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(a) On January \(22,1943,\) the temperature in Spearfish, South Dakota, rose from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\) in just 2 minutes. What was the temperature change in Celsius degrees? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January \(23,1916 .\) The next day the temperature plummeted to \(-56^{\circ} \mathrm{C}\) . What was the temperature change in Celsius degrees?

Short Answer

Expert verified
(a) South Dakota: 27.2°C; (b) Montana: 62.7°C.

Step by step solution

01

Identify Temperature Changes

First, identify the initial and final temperatures for both events. For South Dakota, the temperature changed from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\). For Montana, the temperature changed from \(44.0^{\circ} \mathrm{F}\) to \(-56^{\circ} \mathrm{C}\).
02

Convert Fahrenheit to Celsius

Use the formula \(T(\mathrm{C}) = \frac{5}{9}(T(\mathrm{F}) - 32)\) to convert the temperatures from Fahrenheit to Celsius. Calculate the initial and final temperatures in South Dakota: \(-4.0^{\circ} \mathrm{F}\) becomes \(-20.0^{\circ} \mathrm{C}\) and \(45.0^{\circ} \mathrm{F}\) becomes \(7.2^{\circ} \mathrm{C}\).
03

Calculate Temperature Change (South Dakota)

Now that both temperatures have been converted to Celsius, calculate the change by subtracting the initial temperature from the final temperature: \(7.2^{\circ} \mathrm{C} - (-20.0^{\circ} \mathrm{C}) = 27.2^{\circ} \mathrm{C}\).
04

Convert Fahrenheit to Celsius (Montana)

Convert the initial temperature from Fahrenheit to Celsius for Browning, Montana, using the same conversion formula: \(44.0^{\circ} \mathrm{F}\) becomes \(6.7^{\circ} \mathrm{C}\).
05

Calculate Temperature Change (Montana)

With the temperatures in Celsius, calculate the change by subtracting the final temperature from the initial temperature: \(6.7^{\circ} \mathrm{C} - (-56.0^{\circ} \mathrm{C}) = 62.7^{\circ} \mathrm{C}\).

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Key Concepts

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

Celsius to Fahrenheit conversion
Converting temperatures from Celsius to Fahrenheit is a basic yet essential skill when dealing with thermodynamics and weather data. The formula to convert temperatures from Fahrenheit (\(T(\mathrm{F})\)) to Celsius (\(T(\mathrm{C})\)) is:\[T(\mathrm{C}) = \frac{5}{9} (T(\mathrm{F}) - 32)\]This formula involves subtracting 32 from the Fahrenheit value, which adjusts for the offset between the two scales since 32°F corresponds to 0°C. Multiplying by \(\frac{5}{9}\) scales the temperature change to reflect the different unit sizes.
For example, let's convert \(-4.0^{\circ} \mathrm{F}\): - Start by subtracting 32: \(-4.0 - 32 = -36.0\).- Multiply by \(\frac{5}{9}\): \(-36.0 \times \frac{5}{9} = -20.0^{\circ} \mathrm{C}\).This easy conversion method provides a clear understanding of the temperature differences, essential for accurately interpreting weather phenomena.
Remember, whenever temperatures are presented in a weather report, applying this conversion can help place them within a familiar context.
Temperature change calculation
Calculating temperature changes is a straightforward process once you have both readings in the same unit. When given a temperature change in Fahrenheit, it must first be converted to Celsius for a proper comparison. This ensures consistency and accuracy in interpretation.
In the 1943 South Dakota event, the transformation was from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\). After converting these values to Celsius (-20.0°C and 7.2°C, respectively), you subtract the initial temperature from the final temperature for the change:\[7.2^{\circ} \mathrm{C} - (-20.0^{\circ} \mathrm{C}) = 27.2^{\circ} \mathrm{C}\]This calculation reveals a remarkable temperature jump in a short time frame.
In Montana's case from 1916, the temperatures went from \(44.0^{\circ} \mathrm{F}\) to \(-56.0^{\circ} \mathrm{C}\). By converting 44.0°F to 6.7°C, the temperature change can be calculated as follows:\[6.7^{\circ} \mathrm{C} - (-56.0^{\circ} \mathrm{C}) = 62.7^{\circ} \mathrm{C}\]Identifying these temperature changes helps understand the intensity and impact of such rapid atmospheric shifts visually.
Quick calculations like these are vital when studying weather patterns.
Thermodynamics in meteorology
Thermodynamics plays a crucial role in meteorology, as it governs the behavior of atmospheric systems and temperature variations. Understanding how energy transfers within the atmosphere helps explain various weather phenomena, including rapid temperature changes observed in certain events.
The atmosphere is a dynamic system where energy transfer often leads to sudden temperature shifts. For instance:
  • Rapid temperature changes can be driven by the interaction of warm and cold air masses.
  • Energy exchange through radiation, conduction, and convection affects local temperatures.
In Spearfish, South Dakota, on a remarkable January day, thermodynamic principles underpinned the swift temperature rise from -4°F to 45°F in just two minutes. Such events often occur with chinook winds, where warm air rapidly descends, warming the surface quickly.
In colder regions like Montana, significant drops in temperature, such as the shift from 44°F to -56°C, showcase the extreme outcomes of such dynamics, reinforcing the importance of thermodynamics in predicting and explaining weather patterns.
By connecting thermodynamic concepts to real-world weather situations, one can appreciate the complex mechanisms governing our planet’s climate and weather systems.

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Most popular questions from this chapter

(a) You feel sick and are told that you have a temperature of \(40.2^{\circ} \mathrm{C}\) . What is your temperature in "F? Should you be concerned? (b) The morning weather report in Sydney gives a current temperature of \(12^{\circ} \mathrm{C}\) . What is this temperature in \(^{\circ} \mathrm{F} ?\)

The rate at which radiant energy from the sun reaches the earth's upper atmosphere is about 1.50 \(\mathrm{kW} / \mathrm{m}^{2}\) . The distance from the earth to the sun is \(1.50 \times 10^{11} \mathrm{m},\) and the radius of the sun is \(6.96 \times 10^{8} \mathrm{m}\) (a) What is the rate of radiation of energy per unit area from the sun's surface? (b) If the sun radiates as an ideal black-body, what is the temperature of its surface?

While running, a \(70-\mathrm{kg}\) student generates thermal energy at a rate of 1200 \(\mathrm{W}\) . To maintain a constant body temperature of \(37^{\circ} \mathrm{C},\) this energy must be removed by perspiration or other mechanisms. If these mechanisms failed and the heat could not flow out of the student's body, for what amount of time could a student run before irreversible body damage occurred? (Note: Protein structures in the body are irreversibly damaged if body temperature rises to \(44^{\circ} \mathrm{C}\) or higher. The specific heat of a typical human body is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) , shightly less than that of water. The difference is due to the presence of protein, fat, and minerals, which have lower specific heats.)

An asteroid with a diameter of 10 \(\mathrm{km}\) and a mass of \(2.60 \times 10^{15} \mathrm{kg}\) impacts the earth at a speed of 32.0 \(\mathrm{km} / \mathrm{s}\) , landing in the Pacific Ocean. If 1.00\(\%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(10.0^{\circ} \mathrm{C} )\) , what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{kg}\) .)

You put a bottle of soft drink in a refrigerator and leave it until its temperature has dropped 10.0 \(\mathrm{K}\) . What is its temperature change in (a) \(\mathrm{F}^{\circ}\) and \((\mathrm{b}) \mathrm{C}^{\circ}\) ?

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