Substitute $t=0$ into $T(t)=22+68(0.88)^t$. What does $(0.88)^0$ equal, and what temperature do you get?

The room is at a constant $22^{\text{o}}\text{C}$. How much hotter than $22^{\text{o}}\text{C}$ is the soup right when it is removed from heat, and where does that number appear in the formula?

Type T(t) = 22 + 68*(0.88)^t into Desmos so it defines the function $T(t)$.

Either make a table for $T(t)$ and include $t=0$, or type T(0) on a new line. Observe the value of $T(0)$; this is the soup’s temperature right when it is removed from heat.

On a new line, type T(0) - 22 to find how many degrees hotter than the $22^{\text{o}}\text{C}$ room temperature the soup is at $t=0$. Notice that this difference matches the coefficient from the original formula, and then select the answer choice that describes that difference in words.

The model is

Here:

• $T(t)$ is the soup temperature (in $^{\text{o}}\text{C}$) at time $t$ hours.
• $22$ is the constant room temperature (given in the problem).
• $68(0.88)^t$ is how much hotter than the room the soup is at time $t$, and that amount shrinks over time because $0.88<1$ (cooling).

The soup is removed from heat at time $t=0$.

Compute $T(0)$:

Since $0.88^0=1$,

So the initial soup temperature is $22+68=90^{\text{o}}\text{C}$. This is the temperature at the instant cooling begins.

We know:

• Initial soup temperature: $90^{\text{o}}\text{C}$.
• Room temperature: $22^{\text{o}}\text{C}$.

The difference (how many degrees hotter than the room) at $t=0$ is

So the number $68$ in the formula is exactly the initial difference between the soup’s temperature and the room temperature.

Therefore, the correct interpretation is: The initial temperature of the soup is 68 degrees Celsius above the room temperature.