Based on the thermodynamic properties provided for water, determine the energy change when the temperature of 0.650 kg of water decreased from 101 °C to 49.0 °C.

Question

Property	Value	Units
Melting point	0	°C
Boiling point	100.0	°C
ΔHfus	6.01	kJ/mol
ΔHvap	40.67	kJ/mol
cp (s)	37.1	J/mol · °C
cp (l)	75.3	J/mol · °C
cp (g)	33.6	J/mol · °C

Answer 1

Since the sample of water 0.650 kg) is beginning at 101ºC, it is a gas (vapor) because the boiling point is 100.0ºC. The final temperature is 49.0ºC, so it will then be a liquid. Thus, there will be 3 steps that we need to consider in going from 101º to 49º. And since all of the thermodynamic data are given on a per mole basis, we will convert 0.650 kg to moles.

0.650 kg H₂O x 1000 g / kg x 1 mol H₂O / 18.0 g = 36.1 moles H₂O

Step 1: heat given off going from 101º to 100º

q = mC∆T where q = heat; m = mass; C = sp.heat of gas = 33.6 J/moleº; ∆T = change in temp = 1º

q = (36.1 mols)(33.6 J/molº)(1º) = 1213 J = 1.213 kJ

Step 2: heat given off going from gas to liquid @ 100º (phase change, no change in temperature)

q = m∆Hvap where q = heat; m = mass; ∆Hvap = heat of vaporization = 40.67 kJ/mol

q = (36.1 mols)(40.67 kJ/mol) = 1468 kJ

Step 3: heat given off going from 100º to 49.0º

q = mC∆T where q = heat; m = mass; C = sp. heat of liquid = 75.3 J/molº; ∆T = change in temp = 51º

q = (36.1 mols)(75.3 J/molº)(51º) = 138,635 J = 138.6 kJ

Now we simply add up all the energy given off (negative values) to get our final answer...

Energy change = -1.213 kJ + -1468 kJ + -138.6 kJ = -1608 kJ = -1610 kJ (3 significant figures)