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!
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!
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