Do you want this in BTU?
q1 = heat to move from 180 to 212.
q1 = mass x specific heat x (212-180)
q2 = heat to vaporize water at 212 to steam at 212 F.
q2 = mass x heat vaporization.
Total heat = q1 + q2.
Ten pounds of liquid water is at a temperature of 180 degrees F at standard atmospheric conditions.
Determine the energy required to convert this to water vapor (steam) at a temperature of 212 degrees F.
3 answers
I don't know how to find the specific heat in q1.
I have q1 = 320 (specific heat)
#2) How do I find heat vaporitzation for q2?
I have q1 = 320 (specific heat)
#2) How do I find heat vaporitzation for q2?
You don't have a table in your text/notes?
specific heat H2O = 1.00 BTU/lb*F
heat vaporization = 970.4 BTU/lb
specific heat H2O = 1.00 BTU/lb*F
heat vaporization = 970.4 BTU/lb