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Chapter 19: Problem 57

Chinook. During certain seasons strong winds called chinooks blow from the west across the eastern slopes of the Rockies and downhill into Denver and nearby areas. Although the mountains are cool, the wind in Denver is very hot; within a few minutes after the chinook wind arrives, the temperature can climb 20 \(\mathrm{C}^{\circ}\) ("chinook" is a Native American word meaning "snow cater"). Similar winds occur in the Alps (called fochns) and in southern California (called Santa Anas). (a) Explain why the temperature of the chinook wind rises as it descends the slopes. Why is it important that the wind be fast moving? (b) Suppose a strong wind is blowing toward Denver (elevation 1630 \(\mathrm{m} )\) from Grays Peak \((80 \mathrm{km} \text { west of Denver, at an elevation of } 4350 \mathrm{m})\) , where the air pressure is \(5.60 \times 10^{4} \mathrm{Pa}\) and the air temperature is \(-15.0 ^{\circ} \mathrm{C}\) . The temperature and pressure in Denver before the wind arrives are \(20^{\circ} \mathrm{C}\) and \(8.12 \times 10^{4} \mathrm{Pa}\) . By how many Celsius degrees will the temperature in Denver rise when the chinook arrives?

Short Answer

Expert verified
The temperature in Denver will rise by approximately 26.656掳C.

Step by step solution

01

Understanding the Adiabatic Process

As the chinook wind descends from Grays Peak to Denver, it undergoes an adiabatic warming process. When air descends, it is compressed due to the increase in air pressure, which raises the temperature of the air.
02

Calculate the Elevation Difference

The elevation difference between Grays Peak and Denver is given by the difference in their elevations. Grays Peak is at 4350 m, while Denver is at 1630 m. Thus, the elevation difference is 4350 m - 1630 m = 2720 m.
03

Identify the Adiabatic Lapse Rate

For unsaturated (dry) air, the dry adiabatic lapse rate is approximately 9.8掳C per 1000 m. This will be used to calculate the temperature change due to the elevation drop.
04

Calculate the Temperature Increase Due to Descent

Given the elevation drop of 2720 m and the dry adiabatic lapse rate of 9.8掳C per 1000 m, the temperature increase can be calculated as follows: \[ \Delta T = 9.8 \frac{\text{掳C}}{1000 \text{m}} \times 2720 \text{m} = 26.656 \text{掳C} \]
05

Determine Temperature Change in Denver

The initial temperature in Denver before the chinook wind arrives is 20掳C. With an increase of approximately 26.656掳C due to the chinook effect, the final temperature in Denver can be calculated: \[ T_{\text{final}} = 20\text{掳C} + 26.656\text{掳C} = 46.656\text{掳C} \]
06

Calculate the Temperature Rise in Denver

The temperature in Denver increases by 26.656掳C when the chinook wind arrives, as calculated in the previous step.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chinook Winds
Chinook winds are fascinating meteorological phenomena that dramatically affect the climate of regions like the eastern slopes of the Rockies. These winds, also known as "snow eaters," quickly raise temperatures, sometimes by 20掳C or more. Their high speed is critical because it minimizes the time air has to lose heat to the surroundings as it travels down the mountain slopes. By maintaining momentum, the hot air reaches areas like Denver before cooling, rapidly increasing local temperatures. Chinook winds share similarities with other local wind phenomena like the F枚hn in the Alps and Santa Ana winds in Southern California, which also produce swift temperature changes.
Adiabatic Lapse Rate
The adiabatic lapse rate describes how the temperature of air changes with altitude, specifically when it is moving up or down without exchanging heat with the environment. For dry air, the rate of temperature change is approximately 9.8掳C per 1000 meters of altitude change. This is known as the dry adiabatic lapse rate. When chinook winds descend the slopes of the Rockies, they undergo adiabatic heating. This lapse rate helps us calculate how much warmer the air becomes as it drops in altitude, leading to their heating effect at lower elevations. Understanding this concept is key for meteorologists in predicting weather patterns.
Temperature Compression
Temperature compression occurs when air descends and is compressed due to higher atmospheric pressure closer to the earth's surface. This compression increases the air's kinetic energy, raising its temperature. In the case of chinook winds, as they descend from high mountain elevations to lower areas like Denver, the increase in atmospheric pressure causes the air temperature to rise quickly. This process explains why these winds bring warmth even when originating from cooler mountain tops. Temperature compression is a vital concept in explaining how local weather patterns, like chinook winds, can lead to swift temperature changes.
Atmospheric Pressure Changes
Atmospheric pressure changes are crucial in understanding how chinook winds affect weather conditions. As air moves from higher elevations, where pressure is lower, to lower elevations like Denver, where pressure is higher, it compresses and heats up. This change in pressure is instrumental in the adiabatic processes, causing the temperature to rise rapidly. Atmospheric pressure varies significantly with altitude, and these changes are a driving force behind weather phenomena. Recognizing the relationship between pressure variations and wind behavior is essential in meteorology, particularly when forecasting the impact of winds like chinooks on local climates.

