BuyFindarrow_forward

Fundamentals of Physical Geography

2nd Edition
James Petersen
ISBN: 9781133606536

Solutions

Chapter
Section
BuyFindarrow_forward

Fundamentals of Physical Geography

2nd Edition
James Petersen
ISBN: 9781133606536
Textbook Problem

As air rises, it expands and cools. The cooling, at the dry adiabatic lapse rate, is 10°C per 1,000 meters. (Descending air always warms at this rate.) In addition, the dew point temperature decreases about 2°C per 1,000 meters within a rising parcel of air. At the height at which the dew point temperature is reached, condensation begins and the wet adiabatic lapse rate of 5°C per 1,000 meters becomes operational. When the wet adiabatic lapse rate is in operation, the dew point temperature is equal to the air temperature. When a parcel of air descends, its dew point temperature increases 2°C per 1,000 meters. The height at which condensation begins, termed the lifting condensation level (LCL), can be determined by using the formula in this chapter’s special section entitled The Lifting Condensation Level.

  1. a. An air parcel has a temperature of 25°C and a dew point temperature of 14°C. What is the height of the LCL? If that parcel rises to 4,000 meters, what would be its temperature?

A parcel of air at 6,000 meters has a temperature of –5°C and a dew point of –10°C. If it descended to 2,000 meters, what would be its temperature and dew point temperature?

(a)

To determine

The height of the LCL if the air parcel has a temperature of 25°C and dew point temperature of 14°C and the temperature of the air parcel if it rises to 4,000 meters.

Explanation

The height of an air parcel at which the condensation begins is termed as Lifting Condensation Level (LCL). The formula to determine the LCL is as follows:

Lifting Condensation Level=125meters×(Celcius temperatureCelcius dew point)

Substituting the values of Celsius temperature as 25°C and Celsius dew point as 14°C,

LiftingCondensationLevel=125meters×(Celcius temperatureCelcius dew point)=125meters(2514)=125meters(11)=1375meters

To identify the temperature of air parcel at 4000meters,

Given info:

The temperature of the air parcel = 25°C

The height of the Lifting Condensation Level (LCL) =1375meters

To identify the distance between the starting point and the point of condensation, subtract the point of condensation with the starting point.

Distance between the starting point and the point of condensation} = The point of condensation - The starting point

Substituting the values 1375meters in the point of condensation and 0 meter in the starting point in the above expression,

Distance between the starting pointand the point of condensation} = 1375m - 0m                                                           = 1375m

Now, divide 1375meters by 100meters and multiply by 1°C. The reason for dividing by 100m and multiplying by 1°C is because the air parcel rises for every 100meters and the air temperature drops by 1°C.

1375meters100meters × 1°C = 13.75°C ----------------- (1)

Now, subtract the starting temperature (25°C) with the above temperature (13.75°C), as the above temperature is not the actual temperature at the point of condensation as it is in fact the total drop in temperature from the starting point to the point of the condensation level

(b)

To determine

The calculation of temperature and dew point temperature of an air parcel that has descended to 2,000 meters, which was previously at 6,000 meters having a temperature of –5°C and a dew point of –10°C.

Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

Additional Science Solutions

Find more solutions based on key concepts

Show solutions add

To prevent both muscle cramps and hyponatremia, endurance athletes who compete and sweat heavily for four or mo...

Nutrition: Concepts and Controversies - Standalone book (MindTap Course List)

26-67 How are epigenetic change related to diseases?

Introduction to General, Organic and Biochemistry

What are sister chromatids?

Human Heredity: Principles and Issues (MindTap Course List)

Overall, the normal V/Q ratio is about 0.2 0.4 0.6 0.8

Cardiopulmonary Anatomy & Physiology

Do heterotrophs (zooplankton) have a compensation depth?

Oceanography: An Invitation To Marine Science, Loose-leaf Versin

A charged cork ball of mass m is suspended on a light string in the presence of a uniform electric field as sho...

Physics for Scientists and Engineers, Technology Update (No access codes included)