Question

Calculate the amount of heat ( in kJ) required to convert 344.0 g of liquid water...

Calculate the amount of heat ( in kJ) required to convert 344.0 g of liquid water at 22.5 oC into steam at 145.0 °C. ( Heat of vaporization of water at its boiling point = 40.7 kJ/mol., specific heats of water and steam are 4.184 J/g °C and 2.01 J/g °C, respectively. )

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Homework Answers

Answer #1

Ti = 22.5 oC

Tf = 145.0 oC

here

Cl = 4.184 J/g.oC

Heat required to convert liquid from 22.5 oC to 100.0 oC

Q1 = m*Cl*(Tf-Ti)

= 344 g * 4.184 J/g.oC *(100-22.5) oC

= 111545.44 J

Lv = 40.7KJ/mol =

40700J/mol

Lets convert mass to mol

Molar mass of H2O = 18.016 g/mol

number of mol

n= mass/molar mass

= 344.0/18.016

= 19.0941 mol

Heat required to convert liquid to gas at 100.0 oC

Q2 = n*Lv

= 19.0941 mol *40700 J/mol

= 777131.4387 J

Cg = 2.01 J/g.oC

Heat required to convert vapour from 100.0 oC to 145.0 oC

Q3 = m*Cg*(Tf-Ti)

= 344 g * 2.01 J/g.oC *(145-100) oC

= 31114.8 J

Total heat required = Q1 + Q2 + Q3

= 111545.44 J + 777131.4387 J + 31114.8 J

= 919792 J

= 920. KJ

Answer: 920. KJ

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