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.
(b) Fahrenheit and Kelvin temperatures are related by T
F
= (9/5)T
C
+ 32
◦
= (9/5)(T
− 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.