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?
Well, let me calculate that for you!
Using the equation \(\frac{T-A}{T_0 - A} = e^{-kt}\), and some clown magic, I can rearrange it to solve for t:
\(t = -\frac{1}{k} \cdot ln\left(\frac{T-A}{T_0 - A}\right)\)
Now, plugging in the values we have:
\(T = 77\) degrees F
\(T_0 = 45\) degrees F
\(A = 93\) degrees F
\(k\) is still a mystery to us.
Since we don't know the value of "k," I'm going to need you to provide that information, or we can just make it up! But remember, I'm just a clown bot and not a math bot, so my made-up value might not be very accurate.
To find out how long it will take the pop to reach a temperature of 77 degrees F, we can use the equation you provided:
\(\frac{T-A}{T_0-A} = e^{-kt}\)
Let's substitute the known values into the equation:
\(T_0 = 45\) degrees F (initial temperature of the pop)
\(A = 93\) degrees F (temperature in the car)
\(T = 77\) degrees F (target temperature)
\(k\) is the constant that corresponds to the warming rate (which we don't know yet)
\(t\) is the length of time that the pop has been warming up (which we want to find)
We can rearrange the equation to solve for \(t\):
\(\frac{T-A}{T_0-A} = e^{-kt}\)
\(e^{-kt} = \frac{T-A}{T_0-A}\)
\(-kt = \ln{\left(\frac{T-A}{T_0-A}\right)}\)
\(t = -\frac{\ln{\left(\frac{T-A}{T_0-A}\right)}}{k}\)
Now, we need to find the value of \(k\) to solve for \(t\). We can use the information provided that the pop's temperature reaches 60 degrees F after 23 minutes:
\(T = 60\) degrees F
\(t = 23\) minutes
We can substitute these values into the rearranged equation:
\(23 = -\frac{\ln{\left(\frac{60-93}{45-93}\right)}}{k}\)
Simplifying the equation further, we have:
\(23k = -\ln{\left(\frac{-33}{-48}\right)}\)
\(23k = -\ln{\left(\frac{11}{16}\right)}\)
Finally, we can solve for \(k\):
\(k = -\frac{\ln{\left(\frac{11}{16}\right)}}{23}\)
After finding the value of \(k\), we can substitute it back into the equation for \(t\) to find how long it will take the pop to reach a temperature of 77 degrees F.
To find out how long it will take the pop to reach a temperature of 77 degrees F, we can use the formula:
(T - A)/(T0 - A) = e^(-kt)
Let's substitute the given values into the formula:
(T - 93)/(45 - 93) = e^(-k * 23)
Simplifying this equation:
(T - 93)/(-48) = e^(-23k)
To find the value of k, we need to solve for it. We can rearrange the equation:
e^(-23k) = (T - 93)/(-48)
Taking the natural logarithm (ln) of both sides:
ln(e^(-23k)) = ln((T - 93)/(-48))
This simplifies to:
-23k = ln((T - 93)/(-48))
Now we can solve for k by dividing both sides of the equation by -23:
k = ln((T - 93)/(-48))/(-23)
Now, we'll use this value of k to find how long it will take for the pop to reach a temperature of 77 degrees F. We'll substitute the given values into the original formula:
(77 - 93)/(45 - 93) = e^(-k * t)
Substituting the value of k:
(77 - 93)/(45 - 93) = e^(ln((T - 93)/(-48))/(-23) * t)
Simplifying this equation:
-16/(-48) = e^(ln((T - 93)/(-48))/(-23) * t)
Simplifying further:
1/3 = e^(ln((T - 93)/(-48))/(-23) * t)
To isolate t, we need to take the natural logarithm of both sides:
ln(1/3) = ln(e^(ln((T - 93)/(-48))/(-23) * t))
ln(1/3) = (ln((T - 93)/(-48))/(-23)) * t
Now, we can solve for t:
t = ln(1/3) / ((ln((T - 93)/(-48)) / (-23)))
Finally, plug in the desired temperature of 77 degrees F to get the time it will take for the pop to reach that temperature.
(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!