Asked by Tyler
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.
Answers
Answered by
DrBob222
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.
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.
Answered by
Tyler
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?
Answered by
Tyler
Also shouldn't there be another phase involving the Cs of ice where the ice is cooled down to the 253k?
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