/*! 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 27 Make the following conversions: ... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 27

Make the following conversions: (a) \(72{ }^{\circ} \mathrm{F}\) to \({ }^{\circ} \mathrm{C}\), (b) \(216.7^{\circ} \mathrm{C}\) to \({ }^{\circ} \mathrm{F}\), (c) \(233^{\circ} \mathrm{C}\) to \(\mathrm{K}\), (d) \(315 \mathrm{~K}\) to \({ }^{\circ} \mathrm{F}\), (e) \(2500^{\circ} \mathrm{F}\) to \(\mathrm{K}\), (f) \(0 \mathrm{~K}\) to \({ }^{\circ} \mathrm{F}\).

Short Answer

Expert verified
(a) \(72^{\circ}\mathrm{F} = 22.22^{\circ}\mathrm{C}\) (b) \(216.7^{\circ}\mathrm{C} = 454.06^{\circ}\mathrm{F}\) (c) \(233^{\circ}\mathrm{C} = 506.15\mathrm{K}\) (d) \(315\mathrm{K} = 107.33^{\circ}\mathrm{F}\) (e) \(2500^{\circ}\mathrm{F} = 1640.65\mathrm{K}\) (f) \(0\mathrm{K} = -459.67^{\circ}\mathrm{F}\)

Step by step solution

01

Apply Fahrenheit to Celsius formula

Use the formula: \(C = \frac{5}{9}(F - 32)\), where F is the temperature in Fahrenheit. Substitute the given temperature, F = 72, into the formula: \(C = \frac{5}{9}(72 - 32)\)
02

Calculate the Celsius temperature

Carry out the calculations: \(C = \frac{5}{9}(40) \Rightarrow C = 22.22^{\circ}\mathrm{C}\) So, \(72{ }^{\circ} \mathrm{F} = 22.22^{\circ} \mathrm{C}\). (b) Convert \(216.7{ }^{\circ} \mathrm{C}\) to \(^{\circ}\mathrm{F}\)
03

Apply Celsius to Fahrenheit formula

Use the formula: \(F = \frac{9}{5}C + 32\), where C is the temperature in Celsius. Substitute the given temperature, C = 216.7, into the formula: \(F = \frac{9}{5}(216.7) + 32\)
04

Calculate the Fahrenheit temperature

Carry out the calculations: \(F = 422.06 + 32 \Rightarrow F = 454.06^{\circ}\mathrm{F}\) So, \(216.7^{\circ}\mathrm{C} = 454.06^{\circ}\mathrm{F}\). (c) Convert \(233^{\circ} \mathrm{C}\) to \(K\)
05

Apply Celsius to Kelvin formula

Use the formula: \(K = C + 273.15\), where C is the temperature in Celsius. Substitute the given temperature, C = 233, into the formula: \(K = 233 + 273.15\)
06

Calculate the Kelvin temperature

Carry out the calculations: \(K = 506.15\mathrm{K}\) So, \(233^{\circ} \mathrm{C} = 506.15 \mathrm{K}\). (d) Convert \(315 \mathrm{~K}\) to \(^{\circ} \mathrm{F}\)
07

Apply Kelvin to Fahrenheit formula

Use the formula: \(F = \frac{9}{5}(K - 273.15) + 32\), where K is the temperature in Kelvin. Substitute the given temperature, K = 315, into the formula: \(F = \frac{9}{5}(315 - 273.15) + 32\)
08

Calculate the Fahrenheit temperature

Carry out the calculations: \(F = \frac{9}{5}(41.85) + 32 \Rightarrow F = 107.33^{\circ}\mathrm{F}\) So, \(315 \mathrm{K} = 107.33^{\circ} \mathrm{F}\). (e) Convert \(2500{ }^{\circ} \mathrm{F}\) to \(\mathrm{K}\)
09

Apply Fahrenheit to Kelvin formula

Use the formula: \(K = \frac{5}{9}(F - 32) + 273.15\), where F is the temperature in Fahrenheit. Substitute the given temperature, F = 2500, into the formula: \(K = \frac{5}{9}(2500 - 32) + 273.15\)
10

Calculate the Kelvin temperature

Carry out the calculations: \(K = \frac{5}{9}(2468) + 273.15 \Rightarrow K = 1640.65\mathrm{K}\) So, \(2500{ }^{\circ} \mathrm{F} = 1640.65 \mathrm{K}\). (f) Convert \(0 \mathrm{~K}\) to \(^{\circ} \mathrm{F}\)
11

Apply Kelvin to Fahrenheit formula

Use the formula: \(F = \frac{9}{5}(K - 273.15) + 32\), where K is the temperature in Kelvin. Substitute the given temperature, K = 0, into the formula: \(F = \frac{9}{5}(0 - 273.15) + 32\)
12

Calculate the Fahrenheit temperature

Carry out the calculations: \(F = \frac{9}{5}(-273.15) + 32 \Rightarrow F = -459.67^{\circ}\mathrm{F}\) So, \(0 \mathrm{K} = -459.67^{\circ} \mathrm{F}\).

