Calculate the heat required in Joules to convert 18.0 grams of water ice at a temperature of -20° C to liquid water at the normal boiling point of water.

Question

Calculate the heat required in Joules to convert 18.0 grams of water ice at a temperature of -20° C to liquid water at the normal boiling point of water.

Given: -specific heat of ice = 2.09 J/g°C

-specific heat of liquid water = 4.184 J/g°C

-specific heat of water vapor = 2.03 J/g°C

-molar heat of fusion of water = 6.02 kJ/mol

-molar heat of vaporization of water = 40.7 kJ/mol

Answer 1

It's easiest to do these types of problems in "steps", calculating heat energy needed at each step.

(1). ice @ -20º --> ice @ 0º

(2). ice @ 0º --> liquid @ 0º (phase change)

(3). liquid @ 0º --> liquid @ 100º (boiling point)

(1). q = mC∆T

q = (18.0 g)(2.09 J/gº)(20º)

q = 752 J

(2). q = m∆Hfusion

q = (18.0 g)(1 mol/18 g)(6.02 kJ/mol)(1000 J/kJ)

q = 6020 J

(3). q = mC∆T

q = (18.0 g)(4.184 J/gº)(100º)

q = 75,312 J

Total heat energy = 752 J + 6020 J + 75,312 J = 82,084 J