Question

Given that the specific heat capacities of ice and steam are 2.06 J/g°C and 2.03 J/g°C, the molar heats of fusion and vaporization for water are 6.02 kJ/mol and 40.6 kJ/mol, respectively, and the specific heat capacity of water is 4.18 J/g°C, calculate the total quantity of heat evolved when 13.2 g of steam at 192°C is condensed, cooled, and frozen to ice at -50.°C.

Answer 1

Ti = 192.0 oC

Tf = -50.0 oC

Cg = 2.03 J/g.oC

Heat released to convert vapour from 192.0 oC to 100.0 oC

Q1 = m*Cg*(Ti-Tf)

= 13.2 g * 2.03 J/g.oC *(192-100) oC

= 2465 J

Lv = 40.6KJ/mol = 40600J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 13.2/18.016

= 0.7327 mol

Heat released to convert gas to liquid at 100.0 oC

Q2 = n*Lv

= 0.7327 mol *40600 J/mol

= 29747 J

Cl = 4.184 J/g.oC

Heat released to convert liquid from 100.0 oC to 0.0 oC

Q3 = m*Cl*(Ti-Tf)

= 13.2 g * 4.184 J/g.oC *(100-0) oC

= 5523 J

Lf = 6.02KJ/mol = 6020J/mol

Heat released to convert liquid to solid at 0.0 oC

Q4 = n*Lf

= 0.7327 mol *6020 J/mol

= 4411 J

Cs = 2.06 J/g.oC

Heat released to convert solid from 0.0 oC to -50.0 oC

Q5 = m*Cs*(Ti-Tf)

= 13.2 g * 2.06 J/g.oC *(0--50) oC

= 1360 J

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

= 2465 J + 29747 J + 5523 J + 4411 J + 1360 J

= 43505 J

= 43.5 KJ

Answer: 43.5 KJ