Question

A sample of steam with a mass of 0.520 g and at a temperature of 100 ∘C condenses into an insulated container holding 4.45 g of water at 5.0 ∘C.( ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C)

Assuming that no heat is lost to the surroundings, what is the final temperature of the mixture?

Answer 1

m1= 0.52g T1= 100^oC

m2= 4.45g T2 = 5^oC

ΔH∘vap=40.7 kJ/mol, Cwater=4.18 J/g⋅∘C

0.52 g steam condense at 100^oC ......1

0.52 g water cools from 100^oC to final temp t ......2

4.45 g of water warm up from 5^oC to final temp t ......3

heat is liberated in process 1&2 , this heat warm the water in process3

(0.52g / 18g/mol x 40700j/mol) + 0.52 g x (100 - t) (4.18) = 4.45 g x(t - 5.0) (4.18)

1175.77 +217.36 - 2.1736t = 18.60t - 93

1486.13 = 20.7736t

t = 71.54^oC