Problem 6 - University Physics with Modern Physics

Question

(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?

Step-by-Step Solution

Answer

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

Step 1: 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 \(-56^{\circ} \mathrm{C}\).

Step 2: 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 \(7.2^{\circ} \mathrm{C}\).

Step 3: 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: \(7.2^{\circ} \mathrm{C} - (-20.0^{\circ} \mathrm{C}) = 27.2^{\circ} \mathrm{C}\).

Step 4: 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 \(6.7^{\circ} \mathrm{C}\).

Step 5: Calculate Temperature Change (Montana)

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

Key Concepts

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.