Temperature Conversions


Alignments to Content Standards: F-BF.A.1.c F-BF.B.4.a

Task

Let $f$ be the function that assigns to a temperature in degrees Celsius its equivalent in degrees Fahrenheit. Let $g$ be the function that assigns to a temperature in degrees Kelvin its equivalent in degrees Celsius.

  1. Explain what $x$ and $f(g(x))$ represent in terms of temperatures, or explain why there is no reasonable representation.
  2. Explain what $x$ and $g(f(x))$ represent in terms of temperatures, or explain why there is no reasonable representation.
  3. Given that $f(x) = \frac{9}{5}x + 32$ and $g(x) = x - 273$, find an expression for $f(g(x))$.
  4. Find an expression for the function $h$ which assigns to a temperature in degrees Fahrenheit its equivalent in degrees Kelvin.

IM Commentary

Unit conversion problems provide a rich source of examples both for composition of functions (when several successive conversions are required) and inverses (units can always be converted in either of two directions). Note that the conversion function $g(x)$ is an approximation: The exact conversion formula is given by $g(x) = x -273.15$.

Solutions

Solution: 1

 

 

 

 

 

 

 

  1. Given a temperature $x$ in degrees Kelvin, $g(x)$ represents its equivalent in degrees Celsius. Given a temperature $g(x)$ in degrees Celsius, $f(g(x))$ represents its equivalent in degrees Fahrenheit. Combining these two statements, if $x$ represents a temperature in degrees Kelvin, then $f(g(x))$ represents its equivalent in degrees Fahrenheit.
  2. Given a temperature $x$ in degrees Celsius, $f(x)$ represents its equivalent in degrees Fahrenheit. But since $g$ assigns to temperatures in degrees Kelvin their equivalents in degrees Celsius, $g(f(x))$ has no representation in terms of temperatures.
  3. To find an expression for $f (g(x))$ we substitute $g(x)$ for $x$ in the expression for $f(x)$: $$ f (g(x)) = \frac{9}{5}\left(g(x)\right) + 32 = \frac{9}{5}\left(x - 273\right) + 32 = \frac{9}{5}x -\frac{2297}{5}. $$
  4. We remember that the function defined by the equation in c) assigns to a temperature in degrees Kelvin, its equivalent in degrees Fahrenheit. We also notice that $h$ must do the reverse. Since the expression for $f(g(x))$ multiplies $x$ by $\frac{9}{5}$ and then subtracts $\frac{2297}{5}$, an expression for $h(x)$ can be obtained by first adding $\frac{2297}{5}$ to $x$ and then dividing by $\frac{9}{5}$. In other words, \begin{align*} h(x) &= \frac{5}{9}\left( x + \frac{2297}{5} \right) \\ &= \frac{5}{9}x + \frac{2297}{9}. \end{align*}

 

Solution: 2

A second way to solve part d) of the problem is to

1) develop an expression that will convert a temperature in degrees Fahrenheit to its equivalent in degrees Celsius,

2) develop an expression that will convert a temperature in degrees Celsius to its equivalent in degrees Kelvin, and finally

3) use these expressions to develop an expression for $h(x)$.

If $x$ is a temperature in degrees Celsius then $$ a(x) = x + 273 $$ gives its equivalent in degrees Kelvin. If $x$ is a temperature in degrees Fahrenheit then $$ b(x) = \frac{5}{9}(x - 32). $$ gives its equivalent in degrees Celsius. So, if $x$ represents a temperature in degrees Fahrenheit, $a(b(x))$ represents its equivalent in degrees Kelvin. So we find \begin{eqnarray*} h(x) = a(b(x)) &=& \left( \frac{5}{9}\left(x-32\right) \right) + 273 \\ &=& \frac{5}{9}x +255 \times \frac{2}{9}. \end{eqnarray*} Since $255\times\frac{2}{9} = \frac{2297}{9}$ this is the same answer found in solution 1.