Thermochemistry

Question

determine the final temperature if 45.67 kJ of heat energy is removed from 18.5 g of H2O (g) at 122 degrees Celsius

useful information
sp. heat H2O (s) = 2.03 J/g(degree C)
sp heat H2O (l) = 4.18 J/g(degree C)
Sp heat H2O (g) = 2.01 J/g(degree C)

enthalpy H Fus = 334 J/g
enthalpy H vap = 2260 J/g

Answer 1

I would do this in steps.
First, heat released in cooling from 122 to 100.
q = mass x specific heat steam x (deltaT)
q = 18.5 x 2.01 x 22= about 818 approximately. You should confirm ALL of these numbers.
q = heat released on condensing the steam.
q = mass x heat vap = 18.5 x 2260 = about 42000,
42000 + 818 = about 43,000 J.
That's close to what you have at 45,670 J. The difference is 45,670-43000 = about 3,000 (again, redo these numbers)
Then
-3000 = mass H2O x specific heat liquid water x (Tfinal-Tinitial) and solve for Tfinal.

Answer 2

Thank You

for my final temperature I got around
83.1 degrees celsius

Answer 3

How did you know that the temperature dropped down to 100 degrees C? I have the same exact problem, except the heat is being added, not taken away. And the initial temperature is 73.5 degrees C.