Differential Equations Cooling Problem

Thread Starter


Joined Sep 20, 2009
Hello everyone, I'm having trouble figuring out the last step in the diff eq problem. I'll post the problem below, my work, and my question.

Problem: Food, initially at a temperature of 40°F, was placed in an oven preheated to 350°F. After 10 min in the oven, the food had warmed to 120°F. After 20 min, the food was removed from the oven and allowed to cool at room temperature 72°F. If the ideal serving temperature of the the food is 110°F, when should the food be served?

My work:
H'(t) = ks(t) - kH(t)
H'(t) + kH(t) = Ks(t)
Therefore.... p(t) = k .....P(t) = kt ....... µ(t) = e^kt
µ(H'(t) + kH(t)) = µ(ks(t))

(µH) = intergral: (e^kt)ks(t)dt // treating s(t) as a constant and using U substitution I get...

(µH) = s(t)e^kt + C // dividing by µ to solve for H I get....
H = s(t) + C/e^kt // then I impose my initial condition
40 = 350 + C/e^t=0
40 = 350 + C
C = -310

// Now solving for K
at t = 10min
120 = 350 - 310/e^10k
-230 = -310/e^10k
10k = ln -310/-230
k = .0298

// Finding temperature after 20 minutes in the oven
H = 350 - 310/e^20*.0298
H = 179.2

// This is where I get stuck

My Question: At this point I need to figure out how long it would take the food to cool down to 110 degrees. I'm lost on how to setup that equation.
Here is what I tried.

110 = 72 - 310/e^.0298t
38 = -310/e^.0298t
e^.0298t = -310/38
.0298t = ln(-310/38) // This is my problem, you cannot LN a negative number right?

Also intuitively, I feel like the temp of the food as it comes out of the oven needs to come into play here. Do I need to setup a new equation by imposing the new initial condition? Do I have to find a new K? Hopefully my work is clear enough, thanks!
Last edited:


Joined Jun 17, 2014

If you could specify what al your variables were at the start it would be must easier to read your work.
For example,
E is the voltage in volts,
I is the current in amperes,
R is the resistance in Ohms.

Note every variable is mentioned at least once, and a complete description is given with units also.