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Chapter 1: Problem 25

Makethe following conversions: (a) \(62^{\circ} \mathrm{F}\) 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 \({ }^{\circ} \mathrm{F}\), \((\mathrm{e}) 2500^{\circ} \mathrm{F}\) to \(\mathrm{K}\).

Short Answer

Expert verified
(a) \(62^{\circ} \mathrm{F} 鈮� 16.67^{\circ} \mathrm{C}\), (b) \(216.7^{\circ} \mathrm{C} 鈮� 422.06^{\circ} \mathrm{F}\), (c) \(233^{\circ} \mathrm{C} = 506.15 \mathrm{K}\), (d) \(315 \mathrm{K} 鈮� 107.33^{\circ} \mathrm{F}\), (e) \(2500^{\circ} \mathrm{F} 鈮� 1644.26 \mathrm{K}\).

Step by step solution

01

-(a) Fahrenheit to Celsius conversion for 62掳F#

-We are given 62掳F and asked to convert it to Celsius. We will use formula (2) for this conversion. \(C = \frac{5}{9}(F - 32)\) Substitute \(F = 62\): \(C = \frac{5}{9}(62 - 32)\) Calculate the result: \(C 鈮� 16.67\) So, \(62^{\circ} \mathrm{F} 鈮� 16.67^{\circ} \mathrm{C}\).
02

-(b) Celsius to Fahrenheit conversion for 216.7掳C#

-We are given 216.7掳C and asked to convert it to Fahrenheit. We will use formula (1) for this conversion. \(F = \frac{9}{5}C + 32\) Substitute \(C = 216.7\): \(F = \frac{9}{5}\times 216.7 + 32\) Calculate the result: \(F 鈮� 422.06\) So, \(216.7^{\circ} \mathrm{C} 鈮� 422.06^{\circ} \mathrm{F}\).
03

-(c) Celsius to Kelvin conversion for 233掳C#

-We are given 233掳C and asked to convert it to Kelvin. We will use formula (3) for this conversion. \(K = C + 273.15\) Substitute \(C = 233\): \(K = 233 + 273.15\) Calculate the result: \(K = 506.15\) So, \(233^{\circ} \mathrm{C} = 506.15 \mathrm{K}\).
04

-(d) Kelvin to Fahrenheit conversion for 315 K#

-We are given 315 K and asked to convert it to Fahrenheit. We will first convert Kelvin to Celsius using formula (4), and then convert Celsius to Fahrenheit using formula (1). First conversion (Kelvin to Celsius): \(C = K - 273.15\) Substitute \(K = 315\): \(C = 315 - 273.15\) Calculate the result: \(C 鈮� 41.85\) Second conversion (Celsius to Fahrenheit): \(F = \frac{9}{5}C + 32\) Substitute \(C 鈮� 41.85\): \(F = \frac{9}{5}\times 41.85 + 32\) Calculate the result: \(F 鈮� 107.33\) So, \(315 \mathrm{K} 鈮� 107.33^{\circ} \mathrm{F}\).
05

-(e) Fahrenheit to Kelvin conversion for 2500掳F#

-We are given 2500掳F and asked to convert it to Kelvin. We will first convert Fahrenheit to Celsius using formula (2) and then convert Celsius to Kelvin using formula (3). First conversion (Fahrenheit to Celsius): \(C = \frac{5}{9}(F - 32)\) Substitute \(F = 2500\): \(C = \frac{5}{9}(2500 - 32)\) Calculate the result: \(C 鈮� 1371.11\) Second conversion (Celsius to Kelvin): \(K = C + 273.15\) Substitute \(C 鈮� 1371.11\): \(K = 1371.11 + 273.15\) Calculate the result: \(K 鈮� 1644.26\) So, \(2500^{\circ} \mathrm{F} 鈮� 1644.26 \mathrm{K}\).

