What is the total energy transfer(in KJ) when 10.00 pounds of water vapor at 110 degrees C becomes ice at 253K? Assume that the specific gravity of water = 1.00.

Answer 1

Convert 10.00 pounds H2O to grams.

q1 = mass H2O x specific heat steam x (110=100)
q2 = mass H2O x heat vaporization
q3 = mass H2O x specific heat liquid water x (100-0)
q4 = mass water x heat fusion.
Total Q = sum of individual q values.

Answer 2

In q1, did you mean to say (110-100)? I did it assuming that and after I added all the q values I got 13,759.78 KJ as my answer. Is that correct?

Answer 3

Also shouldn't there be another phase involving the Cs of ice where the ice is cooled down to the 253k?