The amount of heat energy needed to convert 866g of ice at –15°C to steam at 146°C is to be calculated. Concept introduction: The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below: q = s m Δ T Here, q is the amount of heat, s is the specific heat, Δ T is the change in temperature, and m is the mass of water. The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows: Δ T = T final − T initial The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows: q = n Δ H vap Here, n is the number of moles and Δ H vap is the molar heat of vaporization. The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows: q = n Δ H fus Here, n is the number of moles and Δ H fus is the molar heat of fusion.

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Answer 1

Question

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

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Answer 2

Question

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

Answer 6

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

Answer 7

Chapter 11, Problem 85QP

Interpretation Introduction

Interpretation:

The amount of heat energy needed to convert 866g of ice at –15°C to steam at $146°C$ is to be calculated.

Concept introduction:

The amount of heatthatis required to raise the temperature having specific heat is calculated by the expression given below:

$q=smΔT$

Here, $q$ is the amount of heat, $s$ is the specific heat, $ΔT$ is the change in temperature, and $m$ is the mass of water.

The change in temperature is the temperature difference between the final and initial temperatures. It is expressed as follows:

$ΔT=Tfinal−Tinitial$

The amount of energy needed for phase change from liquid phase to vaporphase is given by the relation as follows:

$q=nΔHvap$

Here, $n$ is the number of moles and $ΔHvap$ is the molar heat of vaporization.

The amount of energy needed for phase change from solid phase to liquid phase is given by the relation as follows:

$q=nΔHfus$

Here, $n$ is the number of moles and $ΔHfus$ is the molar heat of fusion.

Answer 8

Expert Solution & Answer

Answer to Problem 85QP

Solution: $2.72×103 kJ$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information: Specific heat of ice is $2.03 J/g.∘C$ . Specific heat of steam is $1.99J/g.∘C$ and $m=866g$ . The initial temperature is–15°C. Final temperature is $146°C$

Step $1$ :The melting point of ice is $0°C$

The energy required to warm ice from –15°Cto $0°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting values in the above equation as follows:

$ΔT=(0−(−15))∘C=15∘C$

The specific heat $(s)$ of ice is $2.03 J/g.∘C$ .

Substitute the required values in the above equation as follows:

$q=smΔT =2.03 Jg.°C×866g×15°C=2.637×104J×1kJ1000 J=26.4 kJ$

Step $2$ :The amount of energy needed for phase change from solid ice to liquid water is calculated using the equation as follows:

$q=nΔHfus$

The number of moles of water is calculated by dividing the given mass to the molar mass as follows:

$Number of moles of water (n)=866 g18.02g/mol=48.06 mol$

Substitute $48.06 mol$ for $n$ and $6.01kJ/mol$ for $ΔHfus$ in the above equationto calculate heat energy as follows:

$q=nΔHfus=(48.06 mol)(6.01 kJ/mol)≈289 kJ$

Step $3$ : The energy required to heat water from $0°C$ to $100°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(100−0)∘C=100∘C$

The specific heat of water is $4.184 J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =4.184 Jg.°C×866g×100°C=3.623×105J×1kJ1000 J≈362.3 kJ$

Step $4$ : The amount of energy needed for phase change from liquid water to steam is calculated using the equation as follows:

$q=nΔHfus$

Substitute $48.06 mol$ for $n$ and $40.79kJ/mol$ for $ΔHvap$ in the equation to calculate the heat energy as follows:

$q=nΔHvap=(48.06 mol)(40.79 kJ/mol)=1960.4 kJ$

Step $5$ : The energy required to heat steam from $100°C$ to $146°C$ is calculated using the equation as follows:

$q=smΔT$

Calculate the change in temperatureby using the equation as follows:

$ΔT=Tfinal−Tinitial$

By substituting the values in the above equation as follows:

$ΔT=(146−100)∘C=46∘C$

The specific heat of steam is $1.99J/g.∘C$ .

Substitute the required values in the equation as follows:

$q=smΔT =1.99 Jg.°C×866g×46°C=7.927×104J×1kJ1000 J≈79.3 kJ$

The total amount of energy needed is the sum of all energy changes calculated in the above steps.

It is calculated as follows:

$Total energy =26.4 kJ+288.5 kJ+362.3 kJ+1960.4 kJ+79.3 kJ=2716.9 kJ=2.72×103kJ$

Answer 12

Conclusion

The amount of heat energy needed to convert 866g of ice at –15°Cto steam at $146°C$ is $2.72×103kJ$ .

Answer 13

Ch. 11.1 - Prob. 1PPA Ch. 11.1 - Prob. 1PPB Ch. 11.1 - Prob. 1PPC Ch. 11.1 - Prob. 1CP Ch. 11.1 - Prob. 2CP Ch. 11.2 - Prob. 1PPA Ch. 11.2 - Prob. 1PPB Ch. 11.2 - Prob. 1PPC Ch. 11.2 - Prob. 1CP Ch. 11.2 - 11.2.2 Given the following information for ...

Answer to Problem 85QP

Explanation of Solution

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