A sample of steam with a mass of .550 g and at a temperature of 1 degrees C condenses into an insulated container holding 4.40 g of water at 5.0 degrees C. Assuming that no heat is lost to the surroundings, what will be the final temperature of the mixture?
4 answers
Are you sure that is 1 degree C for the steam? I would feel better if it were 100 C. If it is 1 degree C, do you have a heat of vaporization for water at 1 degree C?
It is 100 degrees, I just mistyped!!
(heat released by condensation of steam) + (heat released by cooling 100 C water to Tfinal) + (heat absorbed by 4.40 g H2O at 5.0 degrees C moving up to Tf) = 0
[0.550 x heat vap] + (0.550 x specific heat water x (Tfinal-Tinitial)] + [4.40 x specific heat water x (Tfinal-Tinitial)]= 0
Ti for the steam is 100; Ti for the 4.40g H2O is 5.0C
[0.550 x heat vap] + (0.550 x specific heat water x (Tfinal-Tinitial)] + [4.40 x specific heat water x (Tfinal-Tinitial)]= 0
Ti for the steam is 100; Ti for the 4.40g H2O is 5.0C
Sorry, I never saw this response so I reposted it correctly! But I do have a question. I used 2257 kJ/kg for the heat vap. and 2.08 J/g*K for the specific heat of water. I'm still getting the problem wrong, am I plugging in the wrong numbers??
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