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Key Concepts

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

Fahrenheit to Celsius
To convert a temperature from Fahrenheit to Celsius, we use a straightforward formula:
  • Start with the temperature in Fahrenheit (F).
  • Subtract 32 from this temperature.
  • Multiply the result by \( \frac{5}{9} \).
For example, to convert 62掳F to Celsius:
First, subtract 32 from 62, which gives 30. Then, multiply 30 by \( \frac{5}{9} \), yielding approximately 16.67掳C. Therefore, 62掳F is approximately equivalent to 16.67掳C.
This formula works because Celsius and Fahrenheit scales are based on different zero points and scaling factors. The number 32 appears in the formula because it is the point where water freezes in Fahrenheit. The fraction \( \frac{5}{9} \) is used to adjust the scaling from Fahrenheit to Celsius.
Celsius to Kelvin
Converting Celsius to Kelvin is perhaps the simplest temperature conversion due to its direct relationship.
The formula is:
  • Add \(273.15\) to the temperature in Celsius.
This straightforward process is because both scales start at absolute zero, but they use different units of measurement.
For instance, converting 233掳C to Kelvin is done by adding 273.15, resulting in 506.15 K.
This means 233掳C corresponds to 506.15 K.
The Kelvin scale is particularly useful in scientific calculations because it is an absolute scale based on absolute zero, the theoretical point where particles have minimal thermal motion.
Kelvin to Fahrenheit
Converting Kelvin to Fahrenheit is a two-step process.
First, convert the temperature from Kelvin to Celsius, then from Celsius to Fahrenheit:
  • Subtract \(273.15\) from the Kelvin temperature to get Celsius.
  • Convert this Celsius temperature to Fahrenheit using the formula:
  • \( F = \frac{9}{5}C + 32 \).
For example, to convert 315 K to Fahrenheit, start by subtracting 273.15 to get \(41.85\,^{\circ}C\).
Next, use the Celsius to Fahrenheit formula: \(F = \frac{9}{5} \times 41.85 + 32\), which results in approximately 107.33掳F.
Understanding this method is beneficial because Kelvin is often used in scientific contexts, whereas Fahrenheit might be encountered in everyday contexts in certain regions.
Fahrenheit to Kelvin
To convert Fahrenheit to Kelvin, a two-step conversion process is used.
First, convert Fahrenheit to Celsius, then convert Celsius to Kelvin:
  • Use the formula to convert Fahrenheit to Celsius: \(C = \frac{5}{9}(F - 32)\).
  • Once you have Celsius, add \(273.15\) to convert to Kelvin.
For example, for a temperature of 2500掳F, start with Fahrenheit to Celsius:
Subtract 32 from 2500, then multiply the result by \(\frac{5}{9}\), yielding approximately 1371.11掳C.
Next, convert this Celsius temperature to Kelvin by adding 273.15, resulting in approximately 1644.26 K.
Whether handling extreme temperatures or engaging in scientific work, knowing how to convert between these scales is essential. Each scale serves different purposes, from everyday life to specialized scientific research.

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Most popular questions from this chapter

(a) After the label fell off a bottle containing a clear liquid believed to be benzene, a chemist measured the density of the liquid to verify its identity. A \(25.0-\mathrm{mL}\) portion of the liquid had a mass of \(21.95 \mathrm{~g} .\) A chemistry handbook lists the density of benzene at \(15^{\circ} \mathrm{C}\) as \(0.8787 \mathrm{~g} / \mathrm{mL}\). Is the calculated density in agreement with the tabulated value? (b) An experiment requires \(15.0 \mathrm{~g}\) of cyclohexane, whose density at \(25{ }^{\circ} \mathrm{C}\) is \(0.7781 \mathrm{~g} / \mathrm{mL}\). What volume of cyclohexane should be used? (c) A spherical ball of lead has a diameter of \(5.0 \mathrm{~cm}\). What is the mass of the sphere if lead has a density of \(11.34 \mathrm{~g} / \mathrm{cm}^{3} ?\) (The volume of a sphere is \(\left(\frac{4}{3}\right) \pi r^{3}\) where \(r\) is the radius.)

Indicate which of the following are exact numbers: (a) the mass of a 32 -oz can of coffee, \((b)\) the number of students in your chemistry class, (c) the temperature of the surface of the sun, (d) the mass of a postage stamp, (e) the number of milliliters in a cubic meter of water, (f) the average height of students in your school.

Label each of the following as either a physical process or a chemical process: (a) corrosion of aluminum metal, (b) melting of ice, (c) pulverizing an aspirin, (d) digesting a candy bar, (e) explosion of nitroglycerin.

By using estimation techniques, arrange these items in order from shortest to longest: a \(57-\mathrm{cm}\) length of string, a 14 -in. long shoe, and a \(1.1-\mathrm{m}\) length of pipe.

(a) A sample of carbon tetrachloride, a liquid once used in dry cleaning, has a mass of \(39.73 \mathrm{~g}\) and a volume of \(25.0 \mathrm{mLat} 25^{\circ} \mathrm{C}\). What is its density at this temperature? Will carbon tetrachloride float on water? (Materials that are less dense than water will float.) (b) The density of platinum is \(21.45 \mathrm{~g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). Calculate the mass of \(75.00 \mathrm{~cm}^{3}\) of platinum at this temperature. (c) The density of magnesium is \(1.738 \mathrm{~g} / \mathrm{cm}^{3}\) at \(20^{\circ} \mathrm{C}\). What is the volume of \(87.50 \mathrm{~g}\) of this metal at this temperature?
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