/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 (a) On January \(22,1943,\) the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

(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 \(-56^{\circ} \mathrm{F} .\) What was the temperature change in Celsius degrees?

Short Answer

Expert verified
Spearfish: 27.2°C increase; Browning: 55.6°C decrease.

Step by step solution

01

Convert Initial Fahrenheit Temperature to Celsius (Spearfish)

Use the formula \[ C = \frac{5}{9}(F-32) \] to convert the initial temperature of \(-4.0^{\circ} F\) to Celsius. Substitute: \[ C = \frac{5}{9}(-4.0 - 32) = \frac{5}{9}(-36) = -20^{\circ} C \]
02

Convert Final Fahrenheit Temperature to Celsius (Spearfish)

Convert the final temperature of \(45.0^{\circ} F\) to Celsius using the same formula. Substitute: \[ C = \frac{5}{9}(45.0 - 32) = \frac{5}{9} \times 13 = 7.2^{\circ} C \]
03

Calculate Temperature Change in Celsius (Spearfish)

Subtract the initial Celsius temperature from the final Celsius temperature: \[ 7.2^{\circ} C - (-20^{\circ} C) = 7.2^{\circ} C + 20^{\circ} C = 27.2^{\circ} C \]
04

Convert Initial Fahrenheit Temperature to Celsius (Browning)

Convert the initial temperature of \(44.0^{\circ} F\) to Celsius. Substitute: \[ C = \frac{5}{9}(44.0 - 32) = \frac{5}{9} \times 12 = 6.7^{\circ} C \]
05

Convert Final Fahrenheit Temperature to Celsius (Browning)

Convert the final temperature of \(-56^{\circ} F\) to Celsius. Substitute: \[ C = \frac{5}{9}(-56 - 32) = \frac{5}{9} \times (-88) = -48.9^{\circ} C \]
06

Calculate Temperature Change in Celsius (Browning)

Subtract the final Celsius temperature from the initial Celsius temperature: \[ 6.7^{\circ} C - (-48.9^{\circ} C) = 6.7^{\circ} C + 48.9^{\circ} C = 55.6^{\circ} C \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Fahrenheit to Celsius Conversion
To convert temperatures from Fahrenheit to Celsius, you use a straightforward formula:
  • Start with the Fahrenheit temperature.
  • Subtract 32 from it.
  • Multiply the result by \( \frac{5}{9} \).
This formula arises because of the different scales that Fahrenheit and Celsius use for freezing and boiling points of water. While water freezes at \(0^{\circ}C\), it does so at \(32^{\circ}F\).
In the example of Spearfish, to convert \(-4^{\circ}F\) to Celsius, you do \(-4 - 32 = -36\), then multiply by \(\frac{5}{9}\), which results in \(-20^{\circ}C\).
A similar process converts \(45^{\circ}F\) to \(7.2^{\circ}C\). This simple conversion is essential when comparing temperatures on different scales.
Temperature Change Calculation
Calculating the change in temperature is another simple yet crucial operation. To find out how much the temperature has increased or decreased:
  • Convert both the initial and final temperatures to Celsius, if they are not already.
  • Subtract the initial temperature from the final temperature.
If the temperature rises, the result is positive, indicating a gain. If it falls, the result is negative, indicating a loss.
In Spearfish, the temperature rose from \(-20^{\circ}C\) to \(7.2^{\circ}C\), resulting in a change of \(7.2 - (-20) = 27.2^{\circ}C\), indicating a significant increase. In Browning, a drop from \(6.7^{\circ}C\) to \(-48.9^{\circ}C\) means a fall of \(55.6^{\circ}C\), a major decrease that underscores the extreme weather of the area.
Historical Weather Events
Historical weather events often highlight dramatic temperature changes, which can be astonishing and offer insights into local climate conditions.
For example, Spearfish, South Dakota, holds the record for one of the fastest recorded temperature changes. On January 22, 1943, it went from \(-4^{\circ}F\) to \(45^{\circ}F\) in just two minutes, likely due to a sudden shift in air masses. Similarly, in Browning, Montana, there's another historical instance where temperatures plummeted sharply within a day, highlighting the unpredictable nature of weather, especially during winter.
Extreme events like these are essential for understanding climate phenomena. They offer insights into how local geographical features, like mountain ranges or atmospheric conditions, can influence weather patterns significantly. Such records also help researchers study climate behavior and better prepare for potential extreme weather in the future.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

Like the Kelvin scale, the Rankine scale is an absolute temperature scale: Absolute zero is zero degrees Rankine \(\left(0^{\circ} \mathrm{R}\right)\) However, the units of this scale are the same size as those of the Fahrenheit scale rather than the Celsius scale. What is the numerical value of the triple-point temperature of water on the Rankine scale?

Heat Loss During Breathing. In very cold weather a significant mechanism for heat loss by the human body is energy expended in warming the air taken into the lungs with each breath. (a) On a cold winter day when the temperature is \(-20^{\circ} \mathrm{C},\) what amount of heat is needed to warm to body temperature \(\left(37^{\circ} \mathrm{C}\right)\) the 0.50 \(\mathrm{L}\) of air exchanged with each breath? Assume that the specific heat of air is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that 1.0 \(\mathrm{L}\) of air has mass \(1.3 \times 10^{-3} \mathrm{kg}\) . (b) How much heat is lost per hour if the respiration rate is 20 breaths per minute?

A copper calorimeter can with mass 0.100 \(\mathrm{kg}\) contains 0.160 \(\mathrm{kg}\) of water and 0.0180 \(\mathrm{kg}\) of ice in thermal equilibrium at atmospheric pressure. If 0.750 \(\mathrm{kg}\) of lead at a temperature of \(255^{\circ} \mathrm{C}\) is dropped into the calorimeter can, what is the final temperature? Assume that no heat is lost to the surroundings.

The outer diameter of a glass jar and the inner diameter of its iron lid are both 725 \(\mathrm{mm}\) at room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\) What will be the size of the difference in these diameters if the lid is briefly held under hot water until its temperature rises to \(50.0^{\circ} \mathrm{C}\) without changing the temperature of the glass?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.