Newton’s Law of Cooling

Question

Coffee is poured at 145ºF into a travel cup. After 25 minutes in a 70º room the coffee has cooled to 137ºF. I accidentally left the coffee on the counter when I left. What is the temperature after 4 hours? after 5 hours? how long before the temperature is below 90º?

Show your process.

Answer 1

We can write the equation for a function modeling Newton's LoC by using the difference between the cooling object's temp and room temp as a function of time.

Let D(t): difference in temp (in °F) and t: time (in min)

D(t) = D₀e^-kt where D₀ is the initial temp difference and k is a constant that will determine the rate of cooling (and for which we can solve by using the given value as follows):

D₀ = 75

D(t) = 75e^-kt and since we want D(25) = 67 we have

D(25) = 75e^-25k = 67

e^-25k = .8933333

-25k = ln(.893333)

k ~ .0045

Finally, D(t) = 75e^-.0045t If we want the function for the object's temp, we add the room temp, 70, to get

T(t) = 70 + 75e^-.0045t

We can use this function to find the temp at 4 hrs, T(240), and 5 hrs, T(300), and to find the time at which the temp = 90 by setting the function = 90 and solving for t: 90 = 70 + 75e^-.0045t

-.0045t = ln(20/75) and t ~ 293 mins or 4 hrs 53 mins (which tells us that we should get a value bigger than 90 for T(240) and a value smaller than 90 for T(300).

Answer 2

dT/dt = - k(T - T_∞) T(0) = T₀ This is Newton's Law of Cooling

The solution is T = T_∞ + (T₀- T_∞) e^-kt You can see that T = T₀ at t=0 and goes to T_∞ when t --> ∞

The first problem tells you enough to solve for k = (-1/t) ln ((T-T_∞)/(T₀-T_∞)) You are given a T(t)

After that you solve for T(t) by plugging in the k found.

And then you can solve for a t given a T and the k found.