How many Joules of heat are removed from 500 g of H2O when it is cooled from 190 C to -35C ?

Question

How many Joules of heat are removed from 500 g of H2O when it is cooled from 190 C to -35C ? The specific heat of solid water is 2.09 J/g*C, liquid is 4.18 J/g*C, and gas is 2.02 J/g*C. The molar heat of fusion of water is 6.01 kJ/mol. The molar heat of vaporization of water is 40.65 kJ/mol.

Answer 1

Work this type of problem in steps.

Step 1: cool 500 g water from 190º(gas) to 100º (b.p.)

q = mC∆T = (500 g)(2.02 J/gº)(90º) = 90,900 J

Step 2: convert 500 g of water (gas) to water (liquid), i.e. phase change. Use ∆Hvap.

q = m∆Hvap = (500 g)(1 mol / 18 g)(40.65 kJ / mol) = 1129 kJ = 1,129,000 J

Step 3: cool 500 g liquid water from 100º to 0º (f.p.)

q = mC∆T = (500 g)(4.18 J/gº)(100º) = 209,000 J

Step 4: convert 500 g liquid water @0º to solid water (ice) @0º, i,e. phase change. Use ∆Hfusion

q = m∆Hfusion = (500 g)(1 mol/18 g)(6.01 kJ/mol) = 166.9 kJ = 166,900 J

Step 5: cool 500 g ice at 0º to -35º

q = mC∆T = (500 g)(2.09 J/gº)(35º) = 36,575 J

Step 6: add up all the joules to get the total amount of joules removed from the water:

90,900 +1,129,000 + 209,000 + 166,900 + 36,575 = 1,632,375 J

If you wanted to round to the correct number of significant figures, the answer would be 2,000,000 J since there is only 1 sig.fig. in the mass of 500 g of water.