Chapter 2, Problem 82A

Interpretation Introduction

Interpretation: If water, mercury, and gallium are at $40{}^{\text{o}}\text{C}$ , on decreasing temperature, the substance that will freeze first and freezes last needs to be identified.

Concept Introduction: There are generally three states of matter that is solid, liquid and gas. Substances that are solid at room temperature have high melting and boiling points. Melting point is defined as the temperature at which a solid substance starts melting. The boiling point is defined as the temperature of a liquid when its vapor pressure is equal to the atmospheric pressure. This is the temperature when liquid starts boiling and converts into vapor.

Explanation of Solution

The melting and boiling point for given substances are as follows:

Substances	Melting point	Boiling point
Water	$0{}^{\text{o}}\text{C}$	$100{}^{\text{o}}\text{C}$
Mercury	$-38.83{}^{\text{o}}\text{C}$	$356.7{}^{\text{o}}\text{C}$
Gallium	$29.76{}^{\text{o}}\text{C}$	$2400{}^{\text{o}}\text{C}$

At $40{}^{\text{o}}\text{C}$ , water, mercury, and gallium all are in a liquid state. As temperature drops, the substance with the highest melting point will freeze first. Here, the melting point of gallium is highest that is $29.76{}^{\text{o}}\text{C}$ thus, it will freeze before mercury and water.

At the melting point, the substance starts to melt that means below this temperature the substance is solid. Now, as temperature decreases from $40{}^{\text{o}}\text{C}$ when it reaches to $29.76{}^{\text{o}}\text{C}$ , gallium starts freezing but water and mercury remain in the liquid state.

The melting point of mercury is very low that is $-38.83{}^{\text{o}}\text{C}$ ; thus, it freezes at last when the temperature is dropped to its melting point. The sequence for substances freezing from first to last is as follows:

  $\text{gallium, water, mercury}$