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Key Concepts

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

Celsius to Fahrenheit
Converting from Celsius to Fahrenheit is a common task in science and everyday life, and understanding it can help you interpret temperature data easily. The formula to convert Celsius (\(C\)) to Fahrenheit (\(F\)) is:
  • \( F = \frac{9}{5}C + 32 \)
Let's break it down: you multiply the Celsius temperature by 9/5 and then add 32. This method accounts for the adjustment difference between the two scales' starting points and size of degrees. For example, to convert \(216.7^{\circ}\)C to Fahrenheit, we would calculate:
  • \( F = \frac{9}{5} \, (216.7) + 32 \)
  • \( F = 454.06^{\circ}F \)
This example shows why conversion is important; it allows us to understand temperature in a format we may be more familiar with.
Fahrenheit to Celsius
Converting Fahrenheit to Celsius is essential when you wish to switch from a more localized scale (used mostly in the U.S.) to a globally recognized one. The conversion formula is:
  • \( C = \frac{5}{9} (F - 32) \)
To use this formula, subtract 32 from the Fahrenheit temperature, then multiply by 5/9. This reflects the degree size difference and shifting zero point of the scales. For example, converting \(72^{\circ}\)F to Celsius involves:
  • \( C = \frac{5}{9} \, (72 - 32) \)
  • \( C = 22.22^{\circ}C \)
The conversion is particularly useful in scientific settings where Celsius is the standard unit of measure.
Celsius to Kelvin
In scientific contexts, the Kelvin scale is often used because it is an absolute temperature scale, meaning zero Kelvin represents absolute zero, or the point where particle motion ceases. Converting from Celsius to Kelvin is simple, using the formula:
  • \( K = C + 273.15 \)
Here, you just add 273.15 to the Celsius value, reflecting the difference between the freezing point of water on both scales. For example, converting\(233^{\circ}\)C to Kelvin:
  • \( K = 233 + 273.15 \)
  • \( K = 506.15 \, K \)
This highlights how Kelvin is useful in thermodynamics and other physics applications, providing a scale that works well with various physical laws.
Kelvin to Fahrenheit
Occasionally, you might need to convert Kelvin to Fahrenheit, especially if you're interpreting scientific data for broader audiences or specific applications. The conversion formula combines parts of both Celsius to Fahrenheit and Celsius to Kelvin conversions:
  • \( F = \frac{9}{5} (K - 273.15) + 32 \)
Start by subtracting 273.15 from the Kelvin temperature, multiply by 9/5, and finally add 32. This method corrects the zero-point shift and adjusts the degree size. For instance, to convert\(315\, K\) to Fahrenheit:
  • \( F = \frac{9}{5} \, (315 - 273.15) + 32 \)
  • \( F = 107.33^{\circ}F \)
This conversion is crucial when the absolute scale needs to be communicated in a more popularly understood format.

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Most popular questions from this chapter

(a) Classify each of the following as a pure substance, a solution, or a heterogeneous mixture: a gold coin, a cup of coffee, a wood plank. (b) What ambiguities are there in answering part (a) from the descriptions given?

A copper refinery produces a copper ingot weighing \(150 \mathrm{lb}\). If the copper is drawn into wire whose diameter is \(7.50 \mathrm{~mm}\), how many feet of copper can be obtained from the ingot? The density of copper is \(8.94 \mathrm{~g} / \mathrm{cm}^{3}\). (Assume that the wire is a cylinder whose volume \(V=\pi r^{2} h\), where \(r\) is its radius and \(h\) is its height or length.)

Classify each of the following as a pure substance or a mixture. If a mixture, indicate whether it is homogeneous or heterogeneous: (a) air, (b) tomato juice, (c) iodine crystals, (d) sand.

(a) What is the difference between a hypothesis and a theory? (b) Explain the difference between a theory and a scientific law. Which addresses how matter behaves, and which addresses why it behaves that way?

Ethyl chloride is sold as a liquid (see photo) under pressure for use as a local skin anesthetic. Ethyl chloride boils at \(12^{\circ} \mathrm{C}\) at atmospheric pressure. When the liquid is sprayed onto the skin, it boils off, cooling and numbing the skin as it vaporizes. (a) What changes of state are involved in this use of ethyl chloride? (b) What is the boiling point of ethyl chloride in degrees Fahrenheit? (c) The bottle shown contains \(103.5 \mathrm{~mL}\) of ethyl chloride. The density of ethyl chloride at \(25^{\circ} \mathrm{C}\) is \(0.765 \mathrm{~g} / \mathrm{cm}^{3}\). What is the mass of ethyl chloride in the bottle?
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