Question

If we start with 18.0 g of water at 22.0°C and heat it until we have water vapor at 111°C, how much heat energy will be needed?

Specific Heat Values-
H2O (ice)= 2.09 J/g C
H2O (liquid)= 4.184 J/g C
H2O (steam)= 1.02 J/g C

Latent Heat Values (H2O)-
(Triangle) H= 334 J/g
(Triangle) H= 2260 J/g

Answer 1

q1 = heat to increase T of liquid H2O from 22 C to 100 C
q1 = mass H2O x specific heat liquid H2O x (Tfinal-Tinitial) = ?
q2 = heat to transform liquid water at 100 to vapor(steam) at 100 C.
q2 = mass H2O x heat vaporization = ? Note: heat vap is the 2260 listed.
q3 = heat to increase T of vapor (steam) from100 C to 111 C
q3 = mass H2O x specific heat H2O vapor x (Tfinal-Tinitial) = ?
Total Q = q1 + q2 + q3
Post your work if you get stuck. In the first one (q1), for example, Tfinal will be 100 C and Tinitial is 22 C.