posted by Kate on .
How many kilojoules of heat is required to completely covert 60.0 grams of water at 32.0° C to steam at 100°C?
s = 4.179 J/g °C
Hfusion = 6.01 kJ/mol
Hvap = 40.7 kJ/mol
q1 = heat required to move T from zero C to 100 C.
q1 = mass water x specific heat water x (Tfinal-Tinitial). Tf = 100 C; Ti = 0 C.
q2 = heat to vaporize liquid water @ 100 C to steam at 100 C.
q2 = mass water x delta Hvap.
qtotal = q1 + q2.