Asked by Becca

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?
Answered by DrBob222
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?
Answered by Becca
It is 100 degrees, I just mistyped!!
Answered by DrBob222
(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
Answered by Becca
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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