A 0.513 g sample of steam at 106.2 C is condensed into a container with 4.80 g of water at 15.3 C

Question

What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18 J/g x C, the specific heat of steam is 2.01 J/g x C, and delta H vap = 40.7 kJ/mol.

Answer 1

Heat lost when 0.513 g steam @ 106.2º cools to 100º = mC∆T = (0.513 g)(2.01 J/gº)(6.2º) = 6.39 J

Heat lost when 0.513 g of steam @100º turns to liquid @100º = (0.513 g)(1 mol/18 g)(40.7 kJ/mol) = 1.16 kJ = 1160 J

heat lost when 0.513 g liquid water @100º cools to the final temperature = (0.513 g)(4.18 J/gº)(100 - Tf)

All the heat lost MUST equal the heat gained by the 4.80 g of water at 15.3º

Heat gained by cooler water = (4.80 g)(4.18 J/gº)(Tf - 15.3º)

Setting the heat lost equal to the heat gained, we have...

6.39 J + 1160 J + (0.513 g)(4.18 J/gº)(100 - Tf) = (4.80 g)(4.18 J/gº)(Tf - 15.3º)

1166 + 214 - 21.4Tf = 20Tf - 307

41.4Tf = 1687

Tf = 41º

Be sure to check the math.