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Chapter 1: Problem 20

(a) Convert the following temperatures to kelvin: (i) \(113^{\circ} \mathrm{C},\) the melting point of sulfur, (ii) \(37^{\circ} \mathrm{C}\), the normal body temperature, (iii) \(357^{\circ} \mathrm{C},\) the boiling point of mercury. (b) Convert the following temperatures to degrees Celsius: (i) \(77 \mathrm{~K}\), the boiling point of liquid nitrogen, (ii) \(4.2 \mathrm{~K},\) the boiling point of liquid helium, (iii) \(601 \mathrm{~K}\), the melting point of lead.

Short Answer

Expert verified
The temperatures converted to Kelvin are: 386.15K, 310.15K, and 630.15K. The temperatures converted to Celsius are: -196.15°C, -268.95°C, and 327.85°C.

Step by step solution

01

Convert the temperature from Celsius to Kelvin

To convert from Celsius to Kelvin, we add 273.15 to the given temperature in Celsius. This will be done for each of the temperatures given in (a), which are 113°C, 37°C, and 357°C.
02

Calculate the Kelvin values

For (i), \(113°C + 273.15 = 386.15K\)\nFor (ii), \(37°C + 273.15 = 310.15K\)\nFor (iii), \(357°C + 273.15 = 630.15K\)
03

Convert the temperature from Kelvin to Celsius

To convert from Kelvin to Celsius, we subtract 273.15 from the given temperature in Kelvin. This will be done for each of the temperatures given in (b), which are 77K, 4.2K, and 601K.
04

Calculate the Celsius values

For (i), \(77K - 273.15 = -196.15°C\)\nFor (ii), \(4.2K - 273.15 = -268.95°C\)\nFor (iii), \(601K - 273.15 = 327.85°C\)

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Most popular questions from this chapter

Express the answers to these in scientific notation: (a) \(0.0095+\left(8.5 \times 10^{-3}\right)\) (b) \(653 \div\left(5.75 \times 10^{-8}\right)\) (c) \(850,000-\left(9.0 \times 10^{5}\right)\) (d) \(\left(3.6 \times 10^{-4}\right) \times\left(3.6 \times 10^{6}\right)\)

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The speed of sound in air at room temperature is about \(343 \mathrm{~m} / \mathrm{s}\). Calculate this speed in miles per hour \((\mathrm{mph})\)

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