Question

what is the eventually change during the process in which 100.0g of water at 50.0 degree Celsius is cooled to ice at -30degree celcius.the specific heat of ice,liquid water, and steam are 2.03,4.18,1.84 respectively. heat of fusion is 6.01kJ/mol and heat of vapor is 40.67kj/mol

Answer 1

Ti = 50.0

Tf = -30.0

Cl = 4.184 J/g.oC

Heat released to convert liquid from 50.0 oC to 0.0 oC

Q1 = m*Cl*(Ti-Tf)

= 100 g * 4.184 J/g.oC *(50-0) oC

= 20920 J

Lf = 6.01KJ/mol =

6010J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 100.0/18.016

= 5.5506 mol

Heat released to convert liquid to solid at 0.0 oC

Q2 = n*Lf

= 5.5506 mol *6010 J/mol

= 33359.2362 J

Cs = 2.03 J/g.oC

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

Q3 = m*Cs*(Ti-Tf)

= 100 g * 2.03 J/g.oC *(0--30) oC

= 6090 J

Total heat released = Q1 + Q2 + Q3

= 20920 J + 33359.2362 J + 6090 J

= 60369 J

= 60.4 KJ

Answer: 60.4 KJ