Chemistry 1401 Need assistance

Question

A 0.544 g0.544 g sample of steam at 107.6 ∘C107.6⁢ ∘C is condensed into a container with 4.50 g4.50 g of water at 15.3 ∘C.15.3⁢ ∘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

Heat lost by hot steam MUST EQUAL heat gained by cooler water (conservation of energy)

q = heat; m = mass; C = specific heat; ∆Hvap = heat of vaporization; ∆T = change in temperature

(note that ∆Hvap is given as kJ/mol so we will change mass from g to moles to be consistent)

The heat lost by the steam requires 3 steps: the first being the heat lost when the steam going from 107.6º to 100º (still steam), and then the next step when the steam at 100º loses heat and turns into liquid at 100º (phase change). Finally, the 3rd step is when the liquid at 100º loses heat to the cooler water already present)

1). q = mC∆T = (0.544 g)(2.01 J/gº)(7.6º) = 8.31 J of heat lost going from 107.6 - 100º

2). q = m∆Hvap = (0.544 g)(1 mol / 18 g)(40.7 kJ/mol) = 1.23 kJ = 1230 J of heat lost during phase change

3). q = mC∆T = (0.544 g)(4.18 J/gº)(100º-T_f) = 227 - 2.27T_f

Total heat lost by steam = 8.31 + 1230 + 227 - 2.27T_f

The heat gained by the water (q) = mC∆T

q = (4.50 g)(4.18 J/gº)(T_f - 15.3º) = 18.8T_f - 288

Setting heat lost by steam to heat gained by water and solving for T_f, we have...

8.31 + 1230 + 227 - 2.27T_f = 18.8T_f - 288

21.07T_f = 1753

T_f = 83.2ºC

(as always, be sure to check the math)