Question

the boiling point of water is 100°c and drops by about 3.75°c for each 10 kpa drop in atmospheric pressure. find a linear function g that models the boiling - point of water (in °c) if the atmospheric pressure is 74 kpa.

Explanation:

Step1: Identify the slope and a point

The boiling - point drops by about $3.75^{\circ}C$ for each $10$ kPa drop in pressure. So the slope $m=-\frac{3.75}{10}=- 0.375$. At a pressure of $101.3$ kPa (standard pressure), the boiling - point is $100^{\circ}C$. Let $x$ be the atmospheric pressure in kPa and $y = g(x)$ be the boiling - point in $^{\circ}C$. We can also use the fact that when $x = 101.3$, $y = 100$. But we can start from the given information in a different way. When $x = 100$ kPa, assume the base - case boiling - point is $100^{\circ}C$.
We know that the linear function has the form $y=mx + b$, where $m$ is the slope and $b$ is the y - intercept.

Step2: Find the y - intercept

We know that when $x = 100$ (a reference pressure), assume $y = 100$. Substituting $m=-0.375$, $x = 100$ and $y = 100$ into $y=mx + b$, we get $100=-0.375\times100 + b$.
Solve for $b$:
\[

$$\begin{align*} 100&=-37.5 + b\\ b&=100 + 37.5\\ b&=137.5 \end{align*}$$

\]

Step3: Write the linear function

The linear function $g(x)$ that models the boiling - point of water as a function of atmospheric pressure $x$ (in kPa) is $g(x)=-0.375x + 137.5$.

Answer:

$g(x)=-0.375x + 137.5$