Introductory Chemistry (5th Edition)

Published by Pearson
ISBN 10: 032191029X
ISBN 13: 978-0-32191-029-5

Chapter 3 - Matter and Energy - Exercises - Problems - Page 89: 66

Answer

a) $38.89^{\circ}C$ b) $-459.67^{\circ}F$ c) $-54.4^{\circ}F$ d) $-0.15^{\circ}C$

Work Step by Step

Using the conversion factors below, we can convert from Celsius to Fahrenheit or vice-versa. To find the Celsius temperature we can convert from Fahrenheit using the following formula $(^{\circ}F-32)\times\frac{5}{9}$ and to convert from Celsius to Fahrenheit, the following formula can be used; $(^{\circ}C\times\frac{9}{5})+32$. Additionally to convert to Kelvin from Celsius, we can do so by simply adding 273.15 and to covert from Kelvin to Celsius, we simply subtract 273.15. a) $(102^{\circ}F-32)\times\frac{5}{9} = 38.89^{\circ}C$ b) $0 - 273.15 = -273.15^{\circ}C$ $(-273.15^{\circ}C\times\frac{9}{5})+32 = -459.67^{\circ}F$ c) $(-48^{\circ}C\times\frac{9}{5})+32 = -54.4^{\circ}F$ d) $273 K - 273.15 = -0.15^{\circ}C$
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