The total heat energy transferred to cool a 10.0 g C C l 2 F 2 from 40 ∘ C to − 40 ∘ 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, 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, 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 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −427J−1650 J−97J= −2174J=(−2174J)(1 kJ1000J)=−2.174 kJ.

Therefore, the total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

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Students have asked these similar questions

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Answer 2

Question

Chapter 9, Problem 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −427J−1650 J−97J= −2174J=(−2174J)(1 kJ1000J)=−2.174 kJ.

Therefore, the total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

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Answer 3

Question

Chapter 9, Problem 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −427J−1650 J−97J= −2174J=(−2174J)(1 kJ1000J)=−2.174 kJ.

Therefore, the total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

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Answer 4

Question

Chapter 9, Problem 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −427J−1650 J−97J= −2174J=(−2174J)(1 kJ1000J)=−2.174 kJ.

Therefore, the total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

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

Answer 5

Chapter 9, Problem 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −427J−1650 J−97J= −2174J=(−2174J)(1 kJ1000J)=−2.174 kJ.

Therefore, the total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Answer 6

Chapter 9, Problem 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition:

ΔH∘= −427J−1650 J−97J= −2174J=(−2174J)(1 kJ1000J)=−2.174 kJ.

Therefore, the total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Answer 7

Chapter 9, Problem 97QRT

Interpretation Introduction

Interpretation:

The total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘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, 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, 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 97QRT

The total heat energy evolved to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is 2.174 kJ.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

The specific heat capacities of gas and liquid are .61 Jg−1 ∘C−1 and 0.97 Jg−1 ∘C−1 respectively. Its normal boiling point is −30∘C. The vaporization enthalpy is 165 J g−1.

1. Cool it from 40oC to its boiling point -30oC

The heat energy evolved during the cooling of 10 g of CCl2F2 can be calculated as given below.

ΔH0=m×s×Δθ=(10 g)(.61 Jg−1 ∘C−1)[(−30∘C)−(40∘C)]=− 427J.

Where, m= mass of CCl2F2, s= Specific heat of gas and Δθ= Temperature difference.

2. Condensation of the gas to liquid at -30oC

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

ΔH0=m×ΔvapH0= −10 g× 165 Jg= −1650 J.

3. Cool it further from -30oC to -40oC

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

ΔH0=m×s×Δθ=(10 g)(0.97 Jg−1 ∘C−1)[(−40∘C)−(−30∘C)]=−97 J.

Now, the total heat energy transferred to cool a 10.0 g CCl2F2 from 40∘C to −40∘C is the sum total of all the heat energies required for each transition: