The total heat energy required to melt 36.00 g of ice from − 10 ∘ C to 20 ∘ C has to be calculated. Concept Introduction: Calculation of heat energy: When both the phases are same, Δ H 0 = m × s × Δ θ When two phases are different, Δ H 0 = n × Δ f u s o r v a p H 0

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Answer 1

Question

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

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Answer 2

Question

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

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Answer 3

Question

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

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Answer 4

Question

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

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Answer 5

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Answer 6

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Answer 7

Chapter 9, Problem 36QRT

Interpretation Introduction

Interpretation:

The total heat energy required to melt $36.00 g$ of ice from $−10∘C$ to $20∘C$ has to be calculated.

Concept Introduction:

Calculation of heat energy:

When both the phases are same, $ΔH0=m×s×Δθ$

When two phases are different, $ΔH0=n×Δfus or vapH0$

Answer 8

Expert Solution & Answer

Answer to Problem 36QRT

The total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

The heat capacities of solid ice and liquid water are $2.06 Jg−1 ∘C−1$ and $4.184 Jg−1 ∘C−1$ respectively. The fusion of enthalpy for solid ice is $6.02 kJ/mol$ . The vaporization enthalpy of liquid water is $40.7 kJ/mol$ .

Calculation of no. of moles of ice:

No. of moles $(n)=mM=36 g18.0152 g/mol=1.998 mol.$

1. Heat the ice from $-10oC$ to $0oC$

The heat energy required for $36 g$ of ice to melt can be calculated as given below.

$ΔH0=m×s×Δθ=(36 g)(2.06 Jg−1 ∘C−1)[(0∘C)−(−10∘C)]=741.6 J=741.6 J×1 kJ1000J=0.7416 kJ.$

Where, $m=$ mass of ice, $s=$ Specific heat of solid ice and $Δθ=$ Temperature difference.

2. Melt the ice at $0oC$

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

$ΔH0=n×ΔfusH0= 1.998 mol×6.020 kJmol= 12.02796 kJ.$

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

3. Heat the water from $0oC$ to $20oC$

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

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

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

Now, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ is the sum total of all the heat energies required for each transition:

$ΔH∘= 0.7416 kJ+12.02796 kJ+3.01248 kJ= 15.78204 kJ ≅ 15.7 kJ.$

Therefore, the total heat energy required to melt a $36.00 g$ of ice from $−10∘C$ to $20∘C$ has been calculated to be $15.7 kJ$ .

Answer 12

Ch. 9.1 - Prob. 9.1CE Ch. 9.2 - Prob. 9.2CE Ch. 9.2 - Prob. 9.1PSP Ch. 9.2 - What mass (g) of ethanol, CH3CH2OH(), can be...Ch. 9.3 - Prob. 9.3CE Ch. 9.3 - Prob. 9.4CE Ch. 9.3 - Prob. 9.3PSP Ch. 9.4 - What types of solids are these substances? (a) The...Ch. 9.4 - Prob. 9.5PSP Ch. 9.4 - Prob. 9.5E

The total heat energy required to melt 36.00 g of ice from − 10 ∘ C to 20 ∘ C has to be calculated. Concept Introduction: Calculation of heat energy: When both the phases are same, Δ H 0 = m × s × Δ θ When two phases are different, Δ H 0 = n × Δ f u s o r v a p H 0

Answer to Problem 36QRT

Explanation of Solution

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Chapter 9 Solutions