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Chapter 3: Problem 83

What is \(-20^{\circ} \mathrm{F}\) in degrees Celsius and in kelvins? (3.3)

Short Answer

Expert verified
-20掳F is approximately -28.9掳C and 244.25 K

Step by step solution

01

Understanding the conversion formula from Fahrenheit to Celsius

To convert from Fahrenheit to Celsius, use the formula: \( C = \frac{5}{9}(F - 32) \) where F is the temperature in Fahrenheit and C is the temperature in Celsius.
02

Substitute the given Fahrenheit temperature

Substitute \(-20^{\circ} \mathrm{F}\) into the formula: \( C = \frac{5}{9}(-20 - 32) \)
03

Perform the subtraction inside the parentheses

Calculate \(-20 - 32 = -52 \)
04

Multiply and divide to find Celsius

Complete the calculation: \( C = \frac{5}{9}(-52) = \frac{5 \times -52}{9} = \frac{-260}{9} \approx -28.9^{\circ} \mathrm{C}\)
05

Understanding the conversion formula from Celsius to Kelvin

To convert from Celsius to Kelvin, use the formula: \( K = C + 273.15 \) where K is the temperature in Kelvin and C is the temperature in Celsius.
06

Substitute the Celsius temperature

Substitute \(-28.9^{\circ} \mathrm{C}\) into the formula: \( K = -28.9 + 273.15 \)
07

Perform the addition

Calculate the sum: \( -28.9 + 273.15 = 244.25 \mathrm{K} \)

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Key Concepts

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

Fahrenheit to Celsius Conversion
When converting Fahrenheit (掳F) to Celsius (掳C), it's essential to use the right formula. The formula is: \( C = \frac{5}{9}(F - 32) \). In this formula, **C** represents the temperature in Celsius, and **F** represents the temperature in Fahrenheit.
Let's consider the specific example of \(-20^{\circ} \mathrm{F}\).
1. First, subtract 32 from the given Fahrenheit temperature: \( -20 - 32 = -52 \).2. Then, multiply the result by \(\frac{5}{9}\): \( \frac{5}{9}( -52 ) = \frac{-260}{9} \approx -28.9^{\circ} \mathrm{C} \).Converting temperatures this way helps you understand the relationship between the Fahrenheit and Celsius scales.
Celsius to Kelvin Conversion
After converting temperatures from Fahrenheit to Celsius, you might need to convert Celsius to Kelvin (K). The Kelvin scale is used in scientific settings because it starts at absolute zero. Use the following formula to convert Celsius to Kelvin: \( K = C + 273.15 \). Here, **K** is the temperature in Kelvin, and **C** is the temperature in Celsius.
Using our earlier example, where \(-28.9^{\circ} \mathrm{C}\):
  • Add 273.15 to the Celsius temperature: \( K = -28.9 + 273.15 \).
  • This results in \( 244.25 \mathrm{K} \).
Remember, the Kelvin scale does not use degrees, simply Kelvins (K). Each step clarifies the temperature by removing the dependency on how zero is set in everyday temperature scales.
Temperature Calculations
Understanding temperature conversions involves some essential temperature calculations. This skill is crucial not just for physics, but also for chemistry, engineering, and everyday applications. Here are key points to remember:
  • Subtraction or Addition: You first adjust the temperature by common reference points (like subtracting 32 in Fahrenheit to Celsius conversion).
  • Scaling Factor: For Fahrenheit to Celsius, the scaling factor is \(\frac{5}{9}\).
  • Translation: For Celsius to Kelvin, you simply add \(273.15\) rather than scale.
Understanding and practicing these steps will make temperature calculations second nature. Next time you need to convert temperatures, recall these simple steps 鈥� they turn otherwise complex-seeming problems into straightforward tasks.

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Most popular questions from this chapter

Identify each of the following changes of state as melting, freezing, sublimation or deposition: a. Ice sculptures start breaking down as temperature rises above \(0^{\circ} \mathrm{C}\) b. Ice forms on the strawberry plants on a cold wintry night. c. Water becomes solid at \(0^{\circ} \mathrm{C}\). d. Dry ice on heating becomes gas.

a. Water is heated to \(145^{\circ} \mathrm{F}\). What is the temperature of the hot water in degrees Celsius? b. During extreme hypothermia, a child's temperature dropped to \(20.6{ }^{\circ} \mathrm{C}\). What was his temperature in degrees Fahrenheit?

Classify each of the following mixtures as homogeneous or heterogeneous. a. sugar solution b. aerated drinks c. sand in water d. a mixed vegetable dish e. mango juice with mango slices

a. During one week, Charles swam for a total of \(2.5 \mathrm{~h}\) and walked for a total of \(8.0 \mathrm{~h}\). If Charles expends 340 kcal/h swimming and 160 kcal/h walking, how many total kilocalories did he expend for one week? b. For the amount of exercise that Charles did for one week in part a, if expending 3500 kcal is equal to a loss of \(1.0 \mathrm{lb},\) how many pounds did he lose? c. How many hours would Charles have to walk to lose \(1.0 \mathrm{lb} ?\) d. How many hours would Charles have to swim to lose \(1.0 \mathrm{lb} ?\)

Classify each of the following as a pure substance or a mixture: a. a soft drink b. propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) c. a cheese sandwich d. an iron (Fe) nail e. salt substitute (KCl)
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