Problem 3 - College Physics

Question

\(\bullet\) (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 \(44.0^{\circ} \mathrm{Fon}\) 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

(a) 27.2°C, 27.2 K; (b) -55.6°C, 55.6 K.

Step 1: Calculate Temperature Change in Fahrenheit for Part (a)

To find the change in temperature in Fahrenheit for the first situation, subtract the initial temperature from the final temperature: \[ 45.0^{\circ} \text{F} - (-4.0^{\circ} \text{F}) = 45.0^{\circ} \text{F} + 4.0^{\circ} \text{F} = 49.0^{\circ} \text{F} \]

Step 2: Convert Fahrenheit Change to Celsius for Part (a)

Use the conversion factor between Fahrenheit and Celsius: \[ \Delta T(^{\circ}C) = \frac{\Delta T(^{\circ}F)}{1.8} \]Substitute the Fahrenheit change obtained in Step 1: \[ \Delta T(^{\circ}C) = \frac{49.0}{1.8} = 27.2^{\circ}C \]

Step 3: Convert Celsius Change to Kelvins for Part (a)

Since changes in temperature are the same in both Celsius and Kelvin (1 degree Celsius change is equivalent to 1 Kelvin change), the change in Kelvin is:\[ \Delta T(K) = 27.2 K \]

Step 4: Calculate Temperature Change in Fahrenheit for Part (b)

For the second situation, calculate the change in Fahrenheit: \[ -56.0^{\circ} \text{F} - 44.0^{\circ} \text{F} = -56.0^{\circ} \text{F} + (-44.0^{\circ} \text{F}) = -100.0^{\circ} \text{F} \]

Step 5: Convert Fahrenheit Change to Celsius for Part (b)

Use the same conversion as in Step 2: \[ \Delta T(^{\circ}C) = \frac{-100.0}{1.8} = -55.6^{\circ}C \]

Step 6: Convert Celsius Change to Kelvins for Part (b)

Like in Step 3, the change in Kelvin is the same as the change in Celsius:\[ \Delta T(K) = 55.6 K \]

Key Concepts

Temperature Change

Temperature change refers to the difference in temperature between two points in time. In the case of the exercise, we are asked to calculate this change when temperatures fluctuate in both Fahrenheit and then convert this change into Celsius and Kelvin.

• Spearfish, South Dakota: The temperature increase was from \(-4.0^{\circ} \mathrm{F}\) to \(45.0^{\circ} \mathrm{F}\).
• Browning, Montana: The temperature decrease was from \(44.0^{\circ} \mathrm{F}\) to \(-56.0^{\circ} \mathrm{F}.\)

To compute the change, subtract the initial temperature from the final temperature in Fahrenheit. It's important to consider the signs since they show whether the change is an increase or a decrease. Calculating in Fahrenheit first helps because it is often the unit used in the problem. A positive result means an increase in temperature, while a negative result indicates a decrease.
Understanding temperature change is crucial, as it helps to interpret weather events, like rapid temperature changes, which may occur due to climatic conditions.

Fahrenheit to Celsius Conversion

Converting from Fahrenheit to Celsius requires using the formula that accounts for the different starting points of the two scales. The formula for converting a Fahrenheit temperature change to a Celsius temperature change is given by:\[\Delta T(^{\circ}C) = \frac{\Delta T(^{\circ}F)}{1.8}\]To apply the formula:

• Divide the change in Fahrenheit by \(1.8\).
• This conversion factor comes from the relationship that \(1^{\circ}\mathrm{C} = 1.8^{\circ}\mathrm{F}.\)

For instance, in part (a) of the exercise, a \(49.0^{\circ}\mathrm{F}\) change is converted to \(27.2^{\circ}\mathrm{C}\), while in part (b), a \(-100.0^{\circ}\mathrm{F}\) change becomes \(-55.6^{\circ}\mathrm{C}.\)
Understanding this conversion is vital in science, especially in contexts where Celsius is often more commonly used, such as chemistry and physics.

Celsius to Kelvin Conversion

While converting a change in temperature from Celsius to Kelvin, students often find that it's actually quite direct and simple. The Kelvin scale is essential in scientific calculations because it starts at absolute zero, the lowest possible temperature where particles have minimum thermal energy.

• **Add 273.15**: While this is necessary for converting a specific temperature to Kelvin (e.g., 0°C is 273.15 K), **no addition** is necessary for a temperature change.
• **1°C change = 1 K change**: This is because both Celsius and Kelvin have the same increment size.

In the provided exercise, we calculated the change in Celsius and directly translated that to Kelvin.
So, a \(27.2^{\circ}C\) change is equivalent to a \(27.2\mathrm{K}\) change, and similarly, \(-55.6^{\circ}C\) is simply \(55.6\mathrm{K}\).
This simplicity is particularly useful in thermal physics, where Kelvin is prevalent due to its direct link with the fundamental properties of particles.