Problem 109 - Chemistry and Chemical Reactivity

Question

Suppose that an inch of rain falls over a square mile of ground. (A density of \(1.0 \mathrm{g} / \mathrm{cm}^{3}\) is assumed.) The heat of vaporization of water at \(25^{\circ} \mathrm{C}\) is \(44.0 \mathrm{kJ} / \mathrm{mol} .\) Calculate the quantity of heat transferred to the surroundings from the condensation of water vapor in forming this quantity of liquid water. (The huge number tells you how much energy is "stored" in water vapor and why we think of storms as such great forces of energy in nature. It is interesting to compare this result with the energy given off, \(4.2 \times 10^{6} \mathrm{kJ},\) when a ton of dynamite explodes.)

Step-by-Step Solution

Answer

The heat transferred is approximately \(1.606 \times 10^{10} \text{ kJ}\).

Step 1: Understand the Problem

We need to find the heat released when 1 inch of rain falls on a square mile of ground. This water vaporizes and then condenses, releasing heat into the surroundings. We'll calculate this heat using the heat of vaporization.

Step 2: Convert Units

First, we need to convert the volume of the rain from inches over a square mile to liters: \(1 \text{ mile} = 1,609.34 \text{ meters}\), so \(1 \text{ square mile} = 2,589,988.11 \text{ square meters}\). \(1 \text{ inch} = 2.54 \text{ cm}\) and there are \(10,000 \text{ cm}^2\) in a square meter. Therefore, the volume is \(2.54\text{ cm} \times 2,589,988.11 \text{ m}^2 \times 10,000 \text{ cm}^2/\text{m}^2 = 6,574,516,815.4\text{ cm}^3\). Since \(1 \text{ cm}^3 = 1 \text{ mL}\), this is also 6,574,516.8154 L.

Step 3: Convert to Moles of Water

The density of water is \(1.0 \text{ g/cm}^3\), so the mass of water is approximately \(6,574,516,815.4 \text{ g}\). Using the molecular weight of water \(18.015 \text{ g/mol}\), the amount of water in moles is \(\frac{6,574,516,815.4 \text{ g}}{18.015 \text{ g/mol}} = \approx 365,112,953.92 \text{ mol}\).

Step 4: Calculate the Heat Released

Using the heat of vaporization \(44.0 \text{ kJ/mol}\), the total heat released when the water vapor condenses is \(365,112,953.92 \text{ mol} \times 44.0 \text{ kJ/mol} = \approx 1.606 \times 10^{10} \text{ kJ}\).

Step 5: Compare with Dynamite

The calculated heat is \(1.606 \times 10^{10} \text{ kJ}\). When comparing to the energy released by a ton of dynamite, \(4.2 \times 10^{6} \text{ kJ}\), the heat from the rain is significantly greater, demonstrating the tremendous energy involved in meteorological processes.

Key Concepts

Units Conversion

The challenge begins with a common necessity in science problems: converting units. In this exercise, we convert the volume of rainfall over a square mile from inches into liters, a unit more convenient for subsequent calculations.

Let's break it down. One mile equals 1,609.34 meters, therefore a square mile measures 2,589,988.11 square meters. When an inch of rain falls, it's equivalent to a water column of 2.54 centimeters high. Start by calculating the volume in cubic centimeters:

• Multiply the area in square meters by the height of the water column (in cm), and convert the entire area of one square mile into square centimeters.
• This resulted in 6,574,516,815.4 cm³, since 1 m² contains 10,000 cm².

Thus, with the equivalent of 1 cm³ being 1 mL, this volume equals 6,574,516.8154 liters. Converting units is crucial, as it determines how well one navigates through calculations applicable to the real world.

Moles of Water

Knowing the total mass of the rainwater, we move on to understand how many moles it represents. Given water's density at 1 g/cm³, the total mass in grams equals the volume in cm³.

For this example, we have 6,574,516,815.4 grams of water. To convert grams to moles, divide by the molar mass of water, which is 18.015 g/mol:

• Mass in moles = Total mass divided by molecular weight.
• For this volume, we get approximately 365,112,953.92 moles.

Understanding moles of a substance helps us relate macroscopic volumes to molecular-scale quantities, bridging a crucial gap in chemistry and physics.

Energy Comparison

In the exercise, we explore how much energy is released when water vapor condenses into liquid form. The heat of vaporization depicts the energy needed for a liquid to become vapor, in this case, 44 kJ/mol.

With the moles of water calculated, the total energy release is simply the product of moles and the heat of vaporization:

• Total energy released = Number of moles multiplied by heat of vaporization.
• This equates to approximately 1.606 x 10¹⁰ kJ.

To comprehend this energy's magnitude, compare it to a ton of dynamite, which releases 4.2 x 10⁶ kJ. Clearly, the energy involved in the rain cycle is immense. Such comparisons highlight nature's power, spanning storms to volcanic eruptions.

Density of Water

Placing water's density at the heart of the problem emphasizes its fundamental role in fluid mechanics. Water, assumed here at 1 g/cm³, serves as a benchmark for calculating mass from volume.

This density allows easy conversion between different volume and mass units, such as cm³ to grams directly:

• Because 1 liter equates to 1,000 grams when density is 1 g/cm³, extensive calculations can be simplified.
• The consistency of water's density at standard conditions ensures precise mass estimations in various scientific applications.

Such calculations are critical in diverse fields, from meteorology to chemical engineering, helping accurately determine reactive quantities or environmental effects.