Chapter 11, Problem 11.28QP

Consider a substance X with a ΔH_vap = 20.3 kJ/mol and ΔH_fus = 9.0 kJ/mol. The melting point, freezing point, and heat capacities of both the solid and liquid X are identical to those of water.

1. a If you place one beaker containing 50 g of X at −10°C and another beaker with 50 g of H₂O at −10°C on a hot plate and start heating them, which material will reach the boiling point first?
2. b Which of the materials from part a, X or H₂O, would completely boil away first?
3. c On a piece of graph paper, draw the heating curve for H₂O and X. How do the heating curves reflect your answers from parts a and b?

Interpretation Introduction

Interpretation:

The substance X with ${\text{ΔH}}_{\text{vap}}\text{=20}\text{.3kJ/mol}$ and ${\text{ΔH}}_{fus}\text{=9}\text{.0kJ/mol}$.  Thus substance of both liquid and solid has identical melting point, boiling point and heat capacities to water. Based on this data the given statements has to be explained.

Explanation of Solution

Explanation

To explain: which material will attain the boiling point earlier when heating

Comparing the given values with water

We should need to compare the vaporization of the given substance and heat of fusion with the values of water.  For water, ${\text{ΔH}}_{\text{fus}}\text{= 6}\text{.01kJ/mol}$ and ${\text{ΔH}}_{\text{vap}}\text{= 40}\text{.7kJ/mol}$. In this values, the heat of fusion of X is 1.5 times larger than that of water and heat of vaporization of X is 2.0 times larger than water.

Heating the substance or water from $\text{-10°C}$ to boiling point is three step procedure.

In the first step, solid has heated from $\text{-10°C}$ to $\text{0°C}$(freezing point).

Where,

$\text{Heat required = mass×specific heat capacity×temperature change}$

In second step, the solid is melted to liquid at $\text{0°C}$.

Where,

${\text{Heat required = moles × ΔH}}_{\text{fus}}$

In third step, the liquid has heated from $\text{0°C}$ to $\text{100°C}$.

Where,

$\text{Heat required = mass×specific heat capacity×temperature change}$

The heat capacity, temperature change and mass of the substance X is identical to water, the heat required for first and third step are same for both.  As the heat of fusion is higher for substance X and in step two, heat required for substance X take longer time. So, water will reach the boiling point earlier than substance X.

Interpretation Introduction

Interpretation:

The substance X with ${\text{ΔH}}_{\text{vap}}\text{=20}\text{.3kJ/mol}$ and ${\text{ΔH}}_{fus}\text{=9}\text{.0kJ/mol}$.  Thus substance of both liquid and solid has identical melting point, boiling point and heat capacities to water. Based on this data the given statements has to be explained.

Explanation of Solution

Explanation

To identify: the material which is totally boiled first from part (a)

Comparing the given values with water

We should need to compare the vaporization of the given substance and heat of fusion with the values of water.  For water, ${\text{ΔH}}_{\text{fus}}\text{= 6}\text{.01kJ/mol}$ and ${\text{ΔH}}_{\text{vap}}\text{= 40}\text{.7kJ/mol}$. In this values, the heat of fusion of X is 1.5 times larger than that of water and heat of vaporization of X is 2.0 times larger than water.

Heating the substance or water from $\text{-10°C}$ to boiling point is three step procedure.

In the first step, solid has heated from $\text{-10°C}$ to $\text{0°C}$(freezing point).

Where,

$\text{Heat required = mass×specific heat capacity×temperature change}$

In second step, the solid is melted to liquid at $\text{0°C}$.

Where,

${\text{Heat required = moles × ΔH}}_{\text{fus}}$

In third step, the liquid has heated from $\text{0°C}$ to $\text{100°C}$.

Where,

$\text{Heat required = mass×specific heat capacity×temperature change}$

To entirely boil away the substance a further step is needed (step 4).

In fourth step, liquid is boiled to vapour at $\text{100°C}$. Where,

${\text{Heat required = moles × ΔH}}_{\text{fus}}$

As the values of heat of vaporization is much greater than the heat of fusion values.  In this step needed much more heat than in step 2 for both water and substance X.  Since, heat of vaporisation is less for substance X per mole.  Fourth step will require small heat for X and hence it will take less time.  The total heat required for fourth step is directly proportional to the time taken for entirely boil away the substance X.  However forth step require more time to complete this step.  Hence, substance X will boil way first.

Interpretation Introduction

Interpretation:

The substance X with ${\text{ΔH}}_{\text{vap}}\text{=20}\text{.3kJ/mol}$ and ${\text{ΔH}}_{fus}\text{=9}\text{.0kJ/mol}$.  Thus substance of both liquid and solid has identical melting point, boiling point and heat capacities to water. Based on this data the given statements has to be explained.

Explanation of Solution

Explanation

To draw: heating curve for both water and substance X.

Comparing the given values with water

We should need to compare the vaporization of the given substance and heat of fusion with the values of water.  For water, ${\text{ΔH}}_{\text{fus}}\text{= 6}\text{.01kJ/mol}$ and ${\text{ΔH}}_{\text{vap}}\text{= 40}\text{.7kJ/mol}$.  In this values, the heat of fusion of X is 1.5 times larger than that of water and heat of vaporization of X is 2.0 times larger than water.

Heating the substance or water from $\text{-10°C}$ to boiling point is three step procedure.

In the first step, solid has heated from $\text{-10°C}$ to $\text{0°C}$(freezing point).

Where,

$\text{Heat required = mass×specific heat capacity×temperature change}$

In second step, the solid is melted to liquid at $\text{0°C}$.

Where,

${\text{Heat required = moles × ΔH}}_{\text{fus}}$

In third step, the liquid has heated from $\text{0°C}$ to $\text{100°C}$.

Where,

$\text{Heat required = mass×specific heat capacity×temperature change}$

The heating curve for both water and substance X are illustrated below.

Figure 1

Figure 2