 1 g of water at 100º to steam at 100º (∆Hvap given in kJ/mol, so change g H2O to moles)

q = m∆Hvap = (1.00 g x 1 mol/18g)(40.7 kJ/mol) = 2.26 kJ

Total heat in kJ = 0.1381 kJ + 2.26 kJ = 2.399 kJ = 2.40 kJ (to 3 sig. figs.)