Question 1161289
I bake frozen pizzas for my husband very often, so I can figure this out.
{{{T}}}=temperature at time {{{t}}}
{{{T[0]}}}=temperature at time {{{t=0}}}
{{{C}}}=ambient temperature
The units we use for {{{t}}} and for the temperatures do not matter as long as we are consistent.
At {{{t=0}}} {{{e^kt=e^0=1}}} makes {{{T=C+T[0]-C=T[0]}}}
As {{{t}}} approaches {{{infinity}}} ,
{{{e^kt}}} needs to approach {{{0}}} ,
so that {{{T}}} approaches {{{C}}} ,
so the constant {{{k}}} must be a negative number.
 
a. With time in minutes and temperature in degrees Fahrenheit,
the data gives us the equation {{{315=70+(450-70)*e^(5k)}}} ,
or {{{315=70+380*e^(5k)}}} , if we simplify.
We need to find {{{k}}} , and substitute its value into
{{{T=C+(t[0]-C)*e^kt}}} to find the model
(a formula we can use to figure out how long to wait for that pizza to cool).
{{{315=70+380*e^(5k)}}}
{{{315-70=380*e^(5k)}}}
{{{245=380*e^(5k)}}}
{{{245/380=e^(5k)}}} I could use {{{245/380=0.6447368421050.644736842105}}} (or a rounded version of that) in the next equation, but I will let the calculator remember the number and keep crunching numbers.
{{{ln(245/380)=5k}}}
{{{ln(245/380)/5=k}}}
According to my scientific calculator,
{{{k=-0.08778}}}} (rounded to 4 significant figures).
So, the model is {{{highlight(T=70+380*e^(-0.08778t))}}}

b. After 25 minutes {{{t=25}}} ,
so we substitute that value into our model to find {{{T}}} for that pizza.
From experience, I say we waited too long.
{{{T=70+380*e^(-0.08778*25)}}}
{{{T=70+380*e^(-2.1945)}}}
Using rounded values, {{{e^(-2.1945)=0.1114}}} ,so
{{{T=70+380*0.1114}}} , and continuing to round, {{{T=70+42}}} ,
so {{{highlight(T=112)}}}
After 25 minutes, the temperature of the pizza is {{{112^o}}}{{{F)}}} .
 
c. To find out when the temperature of the pizza will be {{{T=130}}}
(in degrees Fahrenheit), we substitute that into our model, getting
{{{130=70+380*e^(-0.08778t)}}} as the equation to solve.
{{{130-70=380*e^(-0.08778t)}}}
{{{60=380*e^(-0.08778t)}}}
{{{60/380=e^(-0.08778t)}}}
{{{ln(60/380)=-0.08778t}}}
{{{ln(60/380)/(-0.08778)=t}}}
and according to my calculator, {{{t=21}}} .
So, according to our model, to eat the pizza at {{{130^o}}}{{{F}}} , we have to wait {{{highlight(21minutes)}}} .