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Chapter 1: Problem 53

Convert the following Fahrenheit temperatures to the Celsius and Kelvin scales. a. \(-459^{\circ} \mathrm{F}\), an extremely low temperature b. \(-40 .{ }^{\circ} \mathrm{F}\), the answer to a trivia question c. \(68^{\circ} \mathrm{F}\), room temperature d. \(7 \times 10^{7}{ }^{\circ} \mathrm{F}\), temperature required to initiate fusion reactions in the sun

Short Answer

Expert verified
a. \(-459^{\circ} \mathrm{F}\) is equal to \(-273.15^{\circ} \mathrm{C}\) and \(0\) Kelvin. b. \(-40^{\circ} \mathrm{F}\) is equal to \(-40^{\circ} \mathrm{C}\) and \(233.15\) Kelvin. c. \(68^{\circ} \mathrm{F}\) is equal to \(20^{\circ} \mathrm{C}\) and \(293.15\) Kelvin. d. \(7 \times 10^{7}{ }^{\circ} \mathrm{F}\) is equal to approximately \(3.89 \times 10^{7}^{\circ} \mathrm{C}\) and \(3.89 \times 10^{7}\) Kelvin.

Step by step solution

01

a. Convert \(-459^{\circ} \mathrm{F}\) to Celsius and Kelvin.

First, convert the Fahrenheit temperature to Celsius using the formula: \(C = \frac{5}{9}(F - 32)\) \(C = \frac{5}{9}(-459 - 32)\) \(C = -273.15^{\circ}\mathrm{C}\) Next, convert the Celsius temperature to Kelvin using the formula: \(K = C + 273.15\) \(K = -273.15 + 273.15\) \(K = 0\) So, \(-459^{\circ} \mathrm{F}\) is equal to \(-273.15^{\circ} \mathrm{C}\) and \(0\) Kelvin.
02

b. Convert \(-40^{\circ} \mathrm{F}\) to Celsius and Kelvin.

First, convert the Fahrenheit temperature to Celsius using the formula: \(C = \frac{5}{9}(F - 32)\) \(C = \frac{5}{9}(-40 - 32)\) \(C = -40^{\circ}\mathrm{C}\) Next, convert the Celsius temperature to Kelvin using the formula: \(K = C + 273.15\) \(K = -40 + 273.15\) \(K = 233.15\) So, \(-40^{\circ} \mathrm{F}\) is equal to \(-40^{\circ} \mathrm{C}\) and \(233.15\) Kelvin.
03

c. Convert \(68^{\circ} \mathrm{F}\) to Celsius and Kelvin.

First, convert the Fahrenheit temperature to Celsius using the formula: \(C = \frac{5}{9}(F - 32)\) \(C = \frac{5}{9}(68 - 32)\) \(C = 20^{\circ}\mathrm{C}\) Next, convert the Celsius temperature to Kelvin using the formula: \(K = C + 273.15\) \(K = 20 + 273.15\) \(K = 293.15\) So, \(68^{\circ} \mathrm{F}\) is equal to \(20^{\circ} \mathrm{C}\) and \(293.15\) Kelvin.
04

d. Convert \(7 \times 10^{7}{ }^{\circ} \mathrm{F}\) to Celsius and Kelvin.

First, convert the Fahrenheit temperature to Celsius using the formula: \(C = \frac{5}{9}(F - 32)\) \(C = \frac{5}{9}(7 \times 10^{7} - 32)\) \(C \approx 3.89 \times 10^{7}^{\circ}\mathrm{C}\) Next, convert the Celsius temperature to Kelvin using the formula: \(K = C + 273.15\) \(K = 3.89 \times 10^{7} + 273.15\) \(K \approx 3.89 \times 10^{7}\) So, \(7 \times 10^{7}{ }^{\circ} \mathrm{F}\) is equal to approximately \(3.89 \times 10^{7}^{\circ} \mathrm{C}\) and \(3.89 \times 10^{7}\) Kelvin.

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Most popular questions from this chapter

Perform the following mathematical operations, and express each result to the correct number of significant figures. a. \(\frac{0.102 \times 0.0821 \times 273}{1.01}\) b. \(0.14 \times 6.022 \times 10^{23}\) c. \(4.0 \times 10^{4} \times 5.021 \times 10^{-3} \times 7.34993 \times 10^{2}\) d. \(\frac{2.00 \times 10^{6}}{3.00 \times 10^{-7}}\)

What are significant figures? Show how to indicate the number one thousand to 1 significant figure, 2 significant figures, 3 significant figures, and 4 significant figures. Why is the answer, to the correct number of significant figures, not \(1.0\) for the following calculation? $$ \frac{1.5-1.0}{0.50}= $$

At the Amundsen-Scott South Pole base station in Antarctica, when the temperature is \(-100.0^{\circ} \mathrm{F}\), researchers who live there can join the " 300 Club" by stepping into a sauna heated to \(200.0^{\circ} \mathrm{F}\) then quickly running outside and around the pole that marks the South Pole. What are these temperatures in \({ }^{\circ} \mathrm{C}\) ? What are these temperatures in \(\mathrm{K}\) ? If you measured the temperatures only in \({ }^{\circ} \mathrm{C}\) and \(\mathrm{K}\), can you become a member of the " 300 Club" (that is, is there a 300--degree difference between the temperature extremes when measured in \({ }^{\circ} \mathrm{C}\) and \(\mathrm{K}\) )?

The density of an irregularly shaped object was determined as follows. The mass of the object was found to be \(28.90 \mathrm{~g} \pm 0.03 \mathrm{~g}\). A graduated cylinder was partially filled with water. The reading of the level of the water was \(6.4 \mathrm{~cm}^{3} \pm 0.1 \mathrm{~cm}^{3}\). The object was dropped in the cylinder, and the level of the water rose to \(9.8 \mathrm{~cm}^{3} \pm 0.1 \mathrm{~cm}^{3}\). What is the density of the object with appropriate error limits? (See Appendix 1.5.)

A children's pain relief elixir contains \(80 . \mathrm{mg}\) acetaminophen per \(0.50\) teaspoon. The dosage recommended for a child who weighs between 24 and \(35 \mathrm{lb}\) is \(1.5\) teaspoons. What is the range of acetaminophen dosages, expressed in mg acetaminophen/kg body weight, for children who weigh between 24 and \(35 \mathrm{lb}\) ?
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