General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 12, Problem 12.72QP
Interpretation Introduction

Interpretation:

The amount of heat needed for the conversion of ice to steam, has to be calculated.

Concept Introduction:

Two types of phase transitions are involved in the conversion of ice to steam. Phase transition is the transfer of one phase to another. To convert ice to steam, first it is needed to convert it into liquid water. Then the liquid water can be converted into steam. By this way ice undergoes two types of phase transitions such as transfer of solid phase of ice to liquid phase. Then transfer of liquid phase to steam which is the gaseous phase.

Expert Solution & Answer
Check Mark

Answer to Problem 12.72QP

The amount of heat needed to convert 866g of ice to steam is 2317.605kJ.

Explanation of Solution

The amount of heat needed to convert ice to steam can be calculated as shown here:

1st phase transition: Melting of ice to liquid water:

q=m×s×ΔT

q=HeattransferedIt can be released or absorbeds=specificheatofice=2.03J/g.oCGiven.m=massofice=866gGiven.

ΔT=TemperaturedifferenceFinalTemperature - Initialtemperature=T2-T1

The initial temperature is given as -10oC.

In the melting of ice, let the final temperature be its melting point which is 0oC.

T2= 0oCT1=-10oCΔT=0oC-(-10oC)=10oC

Substituting all known values in the formula of heat transferred:

q=m×s×ΔT=866g×2.03J/g.oC×10oC=17579.8J

Converting joules into kilojoules:

1000J=1kJ17579.8J=1kJ1000J×17579.8J=17579.8×10-3kJ=17.580kJ

Therefore, q=17.580kJ. Let this be q1.

The molar heat of fusion of water is ΔHfus=6.01kJ/mol.It is the amount of heat energy required to melt one mole of ice to one mole of liquid water.

It is known that q=ΔHreaction

Converting ΔHreactioninto ΔHfus:

866gofice=866gofH2O1moleofH2O=18gofH2O1moleofH2O18gofH2O×866gofH2O=48.111moleofH2O

Hence,

866gofice=48.111moleofH2O.ΔHfus=6.01kJ/mol6.01kJ1moleofH2O×48.111moleofH2O=289.147kJ

Therefore, ΔHfus=289.147kJ. Let this be q2.

2nd phase transition: Vaporization of liquid water to steam:

q=m×s×ΔT

q=HeattransferedIt can be released or absorbeds=specificheatofwater=4.184J/g.oCm=massofwater=866gGiven.

ΔT=TemperaturedifferenceFinalTemperature - Initialtemperature=T2-T1

Let the initial temperature be 0oC.

In the vaporization of liquid water, let the final temperature be its boiling point which is 100oC.

T2100oCT1=0oCΔT=100oC-0oC=100oC

Substituting all known values in the formula of heat transferred:

q=m×s×ΔT=866g×4.184J/g.oC×100oC=3623.344J

Converting joules into kilojoules:

1000J=1kJ3623.344J=1kJ1000J×3623.344J=3623.344×10-3kJ=3.623kJ

Therefore, q=3.623kJ. Let this be q3.

The molar heat of vaporization of water is ΔHvap=40.79kJ/mol.It is the amount of heat energy required to vaporize one mole of liquid water to one mole of vapor.

It is known that q=ΔHreaction

Converting ΔHreactioninto ΔHfus:

866gofice=866gofH2O1moleofH2O=18gofH2O1moleofH2O18gofH2O×866gofH2O=48.111moleofH2O

Hence,

866gofice=48.111moleofH2O.ΔHvap=40.79kJ/mol40.79kJ1moleofH2O×48.111moleofH2O=1962.448kJ

Therefore, ΔHfus=1962.448kJ. Let this be q4.

Heating water from 100oC to 126oC for converting into steam:

q=m×s×ΔT

q=HeattransferedIt can be released or absorbeds=specificheatofsteam=1.99J/g.oCGiven.m=massofwater=866gGiven.

ΔT=TemperaturedifferenceFinalTemperature - Initialtemperature=T2-T1

Let the initial temperature be 100oC which is the boiling point of water. The final temperature is given as 126oC.

T2=126oCT1=100oCΔT=126oC-100oC=26oC

Substituting all known values in the formula of heat transferred:

q=m×s×ΔT=866g×1.99J/g.oC×26oC=44806.84J

Converting joules into kilojoules:

1000J=1kJ44806.84J=1kJ1000J×44806.84J=3623.344×10-3kJ=44.807kJ

Therefore, q=44.807kJ. Let this be q5.

It is now known that

q1=17.580kJq2=289.147kJ

q3=3.623kJq4=1962.448kJ

q5=44.807kJ

Therefore,

qtotal=q1+q2+q3+q4+q5=17.580kJ+289.147kJ+3.623kJ+1962.448kJ+44.807kJ=2317.605kJ

Hence, the amount of heat needed to conversion of 866g of ice to steam, is 2317.605kJ.

Conclusion

Thus the amount of heat required to convert ice to steam was calculated.

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Chapter 12 Solutions

General Chemistry

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