Question

A 5.00 g sample of H2O is removed from a freezer and heated from an initial temperature of -5.00 oC through to a final temperature of 102.0 oC.

(a) Given the following data, calculate the amount of energy that has to be absorbed by the H2O during the whole process.

Freezing point of H2O = 0.00oC;

Boiling point of H2O = 100.oC

ΔHfusion of H2O = 6.01 kJ mol-1

ΔHevaporation of H2O = 40.7 kJ mol-1

Specific heat capacity of ice = 2.05 J g-1 K-1;

Specific heat capacity of water = 4.18 J g-1 K-1

Specific heat capacity of steam = 2.08 J g-1 K-1

(b) Calculate the percentage of the total energy required for the overall process in (a) that is used just in heating the liquid water.

(c) Calculate the mass of ice at 0.00oC, that can be melted in 30 minutes by a heater that produces 8000 kJ of energy per hour but in such a way that only 70% of the heat produced is absorbed by the ice.

Answer 1

We know that 1 kJ=1000 J

Molar mass of H₂O=2xmolar mass of H+Molar mass of O

=2x1 g/mol+16 g/mol

=2 g/mol+16 g/mol

=18 g/mol

Number of moles in 5.00 g (H₂O)=Given mass/molar mass

=5.00 g/18 g/mol=0.278 mol

Amount of heat energy absorbed by water in the whole process

=Heat absorbed by ice to increase the temperature to 0°C + Heat absorbed to melt ice at 0°C

+ Heat absorbed to increase the temperature of liquid water to 100°C

+ Heat absorbed to vapourize water at 100°C

+Heat absorbed to increase the temperature of vapour to 102°C

=Mass of ice x specific heat of ice x rise in temperature of ice to 0°C

+ number of moles of ice x heat of fusion of ice

+Mass of liquid water x specific heat of liquid water x rise in temperature of water to 100°C

+ number of moles of water x heat of vapourization

+ mass of water vapour x specific heat of water vapour x rise in temperature of vapour to 102°C

=5.00 g x 2.05 J/g°C x (0°C-(-5.0°C))

+0.278 mol x 6.01 kJ/mol

+5.00 g x 4.18 J/g°C x (100°C-0°C)

+0.278 mol x 40.7 kJ/mol

+5.00 g x 2.08 J/g°C x (102°C-100°C)

=5.00 g x 2.05 J/g°C x 5.0°C

+0.278 mol x 6.01 kJ/mol x 1000 J/kJ

+5.00 g x 4.18 J/g°C x 100°C

+0.278 mol x 40.7 kJ/mol x 1000 J/kJ

+5.00 g x 2.08 J/g°C x 2°C

=51.25 J+1670.78 J+2090 J+11314.6 J+20.8 J

=15147.43 J

So amount of heat absorbed in the whole process=15147.43 J

(b) Amount of heat used in only heating the liquid water =mass of water x specific heat of water x rise in temperature of water from 0°C to 100°C

=5.00 g x 4.18 J/g°C x (100°C-0°C)

=5.00 g x 4.18 J/g°C x 100°=2090 J

Percentage of energy used in heating the liquid water out of the total energy used in the process

=(Energy used in heating the water/energy used in overall process) x 100

=(2090 J/15147.43 J)x100

=13.80%

(c)Amount of energy produced by heater in 1 hour i.e. 60 minutes=8000 kJ

Amount of energy produced by heater in 60 minutes /2 i.e. 30 minutes =8000 kJ/2=4000 kJ

Amount of heat absorbed by ice in 30 minutes=70% of 4000 kJ

=(70/100)x4000 kJ

=2800 kJ

Heat of fusion of ice=6.01 kJ/mol

Number of moles of ice that can melt with 2800 kJ energy=2800 kJ/6.01 kJ/mol

=465.89 mol

Mass of ice at 0.00°C that can melt with given amount of energy=Number of moles x molar mass

=465.89 mol x 18 g/mol

=8386.02 g

So 8386.02 g ice at 0.00°C can be melted in 30 minutes by given heater.