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Most popular questions from this chapter

Boiling Water at High Pressure. When water is boiled at a pressure of \(2.00 \mathrm{atm},\) the heat of vaporization is \(2.20 \times 10^{5} \mathrm{J} / \mathrm{kg}\) and the boiling point is \(120^{\circ} \mathrm{C}\) . At this pressure, 1.00 \(\mathrm{kg}\) of water has a volume of \(1.00 \times 10^{-3} \mathrm{m}^{3}\) , and 1.00 \(\mathrm{kg}\) of steam has a volume of \(0.824 \mathrm{m}^{3} .\) (a) Compute the work done when 1.00 \(\mathrm{kg}\) of steam is formed at this temperature. (b) Compute the increase in internal energy of the water.

An experimenter adds \(970 \mathrm{~J}\) of heat to \(1.75 \mathrm{~mol}\) of an ideal gas to heat it from \(10.0^{\circ} \mathrm{C}\) to \(25.0^{\circ} \mathrm{C}\) at constant pressure. The gas does \(+223 \mathrm{~J}\) of work during the expansion. (a) Calculate the change in internal energy of the gas. (b) Calculate \(\gamma\) for the gas.

A cylinder with a movable piston contains 3.00 \(\mathrm{mol}\) of \(\mathrm{N}_{2}\) gas (assumed to behave like an ideal gas). (a) The \(\mathrm{N}_{2}\) is heated at constant volume until 1557 \(\mathrm{J}\) of heat have been added. Calculate the change in temperature. (b) Suppose the same amount of heat is added to the \(\mathrm{N}_{2}\) , but this time the gas is allowed to expand while remaining at constant pressure. Calculate the temperature change. (c) In which case, (a) or \((b),\) is the final internal energy of the \(\mathbf{N}_{2}\) higher? How do you know? What accounts for the difference between the two cases?

Oscillations of a Piston. A vertical cylinder of radins \(r\) contains a quantity of ideal gas and is fitted with a piston with mass \(m\) that is free to move (Fig. 19.34\()\) . The piston and the walls of the cylinder are frictionless and the entire cylinder is placed in aconstant-temperature bath. The outside air pressure is \(p_{0}\) . In equilibrium, the piston sits at a height \(h\) above the bottom of the cylinder. (a) Find the absolute pressure of the gas trapped below the piston when in equilibrium. (b) The piston is pulled up by a small distance and released. Find the net force acting on the piston when its base is a distance \(h+y\) above the bottom of the cylinder, where \(y\) is much less than \(h\) (c) After the piston is displaced from equilibrium and released, it oscillates up and down. Find the frequency of nthese small oscillations. If the displacement is not small, are the oscillations simple harmonic? How can you tell?

A gas in a cylinder is held at a constant pressure of \(2.30 \times 10^{5} \mathrm{Pa}\) and is cooled and compressed from 1.70 \(\mathrm{m}^{3}\) to \(1.20 \mathrm{m}^{3} .\) The internal energy of the gas decreases by \(1.40 \times 10^{5} \mathrm{J}\) (a) Find the work done by the gas. (b) Find the absolute value \(|Q|\) of the heat flow into or out of the gas, and state the direction of the heat flow. (c) Does it matter whether the gas is ideal? Why or why not?
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