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

3. (a) The Lattice enthalpy for the solid ionic compound AgBr is +900. kJ/mole. Write the chemical equation that corresponds to the Lattice Enthalpy for AgBr(s) in the space above. Then explain in your own words why this is a large positive number. (b) The hydration enthalpy for AgBris -821 kJ/mole. Write the chemical equation that corresponds to the Hydration Enthalpy for AgBr(s) in the space above. Then explain in your own words why this is a large negative number. (c) Would you expect this compound to be soluble in water? Why/Why not? Calculate anything you need in order to figure this out, and explain your answer.

(a) Consider a substance where the intermolecular forces hold the molecules in fixed rigid positions. What is the process called when enough heat has been added to the substance so that the molecules begin to flow? (b) Consider a substance where the intermolecular forces hold the molecules in close contact with each other, but the molecules can flow. What is the process called when enough heat has been added to the substance so that the molecules escape each other? (c) Consider a substance where the intermolecular forces hold the molecules in fixed rigid positions. What is the process called when enough heat has been added to the substance so that the molecules can escape each other?

Mercury (Hg) vapor is toxic and readily absorbed from the lungs. At 20.∘C, mercury (ΔHvap = 59.1 kJ/mol) has a vapor pressure of 1.20 x 10-3 torr, which is high enough to be hazardous. To reduce the danger to workers in processing plants, Hg is cooled to lower its vapor pressure. At what temperature would the vapor pressure of Hg be at the safer level of 5.0 x 10-5 torr?

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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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: