(T-A)/(To-A) = e^-kt
We have
(60-93)/(45-93) = e^(-23k)
e^(-23k) = .6875
k = .0163
So, now we want to find when T=77
e^(-.0163t) = (77-93)/(45-93) = 0.333
t = 67.4 minutes
Nice model, but I'm sure that in real life, after more than an hour, that pop is very nearly 93 degrees!
You are taking a road trip in a car without A/C. The temperture in the car is 93 degrees F. You buy a cold pop at a gas station. Its initial temperature is 45 degrees F. The pop's temperature reaches 60 degrees F after 23 minutes.
Given that
\frac{T-A}{T_0 - A} = e^{-kt}
where T = the temperature of the pop at time t.
T_0 = the initial temperature of the pop.
A = the temperature in the car.
k = a constant that corresponds to the warming rate.
and t = the length of time that the pop has been warming up.
How long will it take the pop to reach a temperature of 77 degrees F?
1 answer
3 answers