Problem 3 - College 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 and in kelvins? (b) The temperature in Browning, Montana, was \(44.0^{\circ} \mathrm{F}\) on January 23,1916 , and the next day it plummeted to \(-56.0^{\circ} \mathrm{F}\). What was the temperature change in Celsius degrees and in kelvins?

Step-by-Step Solution

Answer

Spearfish: 27.2°C, 27.2 K; Browning: -55.6°C, -55.6 K.

Step 1: Find the Fahrenheit Temperature Change for Spearfish

The temperature in Spearfish rose from \(-4.0^{\circ} \text{F} \) to \(45.0^{\circ} \text{F} \). Calculate the temperature change in Fahrenheit: \(\Delta T_F = 45.0^{\circ} \text{F} - (-4.0^{\circ} \text{F}) = 49.0^{\circ} \text{F}.\)

Step 2: Convert the Spearfish Temperature Change to Celsius

Use the conversion formula for Fahrenheit to Celsius: \(\Delta T_C = \frac{5}{9}(\Delta T_F) = \frac{5}{9}(49.0^{\circ} \text{F}) \approx 27.2^{\circ} \text{C}.\)

Step 3: Convert the Spearfish Temperature Change to Kelvins

Since a change in Celsius is numerically equal to a change in Kelvin: \(\Delta T_K = \Delta T_C = 27.2 \text{ K}.\)

Step 4: Find the Fahrenheit Temperature Change for Browning

The temperature in Browning dropped from \(44.0^{\circ} \text{F}\) to \(-56.0^{\circ} \text{F}\). Calculate the temperature change in Fahrenheit: \(\Delta T_F = (-56.0^{\circ} \text{F}) - (44.0^{\circ} \text{F}) = -100.0^{\circ} \text{F}.\)

Step 5: Convert the Browning Temperature Change to Celsius

Use the conversion formula for Fahrenheit to Celsius: \(\Delta T_C = \frac{5}{9}(\Delta T_F) = \frac{5}{9}(-100.0^{\circ} \text{F}) \approx -55.6^{\circ} \text{C}.\)

Step 6: Convert the Browning Temperature Change to Kelvins

Since a change in Celsius is numerically equal to a change in Kelvin: \(\Delta T_K = \Delta T_C = -55.6 \text{ K}.\)

Key Concepts

Fahrenheit to Celsius

We often encounter temperature readings in Fahrenheit, particularly in the United States. However, many scientific disciplines and international contexts use Celsius. Therefore, converting between these two scales is essential. The formula to convert a temperature change from Fahrenheit to Celsius is:

• \( \Delta T_C = \frac{5}{9}(\Delta T_F) \)

To apply this, take the difference in Fahrenheit temperatures (\( \Delta T_F \)), and multiply by \( \frac{5}{9} \). For example, if the temperature change is \( 49.0^{\circ} \text{F} \), converting to Celsius would be:

• \( \Delta T_C = \frac{5}{9}(49.0) \approx 27.2^{\circ} \text{C} \)

This formula effectively rescales the measurement from the larger °F increments to the smaller °C ones. Understanding this conversion helps in interpreting temperature changes more widely.

Temperature Change

Temperature change measures how much the temperature rises or falls over a period. It is calculated by subtracting the initial temperature from the final temperature. For instance:

• In Spearfish, the temperature increased from \(-4.0^{\circ} \text{F} \) to \(45.0^{\circ} \text{F} \), making the change \( 49.0^{\circ} \text{F} \).
• In Browning, it fell from \(44.0^{\circ} \text{F} \) to \(-56.0^{\circ} \text{F} \), so the change was \(-100.0^{\circ} \text{F} \).

Even though temperature is typically relative, the change is absolute. When calculated, it indicates the magnitude of increase or decrease. This can help understand the rate of temperature change, especially during unusual weather patterns. In scientific terms, recognizing these changes assists in studies on climate and atmosphere.

Kelvin Scale

The Kelvin scale is often used in scientific research due to its absolute nature. It starts at absolute zero, the point where molecular motion stops. Unlike Celsius and Fahrenheit, Kelvin doesn't have negative numbers.When converting temperature changes, note that an equal change in Celsius is also equal in Kelvin. Therefore, if the change is \(27.2^{\circ} \text{C} \), it would also be \(27.2 \text{ K} \), and similarly for negative changes:

• For Browning's temperature change, \(-55.6^{\circ} \text{C} \) translates directly to \(-55.6 \text{ K} \).

This equivalence simplifies calculations in scientific contexts. Despite its use for absolute temperature readings, Kelvin is highly valuable in scenarios requiring comparison of precise temperature changes, often in physics and engineering.