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.
(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.
-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
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
To solve part (c), we need to find the time it takes for the temperature of the butter to reach 85.25°F, which is the melting point of goat milk butter.
Using the function T(t) that we found in part (a), which is T(t) = -53e^(kt) + 89.5 and the value of k that we found in part (b), which is k = -1.002, we can plug in the melting temperature of 85.25°F into the equation and solve for t.
So, we have:
85.25 = -53e^(-1.002t) + 89.5
To solve for t, we can rearrange the equation as follows:
-53e^(-1.002t) = 85.25 - 89.5
-53e^(-1.002t) = -4.25
Dividing both sides by -53 and taking the natural logarithm of both sides, we get:
e^(-1.002t) = 0.08018867921
Taking the natural logarithm of both sides again, we have:
-1.002t = ln(0.08018867921)
Finally, we can solve for t by dividing both sides by -1.002:
t = ln(0.08018867921) / -1.002 ≈ 3.85 minutes
Therefore, the butter will melt approximately 3.85 minutes after being taken out of the refrigerator.