To find the heat of mixing at 300 K in KJ/kg Concept Introduction: We find the Excess enthalpy of H 2 SO 4 of pure and 35-wt-% from the H-x diagram of H 2 SO 4 Then we find the enthalpy of Pure water. The heat of mixing is as given below, Q = H 3 − x 2 H 2 − x 1 H 1 Where, Mass fraction of H 2 O = x 1 Mass fraction of H 2 SO 4 = x 2 Enthalpy of Pure H 2 O = H 1 Enthalpy of Pure H 2 SO 4 (100-wt-%) = H 2 Enthalpy of 35-wt-% aqueous solution of H 2 SO 4 = H 3

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Answer 1

Question

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

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Answer 2

Question

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

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Answer 3

Question

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

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Answer 4

Question

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

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Answer 5

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

Answer 6

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

Answer 7

Chapter 11, Problem 11.25P

Interpretation Introduction

Interpretation:

To find the heat of mixing at 300 K in KJ/kg

Concept Introduction:

We find the Excess enthalpy of H₂SO₄of pure and 35-wt-% from the H-x diagram of H₂SO₄

Then we find the enthalpy of Pure water.

The heat of mixing is as given below,

Q=H3−x2H2−x1H1

Where,

Mass fraction of H₂O= x₁

Mass fraction of H₂SO₄= x₂

Enthalpy of Pure H₂O = H₁ Enthalpy of Pure H₂SO₄ (100-wt-%) = H₂ Enthalpy of 35-wt-% aqueous solution of H₂SO₄ = H₃

Answer 8

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Answer 9

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Answer 10

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Answer 11

Ch. 11 - Prob. 11.1P Ch. 11 - Prob. 11.2P Ch. 11 - Prob. 11.3P Ch. 11 - Prob. 11.4P Ch. 11 - Prob. 11.5P Ch. 11 - Prob. 11.6P Ch. 11 - Prob. 11.7P Ch. 11 - Prob. 11.8P Ch. 11 - Prob. 11.9P Ch. 11 - Prob. 11.10P

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Chapter 11 Solutions