Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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

Short Answer

Expert verified
(a) 27.22°C, (b) -55.56°C.

Step by step solution

01

Convert Fahrenheit to Celsius Formula

The formula to convert a temperature from Fahrenheit (F) to Celsius (C) is given by: \[ C = \frac{5}{9}(F - 32) \] We will use this formula to compute the temperature changes in Celsius for both parts (a) and (b).
02

Convert Initial and Final Temperatures for Part (a)

Calculate Celsius temperatures for initial and final values in part (a).- Initial temperature: \[ C_{initial} = \frac{5}{9}(-4 - 32) = \frac{5}{9}(-36) = -20 \] - Final temperature: \[ C_{final} = \frac{5}{9}(45 - 32) = \frac{5}{9}(13) \approx 7.22 \]
03

Calculate Temperature Change in Celsius for Part (a)

Subtract the initial Celsius temperature from the final Celsius temperature for part (a) to find the temperature change.\[ \Delta C = C_{final} - C_{initial} = 7.22 - (-20) = 7.22 + 20 \approx 27.22 \] Thus, the temperature change in Celsius is approximately 27.22°C.
04

Convert Initial and Final Temperatures for Part (b)

Calculate Celsius temperatures for initial and final values in part (b).- Initial temperature: \[ C_{initial} = \frac{5}{9}(44 - 32) = \frac{5}{9}(12) \approx 6.67 \] - Final temperature: \[ C_{final} = \frac{5}{9}(-56 - 32) = \frac{5}{9}(-88) = -48.89 \]
05

Calculate Temperature Change in Celsius for Part (b)

Subtract the final Celsius temperature from the initial Celsius temperature for part (b) to determine the temperature change.\[ \Delta C = C_{final} - C_{initial} = -48.89 - 6.67 = -55.56 \] Thus, the temperature change in Celsius is approximately -55.56°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 Vaia!

Key Concepts

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

Fahrenheit to Celsius conversion
Temperature is often measured in either Fahrenheit or Celsius, and it's crucial to be able to convert between these two scales. The formula for converting Fahrenheit to Celsius is quite simple:
  • The formula is \( C = \frac{5}{9}(F - 32) \), where \( C \) represents Celsius and \( F \) represents Fahrenheit.
To convert a Fahrenheit temperature to Celsius, you simply subtract 32 from the Fahrenheit value and then multiply the result by \( \frac{5}{9} \). This method is commonly used in scientific calculations and everyday situations where temperature conversion is necessary.
Whether determining the weather outside or engineering a scientific experiment, understanding how to convert between Fahrenheit and Celsius will aid in comprehending temperature-related data presented in different formats.
Temperature change calculation
Understanding how to calculate temperature change is important in grasping weather fluctuations and their implications. This involves converting specific temperatures from Fahrenheit to Celsius and then finding the difference between the two values in Celsius.
  • Suppose you have an initial temperature \( C_{initial} \) and a final temperature \( C_{final} \).
  • The temperature change \( \Delta C \) is calculated as \( C_{final} - C_{initial} \).
In our example, temperatures initially measured in Fahrenheit are converted to Celsius using the formula discussed earlier. After obtaining the Celsius values, the change is found by simple subtraction.
This process is vital for recognizing the extent of temperature shifts, whether for academic purposes or evaluating historical weather anomalies.
Historical weather events
Historical weather events often provide insight into the extreme capabilities of Earth's climate. Events like the dramatic temperature changes recorded in Spearfish, South Dakota and Browning, Montana, highlight extreme weather phenomena that can occur.
  • For instance, in Spearfish on January 22, 1943, the temperature shot up by about \( 27.22\)°C in just two minutes.
  • In another case, Browning experienced a sudden drop by approximately \(-55.56\)°C overnight from January 23, 1916.
Understanding these events involves recognizing how quickly temperatures can change under specific atmospheric conditions.
Not only do these historical accounts serve as interesting case studies, but they also underscore the importance of accurate temperature conversion and weather monitoring in modern times.

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

Styrofoam bucket of negligible mass contains 1.75 kg of water and 0.450 kg of ice. More ice, from a refrigerator at -15.0\(^\circ\)C, is added to the mixture in the bucket, and when thermal equilibrium has been reached, the total mass of ice in the bucket is 0.884 kg. Assuming no heat exchange with the surroundings, what mass of ice was added?

A U.S. penny has a diameter of 1.9000 cm at 20.0\(^\circ\)C. The coin is made of a metal alloy (mostly zinc) for which the coefficient of linear expansion is \(2.6 \times 10{^-}{^5} K{^-}{^1}\). What would its diameter be on a hot day in Death Valley (48.0\(^\circ\)C)? On a cold night in the mountains of Greenland (-53\(^\circ\)C)?

BIO Conduction Through the Skin. The blood plays an important role in removing heat from the body by bringing this energy directly to the surface where it can radiate away. Nevertheless, this heat must still travel through the skin before it can radiate away. Assume that the blood is brought to the bottom layer of skin at 37.0\(^\circ\)C and that the outer surface of the skin is at 30.0\(^\circ\)C. Skin varies in thickness from 0.50 mm to a few millimeters on the palms and soles, so assume an average thickness of 0.75 mm. A 165-lb, 6-ft-tall person has a surface area of about 2.0 m\(^2\) and loses heat at a net rate of 75 W while resting. On the basis of our assumptions, what is the thermal conductivity of this person’s skin?

A surveyor's 30.0-m steel tape is correct at 20.0\(^\circ\)C. The distance between two points, as measured by this tape on a day when its temperature is 5.00\(^\circ\)C, is 25.970 m. What is the true distance between the points?

Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300 m and the length of the copper section is 0.800 m. Each segment has cross-sectional area 0.00500 m\(^2\). The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?

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.

Sign-up for free