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.