How do I solve this?

Question

A 0.503 g0.503 g sample of steam at 107.1 ∘C107.1⁢ ∘C is condensed into a container with 5.92 g5.92 g of water at 14.5 ∘C.14.5⁢ ∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18  J g⋅ ∘C,4.18  J g⋅ ∘C, the specific heat of steam is 2.01  J g⋅ ∘C,2.01  J g⋅ ∘C, and Δ𝐻vap=40.7 kJ/mol.

Answer 1

A 0.503 g sample of steam at 107.1⁢ ^∘C is condensed into a container with 5.92 g of water at 14.5 ^∘C. What is the final temperature of the water mixture if no heat is lost? The specific heat of water is 4.18  J / g⋅ ^∘C, the specific heat of steam is 2.01  J / g⋅^∘C, and Δ𝐻vap=40.7 kJ/mol

The amount of heat released when 0.503 g vapor are condensed at 100^∘C is

Q = 0.503 g x 7.1^∘C x 2.01  J / g⋅^∘C + 0.503 g x 40.7 kJ/ mol / (18.0 g/mol) = 7.18 J + 1,137 J = 1,144 J

This amount of heat could raise the temperature of the 5.92 g of water initially at 14.5 ^∘C to

14.5 ^∘C + 1,144 J / (4.18 J / g ^∘C x 5.92 g) = 60.7^∘C

Finally we need to calculate what happens if 5.92 g water at 60.7^∘C were mixed with 0.503 g at 100^∘C.

The final temperature is given by

60.7^∘C + (100-60.7) x 0.503 / (5.92 + 0.503) = 60.7^∘C + 3.1^∘C = 63.8 ^∘C