The amount of mass of C C l 2 F 2 must evaporate to freeze 2 mole of water initially at 20 ∘ C has to be calculated. Concept Introduction: Calculation of heat energy: When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, Δ H 0 = m × s × Δ θ . Where, m = mass of substance, s = Specific heat of substance in a particular phase and Δ θ = Temperature difference. When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, Δ H 0 = n × Δ f u s o r v a p H 0 . Where, n = No. of moles of substance , Δ f u s H 0 = Fusion enthalpy of a substance in its solid phase and Δ v a p H 0 = enthalpy of vaporization of a substance in its liquid phase.

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Answer 1

Question

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

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Answer 2

Question

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

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Answer 3

Question

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

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Answer 4

Question

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

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Answer 5

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Answer 6

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Answer 7

Chapter 9, Problem 37QRT

Interpretation Introduction

Interpretation:

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same means a transition is taking place within the same phase like between liquid to liquid. Then, ΔH0=m×s×Δθ.

Where, m= mass of substance, s= Specific heat of substance in a particular phase and Δθ= Temperature difference.

When two phases are different means a transition is taking place between two different phases like between solid to liquid. Then, ΔH0=n×Δfus or vapH0.

Where, n= No. of moles of substance , ΔfusH0= Fusion enthalpy of a substance in its solid phase and ΔvapH0= enthalpy of vaporization of a substance in its liquid phase.

Answer 8

Expert Solution & Answer

Answer to Problem 37QRT

The amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

The specific heat capacity of liquid water is 4.184 Jg−1 ∘C−1. The fusion of enthalpy for solid ice is 6.02 kJ/mol. The vaporization enthalpy of CCl2F2 is 289 J/g.

Calculation of mass of water:

No. of moles (n)=mM⇒m=n×M=2 mol×18.0152 g/mol=36.0304 g≅36 g.

1. Cool the water from 20oC to 0oC

The heat energy evolved during this process can be calculated as given below.

ΔH0=m×s×Δθ=(36 g)(4.184 Jg−1 ∘C−1)[(0∘C)−(20∘C)]=−3012.48 J= −3012.48 J×1 kJ1000J=−3.01248 kJ≅ −3.0 kJ.

Where, m= mass of water, s= Specific heat of liquid water and Δθ= Temperature difference.

2. Freeze the ice at 0oC

The heat energy evolved during this process can be calculated as given below.

ΔcrystH=ΔfreezeH=−ΔfusH=−6.02 kJ/mol

ΔH0=(n)×(−ΔfusH0)= (2 mol)×(−6.020 kJmol)= −12.040 kJ≅ − 12.0 kJ.

Where, n= No. of moles of water and ΔfusH0= Fusion enthalpy of solid ice.

Now, the total heat energy evolved during freezing of 2 mole of water from 20∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −3.0 kJ−12 kJ= -15 kJ.

The heat energy required to evaporate one gram of CCl2F2 is 289 J. So the amount of CCl2F2 will be evaporated by 15 kJ energy can be calculated as given below.

15 kJ×1000 J1 kJ×1 g289 J= 51.9 g.

Therefore, the amount of mass of CCl2F2 must evaporate to freeze 2 mole of water initially at 20∘C is 51.9 g.

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Answer to Problem 37QRT

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