Chemistry 2 problem

Question

A 0.537 g sample of steam at 107.2 °C is condensed into a container with 5.55 g of water at 15.8 °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

The heat lost by the hot steam must equal the heat gained by the cooler water.

The heat lost by the hot steam is the sum of the heat lost going from from 107.2º to 100º and then to condense from steam to liquid at 100º. This is then added to the heat lost by the liquid water at 100º going to the final temperature.

This is written as follows:

heat lost by steam = mass x sp.heat x ∆T + m x ∆Hvap + m x sp.heat x ∆T

The ∆Hvap is given in kJ/mol so we'll convert to J/g: -40.7 kJ/mol x 1000 J/kJ x 1 mol/18 g = -2261 J/g

heat lost by steam = q = (0.537 g)(2.01 J/g/deg)(7.2 deg) + (0.537 g)(2261 J/g) + (0.537 g)(4.18 J/g)(100 - Tf)

heat gained by water = mass x sp.heat x ∆T = (5.55 g)(4.18 J/g/deg)(Tf - 15.8)

Setting these two equal we have...

(0.537 g)(2.01 J/g/deg)(7.2 deg) + (0.537 g)(2261 J/g) + (0.537 g)(4.18 J/g)(100 - Tf) = (5.55 g)(4.18 J/g/deg)(Tf - 15.8)

7.77 + 1214 + 224 + 2.24Tf = 23.2Tf - 367

20.96Tf = 1813

Tf = 86.5º

(Please be sure to check the math)