P19_005.PDF

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Chapter 19 - 19.5
5. (a) Fahrenheit and Celsius temperatures are related by T F =(9 / 5) T C +32 ◦ . T F is numerically equal
to T C if T F =(9 / 5) T F +32 ◦ . The solution to this equation is T F =
−
(5 / 4)(32 ◦ )=
−
40 ◦ F.
273 . 15) +
32 ◦ . The Fahrenheit temperature T F is numerically equal to the Kelvin temperature T if T F =
(9 / 5)( T F −
−
273 . 15) + 32 ◦ . The solution to this equation is
T F = 5
4
9
5 ×
273 . 15
−
32 ◦ = 575 ◦ F .
(c) Since T C = T
−
273 . 15 the Kelvin and Celsius temperatures can never have the same numerical
value.
(b) Fahrenheit and Kelvin temperatures are related by T F =(9 / 5) T C +32 ◦ =(9 / 5)( T

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