Enthalpy for chemistry class

Question

A 0.598 g sample of steam at 106.0 ∘C is condensed into a container with 5.78 g of water at 16.6 ∘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.

Answer 1

I answered a similar question the other day, but it had slightly different masses and temperature. I show it here, but you'll have to make the appropriate changes as needed.

Heat lost by the steam = heat gained by the cooler (17.0º) water

Before beginning calculations, let get our units to be consistent. I'll convert the units of ∆H_vap:

H_vap = 40.7 kJ/mol x 1 mol / 18 g = 2.26 kJ / g = 2260 J / g

Step 1. Lower temp of steam from 103.6º to 100º

q = mC∆T = (0.577 g)(2.01 J/gº)(3.6º) = 4.18 J

Step 2. Convert steam to liquid at 100

q = m∆Hvap = (0.577 g)(2260 J/g) = 1304 J

Step 3. Lower temp of liquid from 100º to the final temp

q = (0577g)(4.18 J/gº)(100º - Tf) = 241 - 2.41Tf

Total heat lost by steam = 4.18 + 1304 + 241 - 2.41Tf

heat gained by cool water

q = mC∆T

q = (5.58 g)(4.18 J/gº)(Tf - 17º) = 23.3Tf - 397

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

4.18 + 1304 + 241 - 2.41Tf = 23.3Tf - 397

1549 - 2.41Tf = 23.3Tf - 397

25.7Tf = 1946

Tf = 75.7º = final temperature

(be sure to check all the math)