Question

How much heat is needed to convert 888 g of ice at -30°C to steam at 136°C? (The specific heats of ice, steam, and liquid are 2.03 J/g · °C, 1.99 J/g · °C, and 4.18 J/g·°C, respectively. The heat of fusion is 6.02 kJ/mol and heat of vaporization is 40.79 kJ/mol.)
________ kJ

Answer 1

Ti = -30.0 oC

Tf = 136.0 oC

here

Cs = 2.03 J/g.oC

Heat required to convert solid from -30.0 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 888 g * 2.03 J/g.oC *(0--30) oC

= 54079.2 J

Lf = 6.02KJ/mol =

6020J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 888.0/18.016

= 49.2895 mol

Heat required to convert solid to liquid at 0.0 oC

Q2 = n*Lf

= 49.2895 mol *6020 J/mol

= 296722.913 J

Cl = 4.18 J/g.oC

Heat required to convert liquid from 0.0 oC to 100.0 oC

Q3 = m*Cl*(Tf-Ti)

= 888 g * 4.18 J/g.oC *(100-0) oC

= 371184 J

Lv = 40.79KJ/mol =

40790J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 888.0/18.016

= 49.2895 mol

Heat required to convert liquid to gas at 100.0 oC

Q4 = n*Lv

= 49.2895 mol *40790 J/mol

= 2010519.5382 J

Cg = 1.99 J/g.oC

Heat required to convert vapour from 100.0 oC to 136.0 oC

Q5 = m*Cg*(Tf-Ti)

= 888 g * 1.99 J/g.oC *(136-100) oC

= 63616.32 J

Total heat required = Q1 + Q2 + Q3 + Q4 + Q5

= 54079.2 J + 296722.913 J + 371184 J + 2010519.5382 J + 63616.32 J

= 2796122 J

= 2796 KJ

Answer: 2796 KJ