Instructions Assume that goat milk butter melts at {85.25}^{\circ} F. On a summer day you take a stick of butter out of the refrigerator, which is set to {36.5}^{\circ} F, and put it on your front porch where the temperature is {89.5}^\circ F.

Question

(a) By Newton's Law of Cooling, which can also be used to describe warming, the temperature T(t) of the butter at time t (measured in minutes) satisfies the initial value problem

T'(t) = k \cdot (T(t)-89.5)\, ,\quad T(0)=36.5\, .

Find the function T(t) by solving this IVP. Use either the method of solving separable or of solving linear DEs. Include k in your calculations as an initially unknown parameter.

T(t) =

T(t) = -53e^(kt) + 89.50

(b) After 4 minutes the temperature of the butter has risen to {54}^F. Use this information to find k as a decimal number.

k = -1.002

(c) How long after being taken out of the refrigerator does the butter melt? Express your answer in minutes and seconds.

t = min , sec

I couldn't figure out part c, I plugged back in using the 85.25, and got 2.519 and then add to the intial 4 mins, I got 6.519, but its saying that isn't the answer.

Answer 1

-1.002 ≈ 1, so let's use
T(t) = -53e^-t + 89.50 = 85.25
53e^-t = 0.25
e^-t = .004717
t = -ln .004717 = 5.3566

Answer 2

I agree with
T(t) = 89.5 - 53 e^(kt)
using your data,
54 = 89.5 - 53 e^(4k)
53 e^(4k) = 35.5
e^4k = 35.5/53 = .6698...
using ln
4k = ln .6693... = -.400759..
k = -.1001898
or
k = -.1002

check you decimal

check on mine:
when t = 4
T(4) = 89.5 - 53 e^(4(-.1002)) = 54.001 , looks like my k is correct

so we have
T(t) = 89.5 - 53 e^(-.1002t)
85.5 = 89.5 -53 e^(-.1002t)
e^(-.100t) = .07547..
-.1002t = ln .07547 = -2.586997...
t = 25.788.. minutes
or appr 26 minutes

you don't add the 4 minutes, the formula assumes your time starts at 0